0.1 First Order Differential Equations
0.1
First Order Differential Equations
1. Solve the following separable differential equations. −2x
(a) y 0 = e y . We start by separating the equation: dy e−2x = dx y y dy = e−2x dx Next, we integrate both sides: Z Z y dy = e−2x dx y2 e−2x =− +C 2 2 Lastly, solving for y: y2 e−2x =− +C 2 2 y 2 = −e−2x + 2C p p y = ± −e−2x + 2C = ± −e−2x + D
D = 2C.
(b) 3y 5 y 0 − x2 (x3 + 5)10 = 0 We start by separating the equation: dy = x2 (x3 + 5)10 dx 3y 5 dy = x2 (x3 + 5)10 dx 3y 5
Next, we integrate both sides: Z Z 5 3y dy = x2 (x3 + 5)10 dx Let u = x3 + 5, so du = 3x2 dx: Z 1 10 y6 = u du 2 3 1 11 = u +C 33 1 3 = (x + 5)11 + C 33 Lastly, solving for y: y6 1 3 = (x + 5)11 + C 2 33 2 3 2 3 y6 = (x + 5)11 + 2C = (x + 5)11 + D 33 33
1
D = 2C.
x
−2 (c) y 0 = ey+2 . We start by separating the equation:
ex − 2 dy = dx y+2 (y + 2) dy = ex − 2 dx Next, we integrate both sides: Z Z (y + 2) dy = ex − 2 dx y2 + 2y = ex − 2x + C 2 Lastly, we solve for y using the quadratic formula: y2 + 2y − (ex − 2x + C) = 0 2 p y = −2 ± 4 − 2(ex − 2x + C) √ = −2 ± 4 − 2ex − 4x + D D = 2C. 1+2y (d) y 0 = − x(1+y) .
We start by separating the equation: dy 1 + 2y =− dx x(1 + y) 1+y 1 dy = − dx 1 + 2y x Next, we integrate both sides: Z Z 1 1+y dy = − dx 1 + 2y x Z 1 Z 1 1 2 (1 + 2y) + 2 dy = − dx 1 + 2y x Z Z 1 1 1 1 + dy = − dx 2 2 1 + 2y x 1 1 y + ln |1 + 2y| = − ln x + C 2 4 Here, we cannot solve for y, so we leave the solution in implicit form.
2
(e) y 0 + x2 y = x2 y 2 . We start by separating the equation: dy = x2 y 2 − x2 y = x2 (y 2 − y) dx dy = x2 dx y2 − y Next, we integrate both sides: Z Z dy = x2 dx. y2 − y We solve the left hand side by partial fractions: y2
1 1 1 =− + , −y y y−1
so − ln |y| + ln |y − 1| = y−1 |= y 1 ln |1 − | = y ln |
x3 +C 3
x3 +C 3 x3 +C 3
Lastly, we solve for y: x3 x3 1 |1 − | = e 3 +C = De 3 y x3 1 1 − = ±De 3 y 1 y= x3 1 ± De 3
3
D = eC
2. Solve the following homogeneous equations. (a) y 0 =
y 5 −x5 y4 x .
Start by checking that the equation is homogeneous: let f (x, y) = y 5 −x5 y 4 x . Then, f (xt, yt) = Let y = vx, so
(yt)5 − (xt)5 t5 (y 5 − x5 ) y 5 − x5 = = = f (x, y). (yt)4 (xt) t5 (y 4 x) y4 x dy dx
v+x
dv = v + x dx . Substitute these into the equation:
v5 − 1 dv ((vx)5 − x5 x5 (v 5 − 1) = = = dx (vx)4 x x5 v 4 v4 5 dv v −1 1 x = − v = − 4. 4 dx v v
The new equation is separable: v 4 dv = − Z
v 4 dv = −
dx Zx
dx x
v5 = − ln |x| + C 5 Let C = − ln |k|, so v5 = − ln |x| − ln |k| = − ln |kx| 5 v 5 = −5 ln |kx| = ln |kx|−5 √ v = 5ln |kx|−5
Lastly, writing the expression in terms of y gives: √ y = vx = x 5ln |kx|−5 .
4
(b) y 0 =
−2y+3x . y
Start by checking that the equation is homogeneous: let f (x, y) = −2y+3x . Then, y f (xt, yt) = Let y = vx, so
dy dx
−2(yt) + 3(xt) −2y + 3x = = f (x, y). (yt) y
dv = v + x dx . Substitute these into the equation:
dv −2(vx) + 3x −2v + 3 = = dx vx v dv −2v + 3 −v 2 − 2v + 3 x = −v = dx v v dx v dv = − 2 Zx Z v + 2v − 3 dx v dv = − 2 v + 2v − 3 x v+x
The right hand side can be done by partial fractions: v2
v 3 1 1 1 = + , + 2v − 3 4v+3 4v−1
so Z 1 1 dx 3 1 + dv = − 4v+3 4v−1 x 3 1 ln |v + 3| + ln |v − 1| = − ln |x| + C 4 4 Z
Let C = − ln |k|, so 3 1 ln |v + 3| + ln |v − 1| = − ln |x| − ln |k| = − ln |kx| 4 4 3 ln |v + 3| + ln |v − 1| = −4 ln |kx| ln |v + 3|3 + ln |v − 1| = ln |kx|−4 ln |v + 3|3 |v − 1| = ln |kx|−4 |v + 3|3 |v − 1| = |kx|−4 y y | + 3|3 | − 1| = |kx|−4 x x This expression cannot be simplified further, so we leave it as is.
5
(c) y 0 =
x2 −y 2 xy .
Start by checking that the equation is homogeneous: let f (x, y) = x2 −y 2 xy . Then, f (xt, yt) = Let y = vx, so
dy dx
x2 − y 2 (xt)2 − (yt)2 = = f (x, y). (xt)(yt) xy
dv = v + x dx . Substitute these into the equation:
dv x2 − (vx)2 1 − v2 = = dx x(vx) v dv 1 − v2 1 − 2v 2 x = −v = dx v v dx v dv = 2 Z 1 − 2v Zx v dx dv = 1 − 2v 2 x
v+x
Let u = 1 − 2v 2 , so du = −4v dv and − du 4 = v dv: Z Z 1 du dx − = 4 u x 1 − ln |u| = ln |x| + C. 4 Let C = ln |k|, so 1 − ln |u| = ln |kx| 4 1 − ln |1 − 2v 2 | = ln |kx| 4 1 ln |1 − 2v 2 |− 4 = ln |kx| 1
|1 − 2v 2 |− 4 = |kx| |1 − 2v 2 | = |kx|−4 1 − 2v 2 = ±|kx|−4 1 v2 = 1 ± |kx|−4 2 r 1 v=± (1 ± |kx|−4 ) 2 r 1 y = ±x (1 ± |kx|−4 ) 2
6
(d) y 0 =
x3 +y 3 xy 2 .
Start by checking that the equation is homogeneous: let f (x, y) = x3 +y 3 xy 2 . Then, f (xt, yt) = Let y = vx, so
dy dx
x3 + y 3 (xt)3 + (yt)3 = = f (x, y). 2 (xt)(yt) xy 2
dv = v + x dx . Substitute these into the equation:
1 + v3 dv x3 + (vx)3 = = dx x(vx)2 v2 dv 1 + v3 1 x = −v = 2 2 dx v v dx 2 v dv = Z Zx dx 2 v dv = x 3 v = ln |x| + C 3
v+x
Let C = ln |k|, so v3 = ln |kx| 3 3 v = 3 ln |kx| = ln |kx|3 p v = 3 ln |kx|3 p y = x 3 ln |kx|3 .
7
(e) y 2 + x2 y 0 = xyy 0 . Start by writing this in the form y 0 = f (x, y): x2 y 0 − xyy 0 = −y 2 (x2 − xy)y 0 = −y 2 y0 = −
y2 x2 − xy
Next, check that the equation is homogeneous: f (xt, yt) = − Let y = vx, so
dy dx
(xt)2
(yt)2 y2 =− 2 = f (x, y). − (xt)(yt) x − xy
dv = v + x dx . Substitute these into the equation:
dv (vx)2 v2 =− 2 = dx x − x(vx) 1−v dv v2 v x = −v = dx 1−v v−1 v−1 dx dv = v Z Zx v−1 dx dv = v x Z Z 1 dx 1 − dv = v x v − ln |v| = ln |x| + C
v+x
Let C = ln |k|, so v − ln |v| = − ln |kx|. This expression cannot be simplfied further, so we simply substitute v = xy : y y − ln | | = ln |kx|. x x
8
3. Find the function y = y(x) such that
dy dx
= y 2 (1 + 2x) and y(0) = −1.
Separating variables and integrating yields: dy = 1 + 2x dx y2 Z Z dy = 1 + 2x dx y2 1 − = x + x2 + C y 1 y=− x + x2 + C Now, we have −1 = y(0) = − so C = 1 and y=−
1 , C
1 . x + x2 + 1
4. Solve the following first-order linear differential equations. (a)
dy dx
+ xy − x5 = 0 This is first-order linear, with P (x) = x1 and Q(x) = x5 , so it has solution Z R R y = e− P (x) dx e P (x) dx Q(x) dx + C Z R R dx dx = e− x e x x5 dx + C Z = e− ln x eln x x5 dx + C Z 1 = x6 dx + C x 1 1 7 x +C = x 7 C 1 = x6 + 7 x
9
(b)
dy dx
+ y = sin x Here, P (x) = 1 and Q(x) = sin x, so Z R R − dx dx y=e e sin x dx + C Z −x x e sin x dx + C =e x e sin x − ex cos x −x =e +C 2 (see section 1.1.5 for this integral.) =
(c)
dy dx
−
y x+1
sin x − cos x + Ce−x 2
= x2
1 Here, P (x) = − 1+x and Q(x) = x2 , so
R 1 e− 1+x dx x2 dx + C Z ln (1+x) − ln (1+x) 2 =e e x dx + C Z x2 = (1 + x) dx + C 1+x Z (x2 − 1) + 1 = (1 + x) dx + C 1+x Z (1 + x)(1 − x) + 1 = (1 + x) dx + C 1+x Z 1 = (1 + x) 1−x+ dx + C 1+x x2 = (1 + x) x − + ln |1 + x| + C 2 R
y=e
1 1+x
dx
10
(d) xy 0 + y = 3x cos 2x. Rearranging, we get y0 +
1 y = 3 cos 2x, x
so P (x) = − x1 and Q(x) = 3 cos 2x. Thus, Z R 1 R 1 y = e− x dx 3e x dx cos 2x dx + C Z = e− ln x 3eln x cos 2x dx + C Z 1 x cos 2x dx + C = x R We integrate x cos 2x dx by parts: u = 3x
dv = cos 2x dx 1 du = 3 dx v = sin 2x 2 Thus, Z Z 3 3 3 3 x cos 2x dx = x sin 2x − sin 2x dx = x sin 2x + cos 2x, 2 2 2 4 so, 1 3 3 y= x sin 2x + cos 2x + C x 2 4 3 3 C = sin 2x + cos 2x + . 2 4x x
11
5. Solve the following linear differential equations. Use the initial conditions given to find all unknown constants. (a) y 0 − 3xy = 15x, y(0) = 5. Here, P (x) = −3x and Q(x) = 15x, so Z R R y = e 3x dx 15xe− 3x dx dx + C Z 3 2 3 2 = e2x 15xe− 2 x dx + C . Let u = − 23 x2 , so du = −3x dx and − du 3 = x dx: Z Z 3 2 15 eu du 15xe− 2 x dx = − 3 = −5eu 3
2
= −5e− 2 x Thus, 3
y = e2x
2
3
2
−5e− 2 x + C 3
= −5 + Ce 2 x
2
Lastly, we use the initial condition y(0) = 5 to find C: 5 = y(0) = −5 + C C = 10 3
2
Thus, y = −5 + 10e 2 x .
12
(b) y 0 + 4xy = x, y(0) = 54 . Here, P (x) = 4x and Q(x) = x, so Z R R − 4x dx − 4x dx y=e xe dx + C Z 2 2 = e−2x xe2x dx + C Let u = 2x2 , so du = 4x dx and du 4 = x dx: Z Z 2 1 eu du xe2x dx = 4 1 = eu 4 1 2 = e2x 4 Thus, y = e−2x =
2
1 2x2 e +C 4
2 1 + Ce−2x 4
Lastly, we use the initial condition y(0) =
5 4
5 1 = y(0) = + C 4 4 C=1 Thus, y =
1 4
2
+ e−2x .
13
to find C:
(c) y 0 + x1 y = ex , y(1) = 2. Here, P (x) = x1 and Q(x) = ex , so Z R 1 1 x int x − x e e dx + C y=e Z − ln x x ln x =e e e dx + C Z 1 x = xe dx + C x R Integrate xex dx by parts: dv = ex dx
u=x
v = ex
du = 1 Thus, Z
x
x
Z
xe dx = xe −
ex dx = xex − ex ,
and 1 (xex − ex + C) x ex = ex + +C x
y=
Lastly, we use the initial condition y(1) = 2 to find C: 2 = y(1) = e − e + C C = 2, so y = ex +
ex x
+ 2.
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6. Make a substitution to solve the following Bernoulli equations. If initial conditions are given, use them to solve for all unknown constants. (a) y 0 + x2 y = x2 y 2 . 0
Let z = y 1−2 = y −1 , so y = z1 and y 0 = − zz2 . Substitute these into the original equation and isolate for z 0 : y 0 + x2 y = x2 y 2 x2 z0 x2 + = z2 z z2 z 0 − x2 z = −x2
−
This is linear, with P (x) = −x2 and Q(x) = −x2 , so Z R 2 R x dx 2 − x2 − x e z=e dx + C Z 1 3 x 2 − 13 x3 3 − x e dx + C =e Let u = − 31 x3 , so du = −x2 dx: Z Z 1 3 1 3 − x2 e− 3 x dx = eu du = eu = e− 3 x , and
1
z = e3x
3
1 3 1 3 e− 3 x + C = 1 + Ce 3 x .
Thus, y=
1 1 = 1 3 . z 1 + Ce 3 x
15
√ (b) y 0 + xy = x 3 y. 1
2
3
1
Let z = y 1− 3 = y 3 , so y = z 2 and y 0 = 23 z 2 z 0 . Substitute these into the original equation and isolate for z 0 : 1
y 0 + xy = xy 3
3 3 1 1 3 1 0 z 2 z + xz 2 = x(z 2 ) 3 = xz 2 2 2 2 z 0 + xz = x 3 3
This is linear, with P (x) = 23 x and Q(x) = 23 x, so Z R 2 2x − 23 x dx 3 z=e xe dx + C 3 Z 2 2 x2 − x3 3 =e xe dx + C 3 Let u =
x2 3 ,
so du = 23 x dx: Z Z x2 2 x2 xe 3 dx = eu du = eu = e 3 , 3
so z = e− and
x2 3
e
x2 3
x2 + C = 1 + Ce− 3
23 2 x2 . y = z 3 = 1 + Ce− 3
16
√ (c) y 0 + 3x2 y = x2 y. 1
1
Let z = y 1− 2 = y 2 , so y = z 2 and y 0 = 2zz 0 . Substitute these into the original equation and isolate for z 0 : 1
y 0 + 3x2 y = x2 y 2
2zz 0 + 3x2 z 2 = x2 z x2 3 z 0 + x2 z = 2 2 This is linear, with P (x) = 23 x2 and Q(x) = x2 , so 3
= e− Let u =
x3 2 ,
x2 3 x2 e 2 dx + C 2 Z 2 x x3 e 2 dx + C 2
2
z = e− 2 x x3 2
dx
so du = 32 x2 and Z
Z
du 3
x2 x3 1 e 2 dx = 2 3
so 3
z=e
− x2
Z
=
x2 2
and 2
y=z =
17
dx:
eu du =
1 x3 e 2 +C 3
=
1 u 1 x3 e = e2, 3 3 x3 1 + Ce− 2 , 3
x3 1 + Ce− 2 3
.
1
(d) y 0 + x4 y = xy 4 , y(1) = 1. 1
3
4
1
Let z = y 1− 4 = y 4 , so y = z 3 and y 0 = 34 z 3 z 0 . Substitute these into the original equation and isolate for z 0 : 1 4 y = xy 4 x 1 4 1 0 4 4 z 3 z + z 3 = xz 3 3 x 3 3 z0 + z = x x 4
y0 +
and Q(x) = 34 x, so Z R 3 3 R 3 z = e− x xe x dx + C 4 Z 3 3 ln x = e−3 ln x xe dx + C 4 Z 3 3 = x−3 xx dx + C 4 Z 3 4 = x−3 x dx + C 4 3 5 = x−3 x +C 20 3 2 = x + Cx−3 20
This is linear with P (x) =
3 x
Thus,
4
y = z3 =
3 2 x + Cx−3 20
43 .
Lastly, we use the initial condition y(1) = 1 to find C: 1 = y(1) =
43 3 3 +C +C 1= 20 20 17 C= 20
Thus, y=
3 2 17 x + 20 20x3
18
43 .
(e) y 0 − x1 y = ex y 2 , y(1) = 12 . 0
Let z = y 1−2 = y −1 , so y = z1 and y 0 = − zz2 . Substitute these into the original equation and isolate for z 0 : 1 y = ex y 2 x 11 z0 1 − 2− = ex 2 z xz z 1 0 x z + z = −e x y0 −
1 x
and Q(x) = −ex , so Z R 1 1 − x dx x x − e e dx + C z=e Z − ln x x ln x =e − e e dx + C Z 1 x = − xe dx + C x
This is linear with P (x) =
We do this by integration by parts, with u=x
dv = ex dx v = ex ,
du = dx so
Z
x
x
xe dx = xe − and
Z
ex dx = xex − ex ,
z=
1 (−xex + ex + C) . x
y=
1 1 = . z −xex + ex + C
Thus,
Lastly, we use the initial condition y(1) =
1 2
to find C:
1 1 1 = y(1) = = 2 −e + e + C C C = 2, so y =
1 −xex +ex +2 .
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7. Solve the following exact differential equations. (a) (y 2 cos x) dx + (y 2 + 2y sin x) dy = 0. Here M (x, y) = y 2 cos x and N (x, y) = y 2 + 2y sin x, and ∂M ∂N = 2y cos x = , ∂y ∂x so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z = y 2 cos x dx + h(y) = y 2 sin x + h(y) Since
∂g ∂y
= N (x, y), we must have
y 2 + 2y sin x =
Thus, h(y) =
y3 3
∂ 2 y sin x + h(y) = 2y sin x + h0 (y) ∂y h0 (y) = y 2
+ C, so g(x, y) = y 2 sin x + y 2 sin x + y 2 sin x +
y3 +C =0 3
y3 =k 3
20
y3 +C 3
k = −C
(b) (3x2 − 2xy + 2) dx + (6y 2 − x2 + 3) = 0 Here M (x, y) = 3x2 − 2xy + 2 and N (x, y) = 6y 2 − x2 + 3, and ∂M ∂N = −2x = , ∂y ∂x so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z = 3x2 − 2xy + 2 dx + h(y) = x3 − x2 y + 2x + h(y) Since
∂g ∂y
= N (x, y), we must have
6y 2 − x2 + 3 =
∂ 3 x − x2 y + 2x + h(y) = −x2 + h0 (y) ∂y h0 (y) = 6y 2 + 3
Thus, h(y) = 2y 3 + 3y + C, so g(x, y) = x3 − x2 y + 2x + 2y 3 + 3y + C x3 − x2 y + 2x + 2y 3 + 3y + C = 0 x3 − x2 y + 2x + 2y 3 + 3y = k
21
k = −C
(c) (ex sin y − 2y sin x) dx + (ex cos y + 2 cos x) dy = 0. Here M (x, y) = ex sin y − 2y sin x and N (x, y) = ex cos y + 2 cos x, and ∂M ∂N = ex cos y − 2 sin x = , ∂y ∂x so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z = ex sin y − 2y sin x dx + h(y) = ex sin y + 2y cos x + h(y) Since
∂g ∂y
= N (x, y), we must have
ex cos y + 2 cos x =
∂ x [e sin y + 2y cos x + h(y)] = ex cos y + 2 cos x + h0 (y) ∂y h0 (y) = 0
Thus, h(y) = C, so g(x, y) = ex sin y + 2y cos x + C ex sin y + 2y cos x + C = 0 x
e sin y + 2y cos x = k
22
k = −C
(d) √
x x2 +y 2
dx + ey + √
Here M (x, y) = √
y x2 +y 2
x x2 +y 2
= 0.
and N (x, y) = ey + √
y , x2 +y 2
and
∂N ∂M xy = − p , 3 = ∂y ∂x x2 + y 2 so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z x p dx + h(y) = 2 x + y2 p = x2 + y 2 + h(y) Since
∂g ∂y
= N (x, y), we must have
ey + p
y x2 + y 2
=
i y ∂ hp 2 x + y 2 + h(y) = p + h0 (y) 2 2 ∂y x +y h0 (y) = ey
Thus, h(y) = ey + C, so p g(x, y) = x2 + y 2 + ey + C p x2 + y 2 + ey + C = 0 p x2 + y 2 + ey = k k = −C
23
(e)
ln y 1+x2
dx +
arctan x y
Here M (x, y) =
+ y ln y dy = 0.
ln y 1+x2
and N (x, y) =
arctan x y
+ y ln y, and
∂M 1 ∂N =− = , ∂y y(1 + x2 ) ∂x so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z ln y = dx + h(y) 1 + x2 = arctan x ln y + h(y) Since
∂g ∂y
= N (x, y), we must have
∂ arctan x arctan x + y ln y = [arctan x ln y + h(y)] = + h0 (y) y ∂y y h0 (y) = y ln y Thus, Z h(y) =
y ln y dy,
which we solve by partial fractions, taking u = ln y du = so
y2 h(y) = ln y − 2
dv = y dy
dy y
Z
v=
y2 2
y2 y2 y dy = ln y − + C. 2 2 4
Thus, y2 y2 ln y − +C 2 4 y2 y2 arctan x ln y + ln y − +C =0 2 4 2 2 y y arctan x ln y + ln y − =k k = −C 2 4 g(x, y) = arctan x ln y +
24
8. Find an integrating factor for the following differential equations and solve. (a) (x sin y) dx + (2x cos y) dy = 0. Here M (x, y) = x sin y and N (x, y) = 2x cos y, so ∂M = x cos y ∂y Since
∂M ∂y
6=
∂N ∂x ,
∂N = 2 cos y. ∂x
and
the equation is not exact. However, 1 N
∂M ∂N − ∂y ∂x
1 1 − 2 x
=
is a function of x alone, so we may solve by introducing an integrating factor 1 R 1 1 1 e2x e 2 − x dx = e 2 x−ln x = . x 1
Thus, if we take M1 (x, y) = 1
e2x x N (x, y)
e2x x M (x, y)
1
= e 2 x sin y and N1 (x, y) =
1
= 2e 2 x cos y then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: 1 ∂N1 ∂M1 = e 2 x cos y = . ∂y ∂x
We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z 1 = e 2 x sin y dx + h(y) 1
= 2e 2 x sin y + h(y). Since
∂g ∂y
= N1 (x, y), we must have 1
2e 2 x cos y =
1 1 ∂ [2e 2 x sin y + h(y)] = 2e 2 x cos y + h0 (y) ∂y h0 (y) = 0
Thus, h(y) = C, so 1
g(x, y) = 2e 2 x sin y + C 1
2e 2 x sin y + C = 0 1
2e 2 x sin y = k,
where k = −C. 25
(b)
1 x
dx +
x y
dy = 0.
Here M (x, y) =
1 x
and N (x, y) = xy , so ∂M =0 ∂y
Since
∂M ∂y
6=
∂N ∂x ,
∂N 1 = . ∂x y
and
the equation is not exact. However, 1 N
∂M ∂N − ∂y ∂x
=−
1 x
is a function of x alone, so we may solve by introducing an integrating factor R 1 1 e − x dx = e− ln x = . x Thus, if we take M1 (x, y) = M (x,y) = x12 and N1 (x, y) = x then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: ∂M1 ∂N1 =0= . ∂y ∂x We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z 1 = dx + h(y) x2 1 = − + h(y). x Since
∂g ∂y
= N1 (x, y), we must have 1 ∂ 1 = [− + h(y)] = h0 (y) y ∂y x
Thus, h(y) = ln y + C, so g(x, y) = − −
1 + ln y + C x
1 + ln y + C = 0 x 1 − + ln y = k, x
where k = −C.
26
N (x,y) x
=
1 y
(c) 2xy 2 dx + 32 x2 y + y dy = 0. Here M (x, y) = 2xy 2 and N (x, y) = 23 x2 y + y, so ∂M = 4xy ∂y Since
∂M ∂y
6=
∂N ∂x ,
∂N = 3xy. ∂x
and
the equation is not exact. However, 1 M
∂M ∂N − ∂y ∂x
=
1 2y
is a function of y alone, so we may solve by introducing an integrating factor R 1 1 1 1 e− 2y dy = e− 2 ln y = y − 2 = √ . y 3
(x,y) Thus, if we take M1 (x, y) = M√ = 2xy 2 and N1 (x, y) = y √ √ 3 2 y + y then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: 2x
∂M1 ∂N1 √ = 3x y = . ∂y ∂x We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z 3 = 2xy 2 dx + h(y) 3
= x2 y 2 + h(y). Since
∂g ∂y
= N1 (x, y), we must have 3 2√ ∂ 2 3 3 √ √ x y+ y= [x y 2 + h(y)] = x2 y + h0 (y) 2 ∂y 2 √ 0 h (y) = y 3
Thus, h(y) = 23 y 2 + C, so 3 2 3 g(x, y) = x2 y 2 + y 2 + C 3 3 2 3 x2 y 2 + y 2 + C = 0 3 3 2 3 x2 y 2 + y 2 = k, 3
where k = −C. 27
N (x,y) √ y
=
(d) (x2 + 4xy) dx + x dy = 0. Here M (x, y) = x2 + 4xy and N (x, y) = x, so ∂M = 4x ∂y Since
∂M ∂y
6=
∂N ∂x ,
∂N = 1. ∂x
and
the equation is not exact. However, 1 N
∂M ∂N − ∂y ∂x
=4−
1 x
is a function of y alone, so we may solve by introducing an integrating factor R 1 e4x . e 4− x dx = e4x−ln x = x 4x
Thus, if we take M1 (x, y) = ex M (x, y) = xe4x +4ye4x and N1 (x, y) = e4x 4x then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: x N (x, y) = e ∂M1 ∂N1 = 4e4x = . ∂y ∂x We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z = xe4x + 4ye4x dx + h(y) = Since
∂g ∂y
1 1 4x xe − e4x + ye4x + h(y). 4 16
= N1 (x, y), we must have
e4x =
1 ∂ 1 4x [ xe − e4x + ye4x + h(y)] = e4x + h0 (y) ∂y 4 16 h0 (y) = 0
Thus, h(y) = C, so 1 4x 1 xe − e4x + ye4x + C 4 16 1 4x 1 xe − e4x + ye4x + C = 0 4 16 1 4x 1 xe − e4x + ye4x = k, 4 16
g(x, y) =
where k = −C. 28
(e) xy dx + (y 2 + x2 ) dy = 0. Here M (x, y) = xy and N (x, y) = y 2 + x2 , so ∂M =x ∂y Since
∂M ∂y
6=
∂N ∂x ,
∂N = 2x. ∂x
and
the equation is not exact. However, 1 M
∂M ∂N − ∂y ∂x
=−
1 y
is a function of y alone, so we may solve by introducing an integrating factor R 1 e y dy = eln y = y Thus, if we take M1 (x, y) = yM (x, y) = xy 2 and N1 (x, y) = yN (x, y) = y 3 + x2 y then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: ∂M1 ∂N1 = 2xy = . ∂y ∂x We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z = xy 2 dx + h(y) = Since
∂g ∂y
1 2 2 x y + h(y). 2
= N1 (x, y), we must have y 3 + x2 y =
∂ 1 2 2 [ x y + h(y)] = x2 y + h0 (y) ∂y 2 h0 (y) = y 3
Thus, h(y) = 41 y 4 + C, so 1 2 2 1 4 x y + y +C 2 4 1 2 2 1 4 x y + y +C =0 2 4 1 2 2 1 4 x y + y = k, 2 4
g(x, y) =
where k = −C.
29
9. Make a substitution to solve the following differential equations. (a) y 0 = (2x + y)2 − 6, y(0) = 4. Let v = 2x + y, so v 0 = 2 + y 0 and y 0 = v − 2. Then, the equation becomes: v0 − 2 = v2 − 6 v0 = v2 − 4 This equation is separable:
Z
dv = 1 dx v2 − Z 4
dv = −4
v2
1 dx = x + C
We can do this integral by partial fractions: 1 A B (A + B)v + (−2A + 2B) = + = , v2 − 4 v+2 v−2 v2 − 4 so A+B =0 −2A + 2B = 1 This pair of linear equations had solution A = − 41 and B = 14 : Z Z dv 1 1 1 = − dv 2 v −4 4 v−2 v+2 1 = ln |v − 2| − ln |v + 2| 4 v−2 1 |. = ln | 4 v+2
30
Thus, we have 1 v−2 ln | |=x+C 4 v+2 v−2 ln | | = 4x + C v+2 v−2 | | = De4x v+2 v−2 = ±De4x v+2 4 = ±De4x 1− v+2 4 −2 v= 1 ∓ De4x 4 y= − 2 − 2x 1 ∓ De4x Lastly, we use the initial condition y(0) = 4 to solve for the constant D: 4 −2 1∓D 1 ∓D = − 3 1 D=± , 3
4 = y(0) =
so, in either case, y=
4 − 2 − 2x. 1 − 13 e4x
(b) y 0 = e2x+y + 1. Let v = 2x + y, so v 0 = 2 + y 0 . Then, the equation becomes v 0 − 2 = y 0 = ev + 1 v 0 = ev + 3 This equation is separable: dv = 1 dx +Z 3
ev Z
dv = ev + 3
31
1 dx = x + C
Let u = ev , so du = ev dv and du u = dv. Then, Z Z dv du = ev + 3 u(u + 3) Z 11 1 1 − = 3u 3u+3 1 1 = ln |u| − ln |u + 3| 3 3 u 1 | = ln | 3 u+3 Thus, 1 u ln | |=x+C 3 u+3 u ln | | = 3x + C u+3 u | | = De3x u+3 u = ±De3x u+3 3 1− = ±De3x u+3 3 −3 u= 1 ± De3x Thus, 3 ev = −3 1 ± De3x 3 v = ln − 3 , 1 ± De3x and 3 2x + y = ln − 3 1 ± De3x 3 − 3 − 2x y = ln 1 ± De3x
32
0.2
Higher Order Differential Equations
1. Find the general solution for the following linear differential equations. (a) y 00 − y 0 − 2y = 0. This has characteristic polynomial λ2 − λ − 2 = (λ + 1)(λ − 2), so it has two distinct real roots, λ1 = −1 and λ2 = 2, so that y = c1 e−x + c2 e2x . (b) 4y 00 + 20y 0 + 25y = 0. This has characteristic polynomial 4λ2 + 20λ + 25 = (2λ + 5)2 , so it has a single real root λ = − 25 . Thus, 5
y = (c1 + c2 x)e− 2 x . (c) y 00 − 4y 0 + 5y = 0. This has characteristic polynomial λ2 − 4λ + 5 = 0, which has roots λ2 = 2 − i
λ1 = 2 + i, so
y = e2x (c1 cos x + c2 sin x). (d) 12y 00 + 16y 0 + 5y = 0.. This has characteristic polynomial 12λ2 + 16λ + 5 = (2λ + 1)(6λ + 5), so it has two distinct real roots, λ1 = − 21 and λ2 = − 65 , so that 1
5
y = c1 e − 2 x + c2 e − 6 x . (e) y 00 + 3y 0 + 5y = 0.. This has characteristic polynomial λ2 + 3λ + 5 = 0, which has roots √ √ 3 11 3 11 λ1 = − + i , λ2 = − − i 2 2 2 2 so
√
y=e
− 23 x
√ 11 11 (c1 cos x + c2 sin x). 2 2
33
(f) y 000 − 3y 0 − 2y = 0. This has characteristic polynomial λ3 − 3λ − 2 = 0. We make a guess for the first root, taking λ1 = −1: (−1)3 − 3(−3) − 2(−1) = 0. Using long division, you can show λ3 − 3λ − 2 = (λ + 1)(λ2 − λ − 2) = (λ + 1)(λ + 1)(λ − 2) so the other roots are λ2 = −1 and λ3 = 2. Thus, the general solution is y = c1 e−x + c2 xe−x + c3 e2x . (g) y 000 − 6y 00 + 12y 0 − 8y = 0. This has characteristic polynomial λ3 − 6λ2 + 12λ − 8 = 0. We make a guess for the first root, taking λ1 = 2: (2)3 − 6(2)2 + 12(2) − 8 = 0. Using long division, you can show λ3 − 6λ2 + 12λ − 8 = (λ − 2)(λ2 − 4λ + 4) = (λ − 2)3 so the other roots are λ2 = 2 and λ3 = 2. Thus, the general solution is y = c1 e2x + c2 xe2x + c3 x2 e2x . (h) y 000 − 7y 00 + 17y 0 − 15y = 0. This has characteristic polynomial λ3 − 7λ2 + 17λ − 15 = 0. We make a guess for the first root, taking λ1 = 3: (3)3 − 7(3)2 + 17(3) − 15 = 0. Using long division, you can show λ3 − 7λ2 + 17λ − 15 = (λ − 3)(λ2 − 4λ + 5) Using the quadratic formula, the other roots are λ2 = 2 + i and λ3 = 2 − i. Thus, the general solution is y = c1 e3x + c2 e2x cos x + c3 e2x sin x.
34
(i) y (6) − y = 0. This has characteristic polynomial λ6 − 1 = (λ3 − 1)(λ3 + 1) = (λ − 1)(λ2 + λ + 1)(λ + 1)(λ2 − λ + 1) Thus λ1 = 1 and λ2 = −1 are roots. The roots of the two remaining quadratic polynomials can be found using the quadratic formula: √ √ 3 1 3 1 and λ3 , λ4 = ± i . λ3 , λ4 = − ± i 2 2 2 2 The general solution is thus √ y = c1 ex + c2 e−x + c3 e
− 12
3 x 2
cos 1
+ c5 e 2 cos
√
!
! 3 + c4 e sin x 2 √ ! √ ! 1 3 3 x + c6 e 2 sin x . 2 2 − 21
(j) y (4) − 8y 00 + 16y = 0. This has characteristic polynomial λ4 − 8λ2 + 16 = (λ2 − 4)2
= (λ + 2)2 (λ − 2)2
Thus, the roots are λ1 = λ2 = −2 and λ3 = λ4 = 2. The general solution is y = c1 e−2x + c2 xe−2x + c3 e2x + c4 xe2x . 2. Solve y 00 − 4y = 0, subject to y(0) = 1, y 0 (0) = −1. Start by finding the general solution. This has characteristic polynomial λ2 − 4 = (λ + 2)(λ − 2), so it has two distinct real roots, λ1 = −2 and λ2 = 2, so that y = c1 e−2x + c2 e2x . Now, y 0 = −2c1 e−2x + 2c2 e2x , so, imposing initial conditions, we have 1 = y(0) = c1 + c2 −1 = y 0 (0) = −2c1 + 2c2 , giving two linear equations for the two constants c1 and c2 . Adding two times the first to the second gives 1 = 4c2 , so c2 = 41 . From the first, we see that c1 = 34 . Thus, 3 1 y = e−2x + e2x . 4 4
35
3. Find the general solution to the following non-homogeneous equations with constant coefficients, using the method of undetermined coefficients. (a) y 00 − 3y 0 − 4y = 5x2 . We start by finding all solutions to the homogeneous equation y 00 − 3y 0 − 4y = 0. This has characteristic equation λ2 − 3λ − 4 = (λ + 1)(λ − 4) = 0, which has distinct real roots λ1 = −1 and λ2 = 4, so the solution to the homogeneous equation is yh = c1 e−x + c2 e4x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A2 x2 +A1 x+A0 , so yp0 = 2A2 x + A1 yp00 = 2A2 Substituting these into y 00 − 3y 0 − 4y = 5x2 , we get (2A2 ) − 3(2A2 x + A1 ) − 4(A2 x2 + A1 x + A0 ) = 5x2 Collecting like terms on the right hand side, we get −4A2 x2 + (−6A2 − 4A1 )x + (−3A1 − 4A0 ) = 5x2 So, equating the coefficients of like terms on the right and left hand sides: −4A2 = 5 −6A2 − 4A1 = 0 −3A1 − 4A0 = 0
95 , A1 = This system of linear equations has solution A0 = − 32 5 A2 = − 4 . Thus, 5 15 95 yp = − x2 + x + − 4 8 32 and 5 15 95 y = yh + yp = c1 e−x + c2 e4x − x2 + x + − . 4 8 32
36
15 8 ,
and
(b) y 00 − 3y 0 − 18y = 4e2x . We start by finding all solutions to the homogeneous equation y 00 − 3y 0 − 18y = 0. This has characteristic equation λ2 − 3λ − 18 = (λ + 3)(λ − 6) = 0, which has distinct real roots λ1 = −3 and λ2 = 6, so the solution to the homogeneous equation is yh = c1 e−3x + c2 e6x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = Ae2x , so yp0 = 2Ae2x yp00 = 4Ae2x Substituting these into y 00 − 3y 0 − 18y = 4e2x , we get (4Ae2x ) − 3(2Ae2x ) − 18(Ae2x ) = 4e2x Collecting like terms on the right hand side, we get −20Ae2x = 4e2x So, equating the coefficients of like terms on the right and left hand sides: −20A = 4 Thus, A = − 51 , yp = − 51 e2x , and 1 y = yh + yp = c1 e−3x + c2 e6x − e2x . 5
37
(c) y 00 − 5y 0 = 2 sin 2x + 5 cos 2x. We start by finding all solutions to the homogeneous equation y 00 − 5y 0 = 0. This has characteristic equation λ2 − 5λ = λ(λ − 5) = 0, which has distinct real roots λ1 = 0 and λ2 = 5, so the solution to the homogeneous equation is yh = c1 + c2 e5x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A sin 2x + B cos 2x, so yp0 = 2A cos 2x − 2B sin 2x yp00 = −4A sin 2x − 4B cos 2x Substituting these into y 00 − 5y 0 = 2 sin 2x + 5 cos 2x, we get (−4A sin 2x−4B cos 2x)−5(2A cos 2x−2B sin 2x) = 2 sin 2x+5 cos 2x Collecting like terms on the right hand side, we get (−4A + 10B) sin 2x + (−4B − 10A) cos 2x = 2 sin 2x + 5 cos 2x So, equating the coefficients of like terms on the right and left hand sides: −4A + 10B = 2 −4B − 10A = 5 This system of linear equations has solution A = − 21 and B = 0. Thus, 1 yp = − sin 2x 2 and 1 y = yh + yp = c1 + c2 e5x − sin 2x. 2
38
(d) y 00 + 2y 0 = 2e2x − 3 sin 3x + 3 cos 3x. We start by finding all solutions to the homogeneous equation y 00 + 2y 0 = 0. This has characteristic equation λ2 + 2λ = λ(λ + 2) = 0, which has distinct real roots λ1 = −2 and λ2 = 0, so the solution to the homogeneous equation is yh = c1 + c2 e−2x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = Ae2x +B sin 3x+ C cos 3x, so yp0 = 2Ae2x + 3B cos 3x − 3C sin 3x yp00 = 4Ae2x − 9B sin 3x − 9C cos 3x Substituting these into y 00 + 2y 0 = 2e2x − 3 sin 3x + 3 cos 3x, we get (4Ae2x − 9B sin 3x − 9C cos 3x) + 2(2Ae2x + 3B cos 3x − 3C sin 3x) = 2e2x − 3 sin 3x + 3 cos 3x. Collecting like terms on the right hand side, we get 8Ae2x +(−9B−6C) sin 3x+(6B−9C) cos 3x = 2e2x −3 sin 3x+3 cos 3x So, equating the coefficients of like terms on the right and left hand sides: 8A = 2 −9B − 6C = 3 6B − 9C = 3 This system of linear equations has solution A = 1 C = − 13 . Thus, yp =
1 4,
B =
5 13 ,
1 2x 5 1 e + sin 3x − cos 3x 4 13 13
and 1 5 1 y = yh + yp = c1 + c2 e−2x + e2x + sin 3x − cos 3x. 4 13 13
39
and
(e) y 00 − 2y 0 − 15y = 32x2 ex + 40xex . We start by finding all solutions to the homogeneous equation y 00 − 2y 0 − 15y = 0. This has characteristic equation λ2 − 2λ − 15 = (λ + 3)(λ − 5) = 0, which has distinct real roots λ1 = −3 and λ2 = 5, so the solution to the homogeneous equation is yh = c1 e−3x + c2 e5x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A2 x2 ex +A1 xex + A0 ex , so yp0 = A2 x2 ex + 2A2 xex + A1 xex + A1 ex + A0 ex yp00 = A2 x2 ex + 4A2 xex + 2A2 ex + A1 xex + 2A1 ex + A0 ex Substituting these into y 00 − 2y 0 − 15y = 32x2 ex + 40xex , we get (A2 x2 ex + 4A2 xex + 2A2 ex + A1 xex + 2A1 ex + A0 ex ) − 2(A2 x2 ex + 2A2 xex + A1 xex + A1 ex + A0 ex ) − 15(A2 x2 ex + A1 xex + A0 ex ) = 32x2 ex + 40xex Collecting like terms on the right hand side, we get −16A2 x2 ex − 16A1 xex + (−16A0 + 2A2 )ex = 32x2 ex + 40xex So, equating the coefficients of like terms on the right and left hand sides: −16A2 = 32 −16A1 = 40 −16A0 + 2A2 = 0
This system of linear equations has solution A0 = − 41 , A1 = − 25 , and A2 = −2. Thus, 5 1 yp = −2x2 ex − xex − ex 2 4 and
1 y = yh + yp = c1 e−3x + c2 e5x − 2x2 ex + ex . 4
40
(f) y 000 − 7y 00 + 17y 0 − 15y = 15x2 − 60x + 27. We start by finding all solutions to the homogeneous equation y 000 − 7y 00 − 17y 0 − 15y = 0. This has characteristic equation λ3 − 7λ2 + 17λ − 15 = (λ − 3)(λ2 − 4λ + 5) = 0. Thus, λ1 = 3 is one root of this equation. We find the other two roots using the quadratic formula: λ2 , λ3 = 2 ± i. The solution to the homogeneous equation is then yh = c1 e3x + c2 e2x cos x + c3 e2x sin x, where c1 , c2 and c3 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A2 x2 +A1 x+A0 , so yp0 = 2A2 x + A1 yp00 = 2A2 yp000 = 0 Substituting these into y 000 − 7y 00 + 17y 0 − 15y = 15x2 − 60x + 27, we get −7(2A2 ) + 17(2A2 x + A1 ) − 15(A2 x2 + A1 x + A0 ) = 15x2 − 60x + 27 Collecting like terms on the right hand side, we get −15A2 x2 + (34A2 − 15A1 )x + (17A1 − 15A0 ) = 15x2 − 60x + 27 So, equating the coefficients of like terms on the right and left hand sides: −15A2 = 15 34A2 − 15A1 = −60 17A1 − 15A0 = 27
This system of linear equations has solution A0 = A2 = −1. Thus, 26 37 yp = −x2 + x + 15 225 and
37 225 ,
y = yh + yp = c1 e3x + c2 e2x cos x + c3 e2x sin x − x2 +
41
A1 =
26 15 ,
and
26 37 x+ . 15 225
(g) y (4) + 3y 000 + 2y 00 = 144e2x . We start by finding all solutions to the homogeneous equation y (4) + 3y 000 + 2y 00 = 0. This has characteristic equation λ4 + 3λ3 + 2λ2 = λ2 (λ + 1)(λ + 2) = 0, which has a repeated root, λ1 = λ2 = 0, and two distinct roots, λ3 = −1 and λ4 = −2. The solution to the homogeneous equation is then yh = c1 + c2 x + c3 e−x + c4 e−2x , where c1 , c2 , c3 and c4 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = Ae2x , so yp0 = 2Ae2x yp00 = 4Ae2x yp000 = 8Ae2x yp(4) = 16Ae2x Substituting these into y (4) + 3y 000 + 2y 00 = 10e2x , we get (16Ae2x ) + 3(8Ae2x ) + 2(4Ae2x ) = 144e2x Collecting like terms on the right hand side, we get 48Ae2x = 144e2x So, A = 3, yp = 3e2x and y = yh + yp = c1 + c2 x + c3 e−x + c4 e−2x + 3e2x
42
4. Solve the following Cauchy-Euler equations. (a) x2 y 00 − 5xy 0 + 13y = 0. Here, a = −5 and b = 13, so the auxiliary equation is λ2 −6λ+13 = 0. Using the quadratic formula, we see this has roots λ1 , λ2 = 3 ± 2i, so the solution to the equation is y = c1 x3 cos (2 ln x) + c2 x3 sin (2 ln x), where c1 and c2 are arbitrary constants. (b) x2 y 00 − xy 0 + 5y = 0. Here, a = −1 and b = 5, so the auxiliary equation is λ2 − 2λ + 5 = 0. Using the quadratic formula, we see this has roots λ1 , λ2 = 1 ± 2i, so the solution to the equation is y = c1 x cos (2 ln x) + c2 x sin (2 ln x), where c1 and c2 are arbitrary constants. (c) x2 y 00 + 6xy 0 + 25y = 0. Here, a = 6 and b = 25, so the auxiliary equation is λ2 + 5λ + 25 = 0. Using the quadratic formula, we see this has roots √ 5 λ1 , λ2 = − ± i5 3, 2 so the solution to the equation is √ √ 5 5 y = c1 x− 2 cos (5 3 ln x) + c2 x− 2 sin (5 3 ln x), where c1 and c2 are arbitrary constants.
43
5. Find the general solution of the following non-homogeneous Cauchy-Euler equations by the method of variation of parameters. (a) x2 y 00 − 2xy 0 + 2y = x3 ex . We start by finding all solutions to the homogeneous equation, x2 y 00 − 2xy 0 + 2y = 0. This is a Cauchy-Euler equation with a = −2 and b = 2 and auxiliary equation λ2 − 3λ + 2 = (λ − 1)(λ − 2) = 0. This has two distinct real roots, λ1 = 1 and λ2 = 2, so the solution to the homogeneous equation is yh = c1 x + c2 x2 , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x and y2 = x2 : x x2 = x2 . W [y1 , y2 ] = 1 2x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where x2 (x3 ex ) dx x2 (x2 )
Z v1 (x) = − Z
xex dx
=−
= −xex + ex
and x(x3 ex ) dx x2 (x2 )
Z v2 (x) = Z =
ex dx
= ex . Thus, yp = x(−xex + ex ) + x2 (ex ) = xex and y = yh + yp = c1 x + c2 x2 − xex .
44
(b) x2 y 00 − 7xy 0 + 15y = x6 cos x. We start by finding all solutions to the homogeneous equation, x2 y 00 − 7xy 0 + 15y = 0. This is a Cauchy-Euler equation with a = −7 and b = 15 and auxiliary equation λ2 − 8λ + 15 = (λ − 3)(λ − 5) = 0. This has two distinct real roots, λ1 = 3 and λ2 = 5, so the solution to the homogeneous equation is yh = c1 x3 + c2 x5 , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x3 and y2 = x5 : 3 x x5 = 2x7 . W [y1 , y2 ] = 2 3x 5x4 The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where x5 (x6 cos x) dx x2 (2x7 )
Z v1 (x) = − Z
x2 cos x dx
=−
= −x2 sin x + 2 sin x − 2x cos x
and Z v2 (x) =
x3 (x6 cos x) dx x2 (2x7 )
Z =
2 cos x dx
= 2 sin x. Thus, yp = x3 (−x2 sin x+2 sin x−2x cos x)+x2 (2 sin x) = x5 sin x+2x3 sin x−2x4 cos x and y = yh + yp = c1 x3 + c2 x5 + x5 sin x + 2x3 sin x − 2x4 cos x.
45
(c) x2 y 00 − 3xy 0 + 4y = x5 We start by finding all solutions to the homogeneous equation, x2 y 00 − 3xy 0 + 4y = 0. This is a Cauchy-Euler equation with a = −3 and b = 4 and auxiliary equation λ2 − 4λ + 4 = (λ − 2)2 = 0. This has two equal real roots, λ1 = λ2 = 2, so the solution to the homogeneous equation is yh = c1 x2 + c2 x2 ln x, where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x2 and y2 = x2 ln x: 2 x x2 ln x = x3 . W [y1 , y2 ] = 2x 2x ln x + x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where Z v1 (x) = − Z =−
x2 ln x(x5 ) dx x2 (x3 ) x2 ln x dx
1 1 = − x3 ln x + x3 3 9 and Z v2 (x) = Z = =
x2 (x5 ) dx x2 (x3 ) x2 dx
1 3 x . 3
Thus, 1 1 1 1 4 yp = x2 (− x3 ln x + x3 ) + x2 ln x( x3 ) = − x5 ln x + x5 3 9 3 3 9 and
4 1 y = yh + yp = c1 x2 + c2 x2 ln x − x5 ln x + x5 . 3 9
46
(d) x2 y 00 − 5xy 0 + 9y = x3 ln x We start by finding all solutions to the homogeneous equation, x2 y 00 − 5xy 0 + 9y = 0. This is a Cauchy-Euler equation with a = −5 and b = 9 and auxiliary equation λ2 − 6λ + 9 = (λ − 3)2 = 0. This has two equal real roots, λ1 = λ2 = 3, so the solution to the homogeneous equation is yh = c1 x3 + c2 x3 ln x, where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x3 and y2 = x3 ln x: 3 x x3 ln x = x5 . W [y1 , y2 ] = 2 3x 3x2 ln x + x2 The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where x3 ln x(x3 ln x) dx x2 (x5 ) Z (ln x)2 =− dx x 1 = − (ln x)3 3 Z
v1 (x) = −
and x3 (x3 ln x) dx x2 (x5 ) Z ln x = dx x 1 = (ln x)2 . 2 Z
v2 (x) =
Thus, 1 1 1 yp = x3 (− (ln x)3 ) + x3 ln x( (ln x)2 ) = (ln x)3 3 2 6 and
1 y = yh + yp = c1 x3 + c2 x3 ln x + (ln x)3 . 6
47
(e) x2 y 00 + 7xy 0 + 9y = x4 We start by finding all solutions to the homogeneous equation, x2 y 00 + 7xy 0 + 9y = 0. This is a Cauchy-Euler equation with a = 7 and b = 9 and auxiliary equation λ2 + 6λ + 9 = (λ + 3)2 = 0. This has two equal real roots, λ1 = λ2 = −3, so the solution to the homogeneous equation is yh = c1 x−3 + c2 x−3 ln x, where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x−3 and y2 = x−3 ln x: −3 x x−3 ln x −7 . W [y1 , y2 ] = −4 −4 −4 = x −3x −3x ln x + x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where Z v1 (x) = − Z =−
x−3 ln x(x4 ) dx x2 (x−7 ) x6 ln x dx
1 1 = − x7 ln x + x7 7 49 and Z v2 (x) = Z = =
x−3 (x4 ) dx x2 (x−7 ) x6 dx
1 7 x . 7
Thus, 1 1 1 1 4 yp = x−3 (− x7 ln x + x7 ) + x−3 ln x( x7 ) = x 7 49 7 49 and y = yh + yp = c1 x−3 + c2 x−3 ln x +
48
1 4 x . 49
6. Solve the following linear equations with constant coefficients by the method of variation of parameters. 2x
(a) y 00 − 4y 0 + 4y = ex . We start by finding all solutions to the homogeneous equation y 00 − 4y 0 + 4y = 0. This has characteristic equation λ2 − 4λ + 4 = (λ − 2)2 = 0, which has repeated roots λ1 = λ2 = 2, so the solution to the homogeneous equation is yh = c1 e2x + c2 xe2x , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = e2x and y2 = xe2x : 2x e xe2x 4x W [y1 , y2 ] = 2x 2x 2x = e . 2e e + 2xe The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where Z xe2x v1 (x) = −
e2x x
e4x
dx
Z =−
1 dx
= −x and Z e2x v2 (x) =
e2x x
e4x
dx
Z
1 dx x = ln x =
Thus, yp = −xe2x + x(ln x)e2x and y = yh + yp = c1 e2x + c2 xe2x − xe2x + x(ln x)e2x .
49
(b) y 00 − 3y 0 − 18y = 27e6x . We start by finding all solutions to the homogeneous equation y 00 − 3y 0 − 18y = 0. This has characteristic equation λ2 − 3λ − 18 = (λ + 3)(λ − 6) = 0, which has distinct real roots λ1 = −3 and λ2 = 6, so the solution to the homogeneous equation is yh = c1 e−3x + c2 e6x , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = e−3x and y2 = e6x : −3x e e6x = 9e3x . W [y1 , y2 ] = −3e−3x 6e6x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where e6x 27e6x dx 9e3x
Z v1 (x) = − Z =− =−
3e3x dx
e9x 3
and e−3x 27e6x dx 9e3x
Z v2 (x) = Z =
3 dx
= 3x Thus, yp = −
e6x e9x −3x e + 3xe6x = − + 3xe6x 3 3
and y = yh + yp = c1 e−3x + c2 e6x −
50
e6x + 3xe6x = c1 e−3x + c2 e6x + 3xe6x . 3
7. Suppose that the solutions to y 000 − 6y 00 + 12y 0 − 8y = 0 are e2x , xe2x , and x2 e2x . Show that these functions are linearly independent. To show they are linearlly independent, we must show that their Wronskian is a non-zero function. 2x e xe2x x2 e2x 2x 2x 2 2x 2x 2x 2x 2x 2 2x = 2e6x 6= 0. e + 2xe 2xe + 2x e W [e , xe , x e ] = 2e 4e2x 4e2x + 4xe2x 2e2x + 8xe2x + 4x2 e2x Therefore, they are linearly independent. 8. Suppose that the solutions to y 00 − 4y 0 + 5y = 0 are e2x cos x and e2x sin x. Show that these functions are linearly independent. To show they are linearlly independent, we must show that their Wronskian is a non-zero function. e2x cos x e2x sin x 2x 2x W [e cos x, e sin x] = 2x 2e cos x − e2x sin x e2x cos x + 2e2x sin x = e4x cos2 x + e4x sin2 x = e4x 6= 0. Therefore, they are linearly independent.
51
9. ex − 1 is a solution to (ex + 1)y 00 − 2y 0 − ex y = 0. Find another linearly independent solution. Let y1 = 3x − 1, sp y10 = y100 = ex . Start by checking that y1 is a solution of the differential equation: (ex + 1)y100 − 2y10 − ex y1 = (ex + 1)ex − 2e− ex (ex − 1) = e2x + ex − 2ex − e2x + ex = 0. Let the unknown linearly independent solution be y2 . Then, we must have y1 y2 W [y1 , y2 ] = 0 y1 y20 = y1 y20 − y2 y10 = (ex − 1)y20 − ex y2 . On the other hand, applying Abel’s formula to the differential equation (ex + 1)y 00 − 2y 0 − ex y = 0, R
W [y1 , y2 ] = Ce
2 ex +1
dx
Let u = ex + 1, so du = ex dx = (u − 1) dx and Z
so W [y1 , y2 ] = C
2 dx = ex + 1
ex ex +1
Z
du u−1
= dx:
2 du u(u − 1) Z 2 2 − du = u u−1 = −2 ln u + 2 ln u − 1 2 u−1 = ln u x 2 e = ln x e +1
2
.
52
Setting these two expressions for the Wronskian equal, we get a linear differential equation for y2 : 2 ex (e − − e y2 = C ex + 1 x e2x e y20 − x y2 = C x . e −1 (e + 1)2 (ex − 1) x
1)y20
x
x
2x
Here, P (x) = − exe−1 and Q(x) = C (ex +1)e2 (ex −1) so R x e2x − exe−1 C x y2 = e dx + C2 e (e + 1)2 (ex − 1) Z x e2x − ln (ex −1) = eln (e −1) C e dx + C 2 (ex + 1)2 (ex − 1) Z e2x = (ex − 1) C dx + C 2 (ex + 1)2 (ex − 1)2 Z e2x = (ex − 1) C dx + C 2 (e2x − 1)2 R
ex ex −1
Z
Let u = e2x − 1, so du = 2e2x dx and Z
du 2
e2x dx = (e2x − 1)2
= e2x dx. Then, Z
1 du 2u2 1 =− 2u 1 1 = − 2x , 2e −1
so C 1 y2 = (ex − 1) − + C 2 2 e2x − 1 x e −1 = C1 2x + C2 (ex − 1), e −1 where C1 = − C2 . Thus,
ex −1 e2x −1
is another linearly independent solution.
53
10. Solve the following differential equations by reduction of order. (a) y 00 + x(y 0 )2 = 0. Let v = y 0 , so v 0 = y 00 and the differential equation becomes v 0 +xv 2 = 0. This is a separable equation: dv = −xv 2 dx dv = −x dx 2 Z v Z dv = − x dx v2 1 x2 − = − + C1 v 2 1 v = x2 2 + C1 Lastly, v = y 0 , so Z y=
v dx Z
1 dx + C1 Z 1 1 dx = x2 C1 1 + 2C 1 =
Let u =
√x , 2C1
so du =
x2 2
√1 2C1
and
√
2C1 du = dx:
Z 1 1 dx x2 C1 1 + 2C 1 √ Z 2C1 1 = du C1 1 + u2 r 2 = arctan u + C2 C1 r 2 x = arctan √ + C2 . C1 2C1
y=
54
(b) y 00 y 0 = 2. Let v = y 0 , so v 0 = y 00 and the differential equation becomes v 0 v = 2. This is separable: dv =2 dx v dv = 2 dx Z Z v dv = 2 dx v
v2 = 2x + C1 2 p v = ± 4x + C1 Lastly, v = y 0 , so Z y= =±
v dx Z p
4x + C1 dx
3 1 = ± (4x + C1 ) 2 = C2 . 6
55
0.3
The Laplace Transform
1. Find the following Laplace transforms of the following functions using the definition. (a) f (t) = 1. The Laplace Transform of f (t) = 1 is Z ∞ F (s) = e−st dt 0
Z = lim
R→∞
R
e−st dt
0
R 1 − e−st R→∞ s 0 1 −sR 1−e = lim R→∞ s 1 = s = lim
if s > 0
and diverges otherwise. (b) f (t) = t2 . The Laplace Transform of f (t) = t2 is Z ∞ F (s) = t2 e−st dt 0
Z = lim
R→∞
R
t2 e−st dt
0
R Z R 1 1 −st 2te dt − te−st + R→∞ s 0 s 0 ! R Z 1 −sR 2 1 −st 1 R −st = lim − Re + − te + e dt R→∞ s s s s 0 0 R 2 1 −sR 2 1 −st −sR − 2 Re = lim − Re + 2 − e R→∞ s s s s 0 1 −sR 2 2 −sR −sR = lim − Re − 2 Re + 3 1−e R→∞ s s s 2 = 3 s
= lim
and diverges otherwise.
56
if s > 0
(c) f (t) = eat . The Laplace Transform of f (t) = eat is Z ∞ F (s) = eat e−st dt 0
Z = lim
R→∞
R
e−(s−a)t dt
0
R 1 −(s−a)t = lim − e R→∞ s−a 0 1 1 − e−(s−a)R = lim − R→∞ s−a 1 = s−a and diverges otherwise.
57
if s > a
(d) f (t) = cos at. The Laplace Transform of f (t) = cos at is Z ∞ F (s) = e−st cos at dt 0
Z = lim
R→∞
We compute R
Z 0
RR 0
R
e−st cos at dt
0
e−st cos at dt by parts:
R Z 1 a R −st e−st cos at dt = − e−st cos at − e sin at dt s s 0 0 1 1 = − e−sR cos aR s s ! R Z a 1 −st a R −st − − e sin at + e cos at dt s s s 0 0 Z 1 a a2 R −st 1 e cos at dt = − e−sR cos aR + 2 e−sR sin aR − 2 s s s s 0
Thus,
a2 1+ 2 s
Z
R
−st
e
R
Z
1 a 1 − e−sR cos aR + 2 e−sR sin aR s s s 1 1 a − e−sR cos aR + 2 e−sR sin aR s s s
e−st cos at dt =
0
cos at dt =
0
1 1+
a2 s2
so Z F (s) = lim
R→∞
=
R
e−st cos at dt
0
1 s
if s > 0
2
1 + as2 s = 2 s + a2
if s > 0
and diverges otherwise.
58
2. Use the properties of the Laplace transform and the table at the front of the book to determine the Laplace Transform of the following functions. (a) 5 + 11t. L {5 + 11t} = 5L {1} + 11L {t} =
5 11 + 2. s s
(b) t4 e−2t . L t4 e−2t =
4! 24 = . 5 (s + 2) (s + 2)5
(c) t3 sin 3t. d3 L t3 sin 3t = (−1)3 3 L {sin 3t} ds 3 d3 = (−1)3 3 2 ds s + 32 72s(s2 − 9) = (s2 + 9)4 (d) e5t sin 4t. L e5t sin 4t =
4 4 = 2 . 2 2 (s − 5) + 4 s − 10s + 41
(e) t2 cos 2t + t3 e−4t . L t2 cos 2t + t3 e−4t = L t2 cos 2t + L t3 e−4t d2 3! L {cos 2t} + ds2 (s + 4)4 d2 s 6 = 2 2 + ds s + 22 (s + 4)4 2s(s2 − 12) 6 = + (s2 + 4)3 (s + 4)4 = (−1)2
59
3. Find the Laplace transform of the following step functions. ( 0 t
First Order Differential Equations
1. Solve the following separable differential equations. −2x
(a) y 0 = e y . We start by separating the equation: dy e−2x = dx y y dy = e−2x dx Next, we integrate both sides: Z Z y dy = e−2x dx y2 e−2x =− +C 2 2 Lastly, solving for y: y2 e−2x =− +C 2 2 y 2 = −e−2x + 2C p p y = ± −e−2x + 2C = ± −e−2x + D
D = 2C.
(b) 3y 5 y 0 − x2 (x3 + 5)10 = 0 We start by separating the equation: dy = x2 (x3 + 5)10 dx 3y 5 dy = x2 (x3 + 5)10 dx 3y 5
Next, we integrate both sides: Z Z 5 3y dy = x2 (x3 + 5)10 dx Let u = x3 + 5, so du = 3x2 dx: Z 1 10 y6 = u du 2 3 1 11 = u +C 33 1 3 = (x + 5)11 + C 33 Lastly, solving for y: y6 1 3 = (x + 5)11 + C 2 33 2 3 2 3 y6 = (x + 5)11 + 2C = (x + 5)11 + D 33 33
1
D = 2C.
x
−2 (c) y 0 = ey+2 . We start by separating the equation:
ex − 2 dy = dx y+2 (y + 2) dy = ex − 2 dx Next, we integrate both sides: Z Z (y + 2) dy = ex − 2 dx y2 + 2y = ex − 2x + C 2 Lastly, we solve for y using the quadratic formula: y2 + 2y − (ex − 2x + C) = 0 2 p y = −2 ± 4 − 2(ex − 2x + C) √ = −2 ± 4 − 2ex − 4x + D D = 2C. 1+2y (d) y 0 = − x(1+y) .
We start by separating the equation: dy 1 + 2y =− dx x(1 + y) 1+y 1 dy = − dx 1 + 2y x Next, we integrate both sides: Z Z 1 1+y dy = − dx 1 + 2y x Z 1 Z 1 1 2 (1 + 2y) + 2 dy = − dx 1 + 2y x Z Z 1 1 1 1 + dy = − dx 2 2 1 + 2y x 1 1 y + ln |1 + 2y| = − ln x + C 2 4 Here, we cannot solve for y, so we leave the solution in implicit form.
2
(e) y 0 + x2 y = x2 y 2 . We start by separating the equation: dy = x2 y 2 − x2 y = x2 (y 2 − y) dx dy = x2 dx y2 − y Next, we integrate both sides: Z Z dy = x2 dx. y2 − y We solve the left hand side by partial fractions: y2
1 1 1 =− + , −y y y−1
so − ln |y| + ln |y − 1| = y−1 |= y 1 ln |1 − | = y ln |
x3 +C 3
x3 +C 3 x3 +C 3
Lastly, we solve for y: x3 x3 1 |1 − | = e 3 +C = De 3 y x3 1 1 − = ±De 3 y 1 y= x3 1 ± De 3
3
D = eC
2. Solve the following homogeneous equations. (a) y 0 =
y 5 −x5 y4 x .
Start by checking that the equation is homogeneous: let f (x, y) = y 5 −x5 y 4 x . Then, f (xt, yt) = Let y = vx, so
(yt)5 − (xt)5 t5 (y 5 − x5 ) y 5 − x5 = = = f (x, y). (yt)4 (xt) t5 (y 4 x) y4 x dy dx
v+x
dv = v + x dx . Substitute these into the equation:
v5 − 1 dv ((vx)5 − x5 x5 (v 5 − 1) = = = dx (vx)4 x x5 v 4 v4 5 dv v −1 1 x = − v = − 4. 4 dx v v
The new equation is separable: v 4 dv = − Z
v 4 dv = −
dx Zx
dx x
v5 = − ln |x| + C 5 Let C = − ln |k|, so v5 = − ln |x| − ln |k| = − ln |kx| 5 v 5 = −5 ln |kx| = ln |kx|−5 √ v = 5ln |kx|−5
Lastly, writing the expression in terms of y gives: √ y = vx = x 5ln |kx|−5 .
4
(b) y 0 =
−2y+3x . y
Start by checking that the equation is homogeneous: let f (x, y) = −2y+3x . Then, y f (xt, yt) = Let y = vx, so
dy dx
−2(yt) + 3(xt) −2y + 3x = = f (x, y). (yt) y
dv = v + x dx . Substitute these into the equation:
dv −2(vx) + 3x −2v + 3 = = dx vx v dv −2v + 3 −v 2 − 2v + 3 x = −v = dx v v dx v dv = − 2 Zx Z v + 2v − 3 dx v dv = − 2 v + 2v − 3 x v+x
The right hand side can be done by partial fractions: v2
v 3 1 1 1 = + , + 2v − 3 4v+3 4v−1
so Z 1 1 dx 3 1 + dv = − 4v+3 4v−1 x 3 1 ln |v + 3| + ln |v − 1| = − ln |x| + C 4 4 Z
Let C = − ln |k|, so 3 1 ln |v + 3| + ln |v − 1| = − ln |x| − ln |k| = − ln |kx| 4 4 3 ln |v + 3| + ln |v − 1| = −4 ln |kx| ln |v + 3|3 + ln |v − 1| = ln |kx|−4 ln |v + 3|3 |v − 1| = ln |kx|−4 |v + 3|3 |v − 1| = |kx|−4 y y | + 3|3 | − 1| = |kx|−4 x x This expression cannot be simplified further, so we leave it as is.
5
(c) y 0 =
x2 −y 2 xy .
Start by checking that the equation is homogeneous: let f (x, y) = x2 −y 2 xy . Then, f (xt, yt) = Let y = vx, so
dy dx
x2 − y 2 (xt)2 − (yt)2 = = f (x, y). (xt)(yt) xy
dv = v + x dx . Substitute these into the equation:
dv x2 − (vx)2 1 − v2 = = dx x(vx) v dv 1 − v2 1 − 2v 2 x = −v = dx v v dx v dv = 2 Z 1 − 2v Zx v dx dv = 1 − 2v 2 x
v+x
Let u = 1 − 2v 2 , so du = −4v dv and − du 4 = v dv: Z Z 1 du dx − = 4 u x 1 − ln |u| = ln |x| + C. 4 Let C = ln |k|, so 1 − ln |u| = ln |kx| 4 1 − ln |1 − 2v 2 | = ln |kx| 4 1 ln |1 − 2v 2 |− 4 = ln |kx| 1
|1 − 2v 2 |− 4 = |kx| |1 − 2v 2 | = |kx|−4 1 − 2v 2 = ±|kx|−4 1 v2 = 1 ± |kx|−4 2 r 1 v=± (1 ± |kx|−4 ) 2 r 1 y = ±x (1 ± |kx|−4 ) 2
6
(d) y 0 =
x3 +y 3 xy 2 .
Start by checking that the equation is homogeneous: let f (x, y) = x3 +y 3 xy 2 . Then, f (xt, yt) = Let y = vx, so
dy dx
x3 + y 3 (xt)3 + (yt)3 = = f (x, y). 2 (xt)(yt) xy 2
dv = v + x dx . Substitute these into the equation:
1 + v3 dv x3 + (vx)3 = = dx x(vx)2 v2 dv 1 + v3 1 x = −v = 2 2 dx v v dx 2 v dv = Z Zx dx 2 v dv = x 3 v = ln |x| + C 3
v+x
Let C = ln |k|, so v3 = ln |kx| 3 3 v = 3 ln |kx| = ln |kx|3 p v = 3 ln |kx|3 p y = x 3 ln |kx|3 .
7
(e) y 2 + x2 y 0 = xyy 0 . Start by writing this in the form y 0 = f (x, y): x2 y 0 − xyy 0 = −y 2 (x2 − xy)y 0 = −y 2 y0 = −
y2 x2 − xy
Next, check that the equation is homogeneous: f (xt, yt) = − Let y = vx, so
dy dx
(xt)2
(yt)2 y2 =− 2 = f (x, y). − (xt)(yt) x − xy
dv = v + x dx . Substitute these into the equation:
dv (vx)2 v2 =− 2 = dx x − x(vx) 1−v dv v2 v x = −v = dx 1−v v−1 v−1 dx dv = v Z Zx v−1 dx dv = v x Z Z 1 dx 1 − dv = v x v − ln |v| = ln |x| + C
v+x
Let C = ln |k|, so v − ln |v| = − ln |kx|. This expression cannot be simplfied further, so we simply substitute v = xy : y y − ln | | = ln |kx|. x x
8
3. Find the function y = y(x) such that
dy dx
= y 2 (1 + 2x) and y(0) = −1.
Separating variables and integrating yields: dy = 1 + 2x dx y2 Z Z dy = 1 + 2x dx y2 1 − = x + x2 + C y 1 y=− x + x2 + C Now, we have −1 = y(0) = − so C = 1 and y=−
1 , C
1 . x + x2 + 1
4. Solve the following first-order linear differential equations. (a)
dy dx
+ xy − x5 = 0 This is first-order linear, with P (x) = x1 and Q(x) = x5 , so it has solution Z R R y = e− P (x) dx e P (x) dx Q(x) dx + C Z R R dx dx = e− x e x x5 dx + C Z = e− ln x eln x x5 dx + C Z 1 = x6 dx + C x 1 1 7 x +C = x 7 C 1 = x6 + 7 x
9
(b)
dy dx
+ y = sin x Here, P (x) = 1 and Q(x) = sin x, so Z R R − dx dx y=e e sin x dx + C Z −x x e sin x dx + C =e x e sin x − ex cos x −x =e +C 2 (see section 1.1.5 for this integral.) =
(c)
dy dx
−
y x+1
sin x − cos x + Ce−x 2
= x2
1 Here, P (x) = − 1+x and Q(x) = x2 , so
R 1 e− 1+x dx x2 dx + C Z ln (1+x) − ln (1+x) 2 =e e x dx + C Z x2 = (1 + x) dx + C 1+x Z (x2 − 1) + 1 = (1 + x) dx + C 1+x Z (1 + x)(1 − x) + 1 = (1 + x) dx + C 1+x Z 1 = (1 + x) 1−x+ dx + C 1+x x2 = (1 + x) x − + ln |1 + x| + C 2 R
y=e
1 1+x
dx
10
(d) xy 0 + y = 3x cos 2x. Rearranging, we get y0 +
1 y = 3 cos 2x, x
so P (x) = − x1 and Q(x) = 3 cos 2x. Thus, Z R 1 R 1 y = e− x dx 3e x dx cos 2x dx + C Z = e− ln x 3eln x cos 2x dx + C Z 1 x cos 2x dx + C = x R We integrate x cos 2x dx by parts: u = 3x
dv = cos 2x dx 1 du = 3 dx v = sin 2x 2 Thus, Z Z 3 3 3 3 x cos 2x dx = x sin 2x − sin 2x dx = x sin 2x + cos 2x, 2 2 2 4 so, 1 3 3 y= x sin 2x + cos 2x + C x 2 4 3 3 C = sin 2x + cos 2x + . 2 4x x
11
5. Solve the following linear differential equations. Use the initial conditions given to find all unknown constants. (a) y 0 − 3xy = 15x, y(0) = 5. Here, P (x) = −3x and Q(x) = 15x, so Z R R y = e 3x dx 15xe− 3x dx dx + C Z 3 2 3 2 = e2x 15xe− 2 x dx + C . Let u = − 23 x2 , so du = −3x dx and − du 3 = x dx: Z Z 3 2 15 eu du 15xe− 2 x dx = − 3 = −5eu 3
2
= −5e− 2 x Thus, 3
y = e2x
2
3
2
−5e− 2 x + C 3
= −5 + Ce 2 x
2
Lastly, we use the initial condition y(0) = 5 to find C: 5 = y(0) = −5 + C C = 10 3
2
Thus, y = −5 + 10e 2 x .
12
(b) y 0 + 4xy = x, y(0) = 54 . Here, P (x) = 4x and Q(x) = x, so Z R R − 4x dx − 4x dx y=e xe dx + C Z 2 2 = e−2x xe2x dx + C Let u = 2x2 , so du = 4x dx and du 4 = x dx: Z Z 2 1 eu du xe2x dx = 4 1 = eu 4 1 2 = e2x 4 Thus, y = e−2x =
2
1 2x2 e +C 4
2 1 + Ce−2x 4
Lastly, we use the initial condition y(0) =
5 4
5 1 = y(0) = + C 4 4 C=1 Thus, y =
1 4
2
+ e−2x .
13
to find C:
(c) y 0 + x1 y = ex , y(1) = 2. Here, P (x) = x1 and Q(x) = ex , so Z R 1 1 x int x − x e e dx + C y=e Z − ln x x ln x =e e e dx + C Z 1 x = xe dx + C x R Integrate xex dx by parts: dv = ex dx
u=x
v = ex
du = 1 Thus, Z
x
x
Z
xe dx = xe −
ex dx = xex − ex ,
and 1 (xex − ex + C) x ex = ex + +C x
y=
Lastly, we use the initial condition y(1) = 2 to find C: 2 = y(1) = e − e + C C = 2, so y = ex +
ex x
+ 2.
14
6. Make a substitution to solve the following Bernoulli equations. If initial conditions are given, use them to solve for all unknown constants. (a) y 0 + x2 y = x2 y 2 . 0
Let z = y 1−2 = y −1 , so y = z1 and y 0 = − zz2 . Substitute these into the original equation and isolate for z 0 : y 0 + x2 y = x2 y 2 x2 z0 x2 + = z2 z z2 z 0 − x2 z = −x2
−
This is linear, with P (x) = −x2 and Q(x) = −x2 , so Z R 2 R x dx 2 − x2 − x e z=e dx + C Z 1 3 x 2 − 13 x3 3 − x e dx + C =e Let u = − 31 x3 , so du = −x2 dx: Z Z 1 3 1 3 − x2 e− 3 x dx = eu du = eu = e− 3 x , and
1
z = e3x
3
1 3 1 3 e− 3 x + C = 1 + Ce 3 x .
Thus, y=
1 1 = 1 3 . z 1 + Ce 3 x
15
√ (b) y 0 + xy = x 3 y. 1
2
3
1
Let z = y 1− 3 = y 3 , so y = z 2 and y 0 = 23 z 2 z 0 . Substitute these into the original equation and isolate for z 0 : 1
y 0 + xy = xy 3
3 3 1 1 3 1 0 z 2 z + xz 2 = x(z 2 ) 3 = xz 2 2 2 2 z 0 + xz = x 3 3
This is linear, with P (x) = 23 x and Q(x) = 23 x, so Z R 2 2x − 23 x dx 3 z=e xe dx + C 3 Z 2 2 x2 − x3 3 =e xe dx + C 3 Let u =
x2 3 ,
so du = 23 x dx: Z Z x2 2 x2 xe 3 dx = eu du = eu = e 3 , 3
so z = e− and
x2 3
e
x2 3
x2 + C = 1 + Ce− 3
23 2 x2 . y = z 3 = 1 + Ce− 3
16
√ (c) y 0 + 3x2 y = x2 y. 1
1
Let z = y 1− 2 = y 2 , so y = z 2 and y 0 = 2zz 0 . Substitute these into the original equation and isolate for z 0 : 1
y 0 + 3x2 y = x2 y 2
2zz 0 + 3x2 z 2 = x2 z x2 3 z 0 + x2 z = 2 2 This is linear, with P (x) = 23 x2 and Q(x) = x2 , so 3
= e− Let u =
x3 2 ,
x2 3 x2 e 2 dx + C 2 Z 2 x x3 e 2 dx + C 2
2
z = e− 2 x x3 2
dx
so du = 32 x2 and Z
Z
du 3
x2 x3 1 e 2 dx = 2 3
so 3
z=e
− x2
Z
=
x2 2
and 2
y=z =
17
dx:
eu du =
1 x3 e 2 +C 3
=
1 u 1 x3 e = e2, 3 3 x3 1 + Ce− 2 , 3
x3 1 + Ce− 2 3
.
1
(d) y 0 + x4 y = xy 4 , y(1) = 1. 1
3
4
1
Let z = y 1− 4 = y 4 , so y = z 3 and y 0 = 34 z 3 z 0 . Substitute these into the original equation and isolate for z 0 : 1 4 y = xy 4 x 1 4 1 0 4 4 z 3 z + z 3 = xz 3 3 x 3 3 z0 + z = x x 4
y0 +
and Q(x) = 34 x, so Z R 3 3 R 3 z = e− x xe x dx + C 4 Z 3 3 ln x = e−3 ln x xe dx + C 4 Z 3 3 = x−3 xx dx + C 4 Z 3 4 = x−3 x dx + C 4 3 5 = x−3 x +C 20 3 2 = x + Cx−3 20
This is linear with P (x) =
3 x
Thus,
4
y = z3 =
3 2 x + Cx−3 20
43 .
Lastly, we use the initial condition y(1) = 1 to find C: 1 = y(1) =
43 3 3 +C +C 1= 20 20 17 C= 20
Thus, y=
3 2 17 x + 20 20x3
18
43 .
(e) y 0 − x1 y = ex y 2 , y(1) = 12 . 0
Let z = y 1−2 = y −1 , so y = z1 and y 0 = − zz2 . Substitute these into the original equation and isolate for z 0 : 1 y = ex y 2 x 11 z0 1 − 2− = ex 2 z xz z 1 0 x z + z = −e x y0 −
1 x
and Q(x) = −ex , so Z R 1 1 − x dx x x − e e dx + C z=e Z − ln x x ln x =e − e e dx + C Z 1 x = − xe dx + C x
This is linear with P (x) =
We do this by integration by parts, with u=x
dv = ex dx v = ex ,
du = dx so
Z
x
x
xe dx = xe − and
Z
ex dx = xex − ex ,
z=
1 (−xex + ex + C) . x
y=
1 1 = . z −xex + ex + C
Thus,
Lastly, we use the initial condition y(1) =
1 2
to find C:
1 1 1 = y(1) = = 2 −e + e + C C C = 2, so y =
1 −xex +ex +2 .
19
7. Solve the following exact differential equations. (a) (y 2 cos x) dx + (y 2 + 2y sin x) dy = 0. Here M (x, y) = y 2 cos x and N (x, y) = y 2 + 2y sin x, and ∂M ∂N = 2y cos x = , ∂y ∂x so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z = y 2 cos x dx + h(y) = y 2 sin x + h(y) Since
∂g ∂y
= N (x, y), we must have
y 2 + 2y sin x =
Thus, h(y) =
y3 3
∂ 2 y sin x + h(y) = 2y sin x + h0 (y) ∂y h0 (y) = y 2
+ C, so g(x, y) = y 2 sin x + y 2 sin x + y 2 sin x +
y3 +C =0 3
y3 =k 3
20
y3 +C 3
k = −C
(b) (3x2 − 2xy + 2) dx + (6y 2 − x2 + 3) = 0 Here M (x, y) = 3x2 − 2xy + 2 and N (x, y) = 6y 2 − x2 + 3, and ∂M ∂N = −2x = , ∂y ∂x so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z = 3x2 − 2xy + 2 dx + h(y) = x3 − x2 y + 2x + h(y) Since
∂g ∂y
= N (x, y), we must have
6y 2 − x2 + 3 =
∂ 3 x − x2 y + 2x + h(y) = −x2 + h0 (y) ∂y h0 (y) = 6y 2 + 3
Thus, h(y) = 2y 3 + 3y + C, so g(x, y) = x3 − x2 y + 2x + 2y 3 + 3y + C x3 − x2 y + 2x + 2y 3 + 3y + C = 0 x3 − x2 y + 2x + 2y 3 + 3y = k
21
k = −C
(c) (ex sin y − 2y sin x) dx + (ex cos y + 2 cos x) dy = 0. Here M (x, y) = ex sin y − 2y sin x and N (x, y) = ex cos y + 2 cos x, and ∂M ∂N = ex cos y − 2 sin x = , ∂y ∂x so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z = ex sin y − 2y sin x dx + h(y) = ex sin y + 2y cos x + h(y) Since
∂g ∂y
= N (x, y), we must have
ex cos y + 2 cos x =
∂ x [e sin y + 2y cos x + h(y)] = ex cos y + 2 cos x + h0 (y) ∂y h0 (y) = 0
Thus, h(y) = C, so g(x, y) = ex sin y + 2y cos x + C ex sin y + 2y cos x + C = 0 x
e sin y + 2y cos x = k
22
k = −C
(d) √
x x2 +y 2
dx + ey + √
Here M (x, y) = √
y x2 +y 2
x x2 +y 2
= 0.
and N (x, y) = ey + √
y , x2 +y 2
and
∂N ∂M xy = − p , 3 = ∂y ∂x x2 + y 2 so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z x p dx + h(y) = 2 x + y2 p = x2 + y 2 + h(y) Since
∂g ∂y
= N (x, y), we must have
ey + p
y x2 + y 2
=
i y ∂ hp 2 x + y 2 + h(y) = p + h0 (y) 2 2 ∂y x +y h0 (y) = ey
Thus, h(y) = ey + C, so p g(x, y) = x2 + y 2 + ey + C p x2 + y 2 + ey + C = 0 p x2 + y 2 + ey = k k = −C
23
(e)
ln y 1+x2
dx +
arctan x y
Here M (x, y) =
+ y ln y dy = 0.
ln y 1+x2
and N (x, y) =
arctan x y
+ y ln y, and
∂M 1 ∂N =− = , ∂y y(1 + x2 ) ∂x so the equation is exact. Thus, Z g(x, y) = M (x, y) dx + h(y) Z ln y = dx + h(y) 1 + x2 = arctan x ln y + h(y) Since
∂g ∂y
= N (x, y), we must have
∂ arctan x arctan x + y ln y = [arctan x ln y + h(y)] = + h0 (y) y ∂y y h0 (y) = y ln y Thus, Z h(y) =
y ln y dy,
which we solve by partial fractions, taking u = ln y du = so
y2 h(y) = ln y − 2
dv = y dy
dy y
Z
v=
y2 2
y2 y2 y dy = ln y − + C. 2 2 4
Thus, y2 y2 ln y − +C 2 4 y2 y2 arctan x ln y + ln y − +C =0 2 4 2 2 y y arctan x ln y + ln y − =k k = −C 2 4 g(x, y) = arctan x ln y +
24
8. Find an integrating factor for the following differential equations and solve. (a) (x sin y) dx + (2x cos y) dy = 0. Here M (x, y) = x sin y and N (x, y) = 2x cos y, so ∂M = x cos y ∂y Since
∂M ∂y
6=
∂N ∂x ,
∂N = 2 cos y. ∂x
and
the equation is not exact. However, 1 N
∂M ∂N − ∂y ∂x
1 1 − 2 x
=
is a function of x alone, so we may solve by introducing an integrating factor 1 R 1 1 1 e2x e 2 − x dx = e 2 x−ln x = . x 1
Thus, if we take M1 (x, y) = 1
e2x x N (x, y)
e2x x M (x, y)
1
= e 2 x sin y and N1 (x, y) =
1
= 2e 2 x cos y then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: 1 ∂N1 ∂M1 = e 2 x cos y = . ∂y ∂x
We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z 1 = e 2 x sin y dx + h(y) 1
= 2e 2 x sin y + h(y). Since
∂g ∂y
= N1 (x, y), we must have 1
2e 2 x cos y =
1 1 ∂ [2e 2 x sin y + h(y)] = 2e 2 x cos y + h0 (y) ∂y h0 (y) = 0
Thus, h(y) = C, so 1
g(x, y) = 2e 2 x sin y + C 1
2e 2 x sin y + C = 0 1
2e 2 x sin y = k,
where k = −C. 25
(b)
1 x
dx +
x y
dy = 0.
Here M (x, y) =
1 x
and N (x, y) = xy , so ∂M =0 ∂y
Since
∂M ∂y
6=
∂N ∂x ,
∂N 1 = . ∂x y
and
the equation is not exact. However, 1 N
∂M ∂N − ∂y ∂x
=−
1 x
is a function of x alone, so we may solve by introducing an integrating factor R 1 1 e − x dx = e− ln x = . x Thus, if we take M1 (x, y) = M (x,y) = x12 and N1 (x, y) = x then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: ∂M1 ∂N1 =0= . ∂y ∂x We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z 1 = dx + h(y) x2 1 = − + h(y). x Since
∂g ∂y
= N1 (x, y), we must have 1 ∂ 1 = [− + h(y)] = h0 (y) y ∂y x
Thus, h(y) = ln y + C, so g(x, y) = − −
1 + ln y + C x
1 + ln y + C = 0 x 1 − + ln y = k, x
where k = −C.
26
N (x,y) x
=
1 y
(c) 2xy 2 dx + 32 x2 y + y dy = 0. Here M (x, y) = 2xy 2 and N (x, y) = 23 x2 y + y, so ∂M = 4xy ∂y Since
∂M ∂y
6=
∂N ∂x ,
∂N = 3xy. ∂x
and
the equation is not exact. However, 1 M
∂M ∂N − ∂y ∂x
=
1 2y
is a function of y alone, so we may solve by introducing an integrating factor R 1 1 1 1 e− 2y dy = e− 2 ln y = y − 2 = √ . y 3
(x,y) Thus, if we take M1 (x, y) = M√ = 2xy 2 and N1 (x, y) = y √ √ 3 2 y + y then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: 2x
∂M1 ∂N1 √ = 3x y = . ∂y ∂x We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z 3 = 2xy 2 dx + h(y) 3
= x2 y 2 + h(y). Since
∂g ∂y
= N1 (x, y), we must have 3 2√ ∂ 2 3 3 √ √ x y+ y= [x y 2 + h(y)] = x2 y + h0 (y) 2 ∂y 2 √ 0 h (y) = y 3
Thus, h(y) = 23 y 2 + C, so 3 2 3 g(x, y) = x2 y 2 + y 2 + C 3 3 2 3 x2 y 2 + y 2 + C = 0 3 3 2 3 x2 y 2 + y 2 = k, 3
where k = −C. 27
N (x,y) √ y
=
(d) (x2 + 4xy) dx + x dy = 0. Here M (x, y) = x2 + 4xy and N (x, y) = x, so ∂M = 4x ∂y Since
∂M ∂y
6=
∂N ∂x ,
∂N = 1. ∂x
and
the equation is not exact. However, 1 N
∂M ∂N − ∂y ∂x
=4−
1 x
is a function of y alone, so we may solve by introducing an integrating factor R 1 e4x . e 4− x dx = e4x−ln x = x 4x
Thus, if we take M1 (x, y) = ex M (x, y) = xe4x +4ye4x and N1 (x, y) = e4x 4x then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: x N (x, y) = e ∂M1 ∂N1 = 4e4x = . ∂y ∂x We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z = xe4x + 4ye4x dx + h(y) = Since
∂g ∂y
1 1 4x xe − e4x + ye4x + h(y). 4 16
= N1 (x, y), we must have
e4x =
1 ∂ 1 4x [ xe − e4x + ye4x + h(y)] = e4x + h0 (y) ∂y 4 16 h0 (y) = 0
Thus, h(y) = C, so 1 4x 1 xe − e4x + ye4x + C 4 16 1 4x 1 xe − e4x + ye4x + C = 0 4 16 1 4x 1 xe − e4x + ye4x = k, 4 16
g(x, y) =
where k = −C. 28
(e) xy dx + (y 2 + x2 ) dy = 0. Here M (x, y) = xy and N (x, y) = y 2 + x2 , so ∂M =x ∂y Since
∂M ∂y
6=
∂N ∂x ,
∂N = 2x. ∂x
and
the equation is not exact. However, 1 M
∂M ∂N − ∂y ∂x
=−
1 y
is a function of y alone, so we may solve by introducing an integrating factor R 1 e y dy = eln y = y Thus, if we take M1 (x, y) = yM (x, y) = xy 2 and N1 (x, y) = yN (x, y) = y 3 + x2 y then M1 (x, y) dx + N1 (x, y) dy = 0 is exact: ∂M1 ∂N1 = 2xy = . ∂y ∂x We may thus solve this differential equation for g(x, y): Z g(x, y) = M1 (x, y) dx + h(y) Z = xy 2 dx + h(y) = Since
∂g ∂y
1 2 2 x y + h(y). 2
= N1 (x, y), we must have y 3 + x2 y =
∂ 1 2 2 [ x y + h(y)] = x2 y + h0 (y) ∂y 2 h0 (y) = y 3
Thus, h(y) = 41 y 4 + C, so 1 2 2 1 4 x y + y +C 2 4 1 2 2 1 4 x y + y +C =0 2 4 1 2 2 1 4 x y + y = k, 2 4
g(x, y) =
where k = −C.
29
9. Make a substitution to solve the following differential equations. (a) y 0 = (2x + y)2 − 6, y(0) = 4. Let v = 2x + y, so v 0 = 2 + y 0 and y 0 = v − 2. Then, the equation becomes: v0 − 2 = v2 − 6 v0 = v2 − 4 This equation is separable:
Z
dv = 1 dx v2 − Z 4
dv = −4
v2
1 dx = x + C
We can do this integral by partial fractions: 1 A B (A + B)v + (−2A + 2B) = + = , v2 − 4 v+2 v−2 v2 − 4 so A+B =0 −2A + 2B = 1 This pair of linear equations had solution A = − 41 and B = 14 : Z Z dv 1 1 1 = − dv 2 v −4 4 v−2 v+2 1 = ln |v − 2| − ln |v + 2| 4 v−2 1 |. = ln | 4 v+2
30
Thus, we have 1 v−2 ln | |=x+C 4 v+2 v−2 ln | | = 4x + C v+2 v−2 | | = De4x v+2 v−2 = ±De4x v+2 4 = ±De4x 1− v+2 4 −2 v= 1 ∓ De4x 4 y= − 2 − 2x 1 ∓ De4x Lastly, we use the initial condition y(0) = 4 to solve for the constant D: 4 −2 1∓D 1 ∓D = − 3 1 D=± , 3
4 = y(0) =
so, in either case, y=
4 − 2 − 2x. 1 − 13 e4x
(b) y 0 = e2x+y + 1. Let v = 2x + y, so v 0 = 2 + y 0 . Then, the equation becomes v 0 − 2 = y 0 = ev + 1 v 0 = ev + 3 This equation is separable: dv = 1 dx +Z 3
ev Z
dv = ev + 3
31
1 dx = x + C
Let u = ev , so du = ev dv and du u = dv. Then, Z Z dv du = ev + 3 u(u + 3) Z 11 1 1 − = 3u 3u+3 1 1 = ln |u| − ln |u + 3| 3 3 u 1 | = ln | 3 u+3 Thus, 1 u ln | |=x+C 3 u+3 u ln | | = 3x + C u+3 u | | = De3x u+3 u = ±De3x u+3 3 1− = ±De3x u+3 3 −3 u= 1 ± De3x Thus, 3 ev = −3 1 ± De3x 3 v = ln − 3 , 1 ± De3x and 3 2x + y = ln − 3 1 ± De3x 3 − 3 − 2x y = ln 1 ± De3x
32
0.2
Higher Order Differential Equations
1. Find the general solution for the following linear differential equations. (a) y 00 − y 0 − 2y = 0. This has characteristic polynomial λ2 − λ − 2 = (λ + 1)(λ − 2), so it has two distinct real roots, λ1 = −1 and λ2 = 2, so that y = c1 e−x + c2 e2x . (b) 4y 00 + 20y 0 + 25y = 0. This has characteristic polynomial 4λ2 + 20λ + 25 = (2λ + 5)2 , so it has a single real root λ = − 25 . Thus, 5
y = (c1 + c2 x)e− 2 x . (c) y 00 − 4y 0 + 5y = 0. This has characteristic polynomial λ2 − 4λ + 5 = 0, which has roots λ2 = 2 − i
λ1 = 2 + i, so
y = e2x (c1 cos x + c2 sin x). (d) 12y 00 + 16y 0 + 5y = 0.. This has characteristic polynomial 12λ2 + 16λ + 5 = (2λ + 1)(6λ + 5), so it has two distinct real roots, λ1 = − 21 and λ2 = − 65 , so that 1
5
y = c1 e − 2 x + c2 e − 6 x . (e) y 00 + 3y 0 + 5y = 0.. This has characteristic polynomial λ2 + 3λ + 5 = 0, which has roots √ √ 3 11 3 11 λ1 = − + i , λ2 = − − i 2 2 2 2 so
√
y=e
− 23 x
√ 11 11 (c1 cos x + c2 sin x). 2 2
33
(f) y 000 − 3y 0 − 2y = 0. This has characteristic polynomial λ3 − 3λ − 2 = 0. We make a guess for the first root, taking λ1 = −1: (−1)3 − 3(−3) − 2(−1) = 0. Using long division, you can show λ3 − 3λ − 2 = (λ + 1)(λ2 − λ − 2) = (λ + 1)(λ + 1)(λ − 2) so the other roots are λ2 = −1 and λ3 = 2. Thus, the general solution is y = c1 e−x + c2 xe−x + c3 e2x . (g) y 000 − 6y 00 + 12y 0 − 8y = 0. This has characteristic polynomial λ3 − 6λ2 + 12λ − 8 = 0. We make a guess for the first root, taking λ1 = 2: (2)3 − 6(2)2 + 12(2) − 8 = 0. Using long division, you can show λ3 − 6λ2 + 12λ − 8 = (λ − 2)(λ2 − 4λ + 4) = (λ − 2)3 so the other roots are λ2 = 2 and λ3 = 2. Thus, the general solution is y = c1 e2x + c2 xe2x + c3 x2 e2x . (h) y 000 − 7y 00 + 17y 0 − 15y = 0. This has characteristic polynomial λ3 − 7λ2 + 17λ − 15 = 0. We make a guess for the first root, taking λ1 = 3: (3)3 − 7(3)2 + 17(3) − 15 = 0. Using long division, you can show λ3 − 7λ2 + 17λ − 15 = (λ − 3)(λ2 − 4λ + 5) Using the quadratic formula, the other roots are λ2 = 2 + i and λ3 = 2 − i. Thus, the general solution is y = c1 e3x + c2 e2x cos x + c3 e2x sin x.
34
(i) y (6) − y = 0. This has characteristic polynomial λ6 − 1 = (λ3 − 1)(λ3 + 1) = (λ − 1)(λ2 + λ + 1)(λ + 1)(λ2 − λ + 1) Thus λ1 = 1 and λ2 = −1 are roots. The roots of the two remaining quadratic polynomials can be found using the quadratic formula: √ √ 3 1 3 1 and λ3 , λ4 = ± i . λ3 , λ4 = − ± i 2 2 2 2 The general solution is thus √ y = c1 ex + c2 e−x + c3 e
− 12
3 x 2
cos 1
+ c5 e 2 cos
√
!
! 3 + c4 e sin x 2 √ ! √ ! 1 3 3 x + c6 e 2 sin x . 2 2 − 21
(j) y (4) − 8y 00 + 16y = 0. This has characteristic polynomial λ4 − 8λ2 + 16 = (λ2 − 4)2
= (λ + 2)2 (λ − 2)2
Thus, the roots are λ1 = λ2 = −2 and λ3 = λ4 = 2. The general solution is y = c1 e−2x + c2 xe−2x + c3 e2x + c4 xe2x . 2. Solve y 00 − 4y = 0, subject to y(0) = 1, y 0 (0) = −1. Start by finding the general solution. This has characteristic polynomial λ2 − 4 = (λ + 2)(λ − 2), so it has two distinct real roots, λ1 = −2 and λ2 = 2, so that y = c1 e−2x + c2 e2x . Now, y 0 = −2c1 e−2x + 2c2 e2x , so, imposing initial conditions, we have 1 = y(0) = c1 + c2 −1 = y 0 (0) = −2c1 + 2c2 , giving two linear equations for the two constants c1 and c2 . Adding two times the first to the second gives 1 = 4c2 , so c2 = 41 . From the first, we see that c1 = 34 . Thus, 3 1 y = e−2x + e2x . 4 4
35
3. Find the general solution to the following non-homogeneous equations with constant coefficients, using the method of undetermined coefficients. (a) y 00 − 3y 0 − 4y = 5x2 . We start by finding all solutions to the homogeneous equation y 00 − 3y 0 − 4y = 0. This has characteristic equation λ2 − 3λ − 4 = (λ + 1)(λ − 4) = 0, which has distinct real roots λ1 = −1 and λ2 = 4, so the solution to the homogeneous equation is yh = c1 e−x + c2 e4x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A2 x2 +A1 x+A0 , so yp0 = 2A2 x + A1 yp00 = 2A2 Substituting these into y 00 − 3y 0 − 4y = 5x2 , we get (2A2 ) − 3(2A2 x + A1 ) − 4(A2 x2 + A1 x + A0 ) = 5x2 Collecting like terms on the right hand side, we get −4A2 x2 + (−6A2 − 4A1 )x + (−3A1 − 4A0 ) = 5x2 So, equating the coefficients of like terms on the right and left hand sides: −4A2 = 5 −6A2 − 4A1 = 0 −3A1 − 4A0 = 0
95 , A1 = This system of linear equations has solution A0 = − 32 5 A2 = − 4 . Thus, 5 15 95 yp = − x2 + x + − 4 8 32 and 5 15 95 y = yh + yp = c1 e−x + c2 e4x − x2 + x + − . 4 8 32
36
15 8 ,
and
(b) y 00 − 3y 0 − 18y = 4e2x . We start by finding all solutions to the homogeneous equation y 00 − 3y 0 − 18y = 0. This has characteristic equation λ2 − 3λ − 18 = (λ + 3)(λ − 6) = 0, which has distinct real roots λ1 = −3 and λ2 = 6, so the solution to the homogeneous equation is yh = c1 e−3x + c2 e6x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = Ae2x , so yp0 = 2Ae2x yp00 = 4Ae2x Substituting these into y 00 − 3y 0 − 18y = 4e2x , we get (4Ae2x ) − 3(2Ae2x ) − 18(Ae2x ) = 4e2x Collecting like terms on the right hand side, we get −20Ae2x = 4e2x So, equating the coefficients of like terms on the right and left hand sides: −20A = 4 Thus, A = − 51 , yp = − 51 e2x , and 1 y = yh + yp = c1 e−3x + c2 e6x − e2x . 5
37
(c) y 00 − 5y 0 = 2 sin 2x + 5 cos 2x. We start by finding all solutions to the homogeneous equation y 00 − 5y 0 = 0. This has characteristic equation λ2 − 5λ = λ(λ − 5) = 0, which has distinct real roots λ1 = 0 and λ2 = 5, so the solution to the homogeneous equation is yh = c1 + c2 e5x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A sin 2x + B cos 2x, so yp0 = 2A cos 2x − 2B sin 2x yp00 = −4A sin 2x − 4B cos 2x Substituting these into y 00 − 5y 0 = 2 sin 2x + 5 cos 2x, we get (−4A sin 2x−4B cos 2x)−5(2A cos 2x−2B sin 2x) = 2 sin 2x+5 cos 2x Collecting like terms on the right hand side, we get (−4A + 10B) sin 2x + (−4B − 10A) cos 2x = 2 sin 2x + 5 cos 2x So, equating the coefficients of like terms on the right and left hand sides: −4A + 10B = 2 −4B − 10A = 5 This system of linear equations has solution A = − 21 and B = 0. Thus, 1 yp = − sin 2x 2 and 1 y = yh + yp = c1 + c2 e5x − sin 2x. 2
38
(d) y 00 + 2y 0 = 2e2x − 3 sin 3x + 3 cos 3x. We start by finding all solutions to the homogeneous equation y 00 + 2y 0 = 0. This has characteristic equation λ2 + 2λ = λ(λ + 2) = 0, which has distinct real roots λ1 = −2 and λ2 = 0, so the solution to the homogeneous equation is yh = c1 + c2 e−2x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = Ae2x +B sin 3x+ C cos 3x, so yp0 = 2Ae2x + 3B cos 3x − 3C sin 3x yp00 = 4Ae2x − 9B sin 3x − 9C cos 3x Substituting these into y 00 + 2y 0 = 2e2x − 3 sin 3x + 3 cos 3x, we get (4Ae2x − 9B sin 3x − 9C cos 3x) + 2(2Ae2x + 3B cos 3x − 3C sin 3x) = 2e2x − 3 sin 3x + 3 cos 3x. Collecting like terms on the right hand side, we get 8Ae2x +(−9B−6C) sin 3x+(6B−9C) cos 3x = 2e2x −3 sin 3x+3 cos 3x So, equating the coefficients of like terms on the right and left hand sides: 8A = 2 −9B − 6C = 3 6B − 9C = 3 This system of linear equations has solution A = 1 C = − 13 . Thus, yp =
1 4,
B =
5 13 ,
1 2x 5 1 e + sin 3x − cos 3x 4 13 13
and 1 5 1 y = yh + yp = c1 + c2 e−2x + e2x + sin 3x − cos 3x. 4 13 13
39
and
(e) y 00 − 2y 0 − 15y = 32x2 ex + 40xex . We start by finding all solutions to the homogeneous equation y 00 − 2y 0 − 15y = 0. This has characteristic equation λ2 − 2λ − 15 = (λ + 3)(λ − 5) = 0, which has distinct real roots λ1 = −3 and λ2 = 5, so the solution to the homogeneous equation is yh = c1 e−3x + c2 e5x , where c1 and c2 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A2 x2 ex +A1 xex + A0 ex , so yp0 = A2 x2 ex + 2A2 xex + A1 xex + A1 ex + A0 ex yp00 = A2 x2 ex + 4A2 xex + 2A2 ex + A1 xex + 2A1 ex + A0 ex Substituting these into y 00 − 2y 0 − 15y = 32x2 ex + 40xex , we get (A2 x2 ex + 4A2 xex + 2A2 ex + A1 xex + 2A1 ex + A0 ex ) − 2(A2 x2 ex + 2A2 xex + A1 xex + A1 ex + A0 ex ) − 15(A2 x2 ex + A1 xex + A0 ex ) = 32x2 ex + 40xex Collecting like terms on the right hand side, we get −16A2 x2 ex − 16A1 xex + (−16A0 + 2A2 )ex = 32x2 ex + 40xex So, equating the coefficients of like terms on the right and left hand sides: −16A2 = 32 −16A1 = 40 −16A0 + 2A2 = 0
This system of linear equations has solution A0 = − 41 , A1 = − 25 , and A2 = −2. Thus, 5 1 yp = −2x2 ex − xex − ex 2 4 and
1 y = yh + yp = c1 e−3x + c2 e5x − 2x2 ex + ex . 4
40
(f) y 000 − 7y 00 + 17y 0 − 15y = 15x2 − 60x + 27. We start by finding all solutions to the homogeneous equation y 000 − 7y 00 − 17y 0 − 15y = 0. This has characteristic equation λ3 − 7λ2 + 17λ − 15 = (λ − 3)(λ2 − 4λ + 5) = 0. Thus, λ1 = 3 is one root of this equation. We find the other two roots using the quadratic formula: λ2 , λ3 = 2 ± i. The solution to the homogeneous equation is then yh = c1 e3x + c2 e2x cos x + c3 e2x sin x, where c1 , c2 and c3 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = A2 x2 +A1 x+A0 , so yp0 = 2A2 x + A1 yp00 = 2A2 yp000 = 0 Substituting these into y 000 − 7y 00 + 17y 0 − 15y = 15x2 − 60x + 27, we get −7(2A2 ) + 17(2A2 x + A1 ) − 15(A2 x2 + A1 x + A0 ) = 15x2 − 60x + 27 Collecting like terms on the right hand side, we get −15A2 x2 + (34A2 − 15A1 )x + (17A1 − 15A0 ) = 15x2 − 60x + 27 So, equating the coefficients of like terms on the right and left hand sides: −15A2 = 15 34A2 − 15A1 = −60 17A1 − 15A0 = 27
This system of linear equations has solution A0 = A2 = −1. Thus, 26 37 yp = −x2 + x + 15 225 and
37 225 ,
y = yh + yp = c1 e3x + c2 e2x cos x + c3 e2x sin x − x2 +
41
A1 =
26 15 ,
and
26 37 x+ . 15 225
(g) y (4) + 3y 000 + 2y 00 = 144e2x . We start by finding all solutions to the homogeneous equation y (4) + 3y 000 + 2y 00 = 0. This has characteristic equation λ4 + 3λ3 + 2λ2 = λ2 (λ + 1)(λ + 2) = 0, which has a repeated root, λ1 = λ2 = 0, and two distinct roots, λ3 = −1 and λ4 = −2. The solution to the homogeneous equation is then yh = c1 + c2 x + c3 e−x + c4 e−2x , where c1 , c2 , c3 and c4 are arbitrary constants. We next look for a particular solution by the method of undetermined coefficients. We look for a solution of the form yp = Ae2x , so yp0 = 2Ae2x yp00 = 4Ae2x yp000 = 8Ae2x yp(4) = 16Ae2x Substituting these into y (4) + 3y 000 + 2y 00 = 10e2x , we get (16Ae2x ) + 3(8Ae2x ) + 2(4Ae2x ) = 144e2x Collecting like terms on the right hand side, we get 48Ae2x = 144e2x So, A = 3, yp = 3e2x and y = yh + yp = c1 + c2 x + c3 e−x + c4 e−2x + 3e2x
42
4. Solve the following Cauchy-Euler equations. (a) x2 y 00 − 5xy 0 + 13y = 0. Here, a = −5 and b = 13, so the auxiliary equation is λ2 −6λ+13 = 0. Using the quadratic formula, we see this has roots λ1 , λ2 = 3 ± 2i, so the solution to the equation is y = c1 x3 cos (2 ln x) + c2 x3 sin (2 ln x), where c1 and c2 are arbitrary constants. (b) x2 y 00 − xy 0 + 5y = 0. Here, a = −1 and b = 5, so the auxiliary equation is λ2 − 2λ + 5 = 0. Using the quadratic formula, we see this has roots λ1 , λ2 = 1 ± 2i, so the solution to the equation is y = c1 x cos (2 ln x) + c2 x sin (2 ln x), where c1 and c2 are arbitrary constants. (c) x2 y 00 + 6xy 0 + 25y = 0. Here, a = 6 and b = 25, so the auxiliary equation is λ2 + 5λ + 25 = 0. Using the quadratic formula, we see this has roots √ 5 λ1 , λ2 = − ± i5 3, 2 so the solution to the equation is √ √ 5 5 y = c1 x− 2 cos (5 3 ln x) + c2 x− 2 sin (5 3 ln x), where c1 and c2 are arbitrary constants.
43
5. Find the general solution of the following non-homogeneous Cauchy-Euler equations by the method of variation of parameters. (a) x2 y 00 − 2xy 0 + 2y = x3 ex . We start by finding all solutions to the homogeneous equation, x2 y 00 − 2xy 0 + 2y = 0. This is a Cauchy-Euler equation with a = −2 and b = 2 and auxiliary equation λ2 − 3λ + 2 = (λ − 1)(λ − 2) = 0. This has two distinct real roots, λ1 = 1 and λ2 = 2, so the solution to the homogeneous equation is yh = c1 x + c2 x2 , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x and y2 = x2 : x x2 = x2 . W [y1 , y2 ] = 1 2x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where x2 (x3 ex ) dx x2 (x2 )
Z v1 (x) = − Z
xex dx
=−
= −xex + ex
and x(x3 ex ) dx x2 (x2 )
Z v2 (x) = Z =
ex dx
= ex . Thus, yp = x(−xex + ex ) + x2 (ex ) = xex and y = yh + yp = c1 x + c2 x2 − xex .
44
(b) x2 y 00 − 7xy 0 + 15y = x6 cos x. We start by finding all solutions to the homogeneous equation, x2 y 00 − 7xy 0 + 15y = 0. This is a Cauchy-Euler equation with a = −7 and b = 15 and auxiliary equation λ2 − 8λ + 15 = (λ − 3)(λ − 5) = 0. This has two distinct real roots, λ1 = 3 and λ2 = 5, so the solution to the homogeneous equation is yh = c1 x3 + c2 x5 , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x3 and y2 = x5 : 3 x x5 = 2x7 . W [y1 , y2 ] = 2 3x 5x4 The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where x5 (x6 cos x) dx x2 (2x7 )
Z v1 (x) = − Z
x2 cos x dx
=−
= −x2 sin x + 2 sin x − 2x cos x
and Z v2 (x) =
x3 (x6 cos x) dx x2 (2x7 )
Z =
2 cos x dx
= 2 sin x. Thus, yp = x3 (−x2 sin x+2 sin x−2x cos x)+x2 (2 sin x) = x5 sin x+2x3 sin x−2x4 cos x and y = yh + yp = c1 x3 + c2 x5 + x5 sin x + 2x3 sin x − 2x4 cos x.
45
(c) x2 y 00 − 3xy 0 + 4y = x5 We start by finding all solutions to the homogeneous equation, x2 y 00 − 3xy 0 + 4y = 0. This is a Cauchy-Euler equation with a = −3 and b = 4 and auxiliary equation λ2 − 4λ + 4 = (λ − 2)2 = 0. This has two equal real roots, λ1 = λ2 = 2, so the solution to the homogeneous equation is yh = c1 x2 + c2 x2 ln x, where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x2 and y2 = x2 ln x: 2 x x2 ln x = x3 . W [y1 , y2 ] = 2x 2x ln x + x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where Z v1 (x) = − Z =−
x2 ln x(x5 ) dx x2 (x3 ) x2 ln x dx
1 1 = − x3 ln x + x3 3 9 and Z v2 (x) = Z = =
x2 (x5 ) dx x2 (x3 ) x2 dx
1 3 x . 3
Thus, 1 1 1 1 4 yp = x2 (− x3 ln x + x3 ) + x2 ln x( x3 ) = − x5 ln x + x5 3 9 3 3 9 and
4 1 y = yh + yp = c1 x2 + c2 x2 ln x − x5 ln x + x5 . 3 9
46
(d) x2 y 00 − 5xy 0 + 9y = x3 ln x We start by finding all solutions to the homogeneous equation, x2 y 00 − 5xy 0 + 9y = 0. This is a Cauchy-Euler equation with a = −5 and b = 9 and auxiliary equation λ2 − 6λ + 9 = (λ − 3)2 = 0. This has two equal real roots, λ1 = λ2 = 3, so the solution to the homogeneous equation is yh = c1 x3 + c2 x3 ln x, where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x3 and y2 = x3 ln x: 3 x x3 ln x = x5 . W [y1 , y2 ] = 2 3x 3x2 ln x + x2 The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where x3 ln x(x3 ln x) dx x2 (x5 ) Z (ln x)2 =− dx x 1 = − (ln x)3 3 Z
v1 (x) = −
and x3 (x3 ln x) dx x2 (x5 ) Z ln x = dx x 1 = (ln x)2 . 2 Z
v2 (x) =
Thus, 1 1 1 yp = x3 (− (ln x)3 ) + x3 ln x( (ln x)2 ) = (ln x)3 3 2 6 and
1 y = yh + yp = c1 x3 + c2 x3 ln x + (ln x)3 . 6
47
(e) x2 y 00 + 7xy 0 + 9y = x4 We start by finding all solutions to the homogeneous equation, x2 y 00 + 7xy 0 + 9y = 0. This is a Cauchy-Euler equation with a = 7 and b = 9 and auxiliary equation λ2 + 6λ + 9 = (λ + 3)2 = 0. This has two equal real roots, λ1 = λ2 = −3, so the solution to the homogeneous equation is yh = c1 x−3 + c2 x−3 ln x, where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = x−3 and y2 = x−3 ln x: −3 x x−3 ln x −7 . W [y1 , y2 ] = −4 −4 −4 = x −3x −3x ln x + x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where Z v1 (x) = − Z =−
x−3 ln x(x4 ) dx x2 (x−7 ) x6 ln x dx
1 1 = − x7 ln x + x7 7 49 and Z v2 (x) = Z = =
x−3 (x4 ) dx x2 (x−7 ) x6 dx
1 7 x . 7
Thus, 1 1 1 1 4 yp = x−3 (− x7 ln x + x7 ) + x−3 ln x( x7 ) = x 7 49 7 49 and y = yh + yp = c1 x−3 + c2 x−3 ln x +
48
1 4 x . 49
6. Solve the following linear equations with constant coefficients by the method of variation of parameters. 2x
(a) y 00 − 4y 0 + 4y = ex . We start by finding all solutions to the homogeneous equation y 00 − 4y 0 + 4y = 0. This has characteristic equation λ2 − 4λ + 4 = (λ − 2)2 = 0, which has repeated roots λ1 = λ2 = 2, so the solution to the homogeneous equation is yh = c1 e2x + c2 xe2x , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = e2x and y2 = xe2x : 2x e xe2x 4x W [y1 , y2 ] = 2x 2x 2x = e . 2e e + 2xe The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where Z xe2x v1 (x) = −
e2x x
e4x
dx
Z =−
1 dx
= −x and Z e2x v2 (x) =
e2x x
e4x
dx
Z
1 dx x = ln x =
Thus, yp = −xe2x + x(ln x)e2x and y = yh + yp = c1 e2x + c2 xe2x − xe2x + x(ln x)e2x .
49
(b) y 00 − 3y 0 − 18y = 27e6x . We start by finding all solutions to the homogeneous equation y 00 − 3y 0 − 18y = 0. This has characteristic equation λ2 − 3λ − 18 = (λ + 3)(λ − 6) = 0, which has distinct real roots λ1 = −3 and λ2 = 6, so the solution to the homogeneous equation is yh = c1 e−3x + c2 e6x , where c1 and c2 are arbitrary constants. Next, we compute the Wronskian of the fundamental solutions y1 = e−3x and y2 = e6x : −3x e e6x = 9e3x . W [y1 , y2 ] = −3e−3x 6e6x The method of variation of parameters tells us that the particular solution is of the form xp = v1 y1 + v2 y2 , where e6x 27e6x dx 9e3x
Z v1 (x) = − Z =− =−
3e3x dx
e9x 3
and e−3x 27e6x dx 9e3x
Z v2 (x) = Z =
3 dx
= 3x Thus, yp = −
e6x e9x −3x e + 3xe6x = − + 3xe6x 3 3
and y = yh + yp = c1 e−3x + c2 e6x −
50
e6x + 3xe6x = c1 e−3x + c2 e6x + 3xe6x . 3
7. Suppose that the solutions to y 000 − 6y 00 + 12y 0 − 8y = 0 are e2x , xe2x , and x2 e2x . Show that these functions are linearly independent. To show they are linearlly independent, we must show that their Wronskian is a non-zero function. 2x e xe2x x2 e2x 2x 2x 2 2x 2x 2x 2x 2x 2 2x = 2e6x 6= 0. e + 2xe 2xe + 2x e W [e , xe , x e ] = 2e 4e2x 4e2x + 4xe2x 2e2x + 8xe2x + 4x2 e2x Therefore, they are linearly independent. 8. Suppose that the solutions to y 00 − 4y 0 + 5y = 0 are e2x cos x and e2x sin x. Show that these functions are linearly independent. To show they are linearlly independent, we must show that their Wronskian is a non-zero function. e2x cos x e2x sin x 2x 2x W [e cos x, e sin x] = 2x 2e cos x − e2x sin x e2x cos x + 2e2x sin x = e4x cos2 x + e4x sin2 x = e4x 6= 0. Therefore, they are linearly independent.
51
9. ex − 1 is a solution to (ex + 1)y 00 − 2y 0 − ex y = 0. Find another linearly independent solution. Let y1 = 3x − 1, sp y10 = y100 = ex . Start by checking that y1 is a solution of the differential equation: (ex + 1)y100 − 2y10 − ex y1 = (ex + 1)ex − 2e− ex (ex − 1) = e2x + ex − 2ex − e2x + ex = 0. Let the unknown linearly independent solution be y2 . Then, we must have y1 y2 W [y1 , y2 ] = 0 y1 y20 = y1 y20 − y2 y10 = (ex − 1)y20 − ex y2 . On the other hand, applying Abel’s formula to the differential equation (ex + 1)y 00 − 2y 0 − ex y = 0, R
W [y1 , y2 ] = Ce
2 ex +1
dx
Let u = ex + 1, so du = ex dx = (u − 1) dx and Z
so W [y1 , y2 ] = C
2 dx = ex + 1
ex ex +1
Z
du u−1
= dx:
2 du u(u − 1) Z 2 2 − du = u u−1 = −2 ln u + 2 ln u − 1 2 u−1 = ln u x 2 e = ln x e +1
2
.
52
Setting these two expressions for the Wronskian equal, we get a linear differential equation for y2 : 2 ex (e − − e y2 = C ex + 1 x e2x e y20 − x y2 = C x . e −1 (e + 1)2 (ex − 1) x
1)y20
x
x
2x
Here, P (x) = − exe−1 and Q(x) = C (ex +1)e2 (ex −1) so R x e2x − exe−1 C x y2 = e dx + C2 e (e + 1)2 (ex − 1) Z x e2x − ln (ex −1) = eln (e −1) C e dx + C 2 (ex + 1)2 (ex − 1) Z e2x = (ex − 1) C dx + C 2 (ex + 1)2 (ex − 1)2 Z e2x = (ex − 1) C dx + C 2 (e2x − 1)2 R
ex ex −1
Z
Let u = e2x − 1, so du = 2e2x dx and Z
du 2
e2x dx = (e2x − 1)2
= e2x dx. Then, Z
1 du 2u2 1 =− 2u 1 1 = − 2x , 2e −1
so C 1 y2 = (ex − 1) − + C 2 2 e2x − 1 x e −1 = C1 2x + C2 (ex − 1), e −1 where C1 = − C2 . Thus,
ex −1 e2x −1
is another linearly independent solution.
53
10. Solve the following differential equations by reduction of order. (a) y 00 + x(y 0 )2 = 0. Let v = y 0 , so v 0 = y 00 and the differential equation becomes v 0 +xv 2 = 0. This is a separable equation: dv = −xv 2 dx dv = −x dx 2 Z v Z dv = − x dx v2 1 x2 − = − + C1 v 2 1 v = x2 2 + C1 Lastly, v = y 0 , so Z y=
v dx Z
1 dx + C1 Z 1 1 dx = x2 C1 1 + 2C 1 =
Let u =
√x , 2C1
so du =
x2 2
√1 2C1
and
√
2C1 du = dx:
Z 1 1 dx x2 C1 1 + 2C 1 √ Z 2C1 1 = du C1 1 + u2 r 2 = arctan u + C2 C1 r 2 x = arctan √ + C2 . C1 2C1
y=
54
(b) y 00 y 0 = 2. Let v = y 0 , so v 0 = y 00 and the differential equation becomes v 0 v = 2. This is separable: dv =2 dx v dv = 2 dx Z Z v dv = 2 dx v
v2 = 2x + C1 2 p v = ± 4x + C1 Lastly, v = y 0 , so Z y= =±
v dx Z p
4x + C1 dx
3 1 = ± (4x + C1 ) 2 = C2 . 6
55
0.3
The Laplace Transform
1. Find the following Laplace transforms of the following functions using the definition. (a) f (t) = 1. The Laplace Transform of f (t) = 1 is Z ∞ F (s) = e−st dt 0
Z = lim
R→∞
R
e−st dt
0
R 1 − e−st R→∞ s 0 1 −sR 1−e = lim R→∞ s 1 = s = lim
if s > 0
and diverges otherwise. (b) f (t) = t2 . The Laplace Transform of f (t) = t2 is Z ∞ F (s) = t2 e−st dt 0
Z = lim
R→∞
R
t2 e−st dt
0
R Z R 1 1 −st 2te dt − te−st + R→∞ s 0 s 0 ! R Z 1 −sR 2 1 −st 1 R −st = lim − Re + − te + e dt R→∞ s s s s 0 0 R 2 1 −sR 2 1 −st −sR − 2 Re = lim − Re + 2 − e R→∞ s s s s 0 1 −sR 2 2 −sR −sR = lim − Re − 2 Re + 3 1−e R→∞ s s s 2 = 3 s
= lim
and diverges otherwise.
56
if s > 0
(c) f (t) = eat . The Laplace Transform of f (t) = eat is Z ∞ F (s) = eat e−st dt 0
Z = lim
R→∞
R
e−(s−a)t dt
0
R 1 −(s−a)t = lim − e R→∞ s−a 0 1 1 − e−(s−a)R = lim − R→∞ s−a 1 = s−a and diverges otherwise.
57
if s > a
(d) f (t) = cos at. The Laplace Transform of f (t) = cos at is Z ∞ F (s) = e−st cos at dt 0
Z = lim
R→∞
We compute R
Z 0
RR 0
R
e−st cos at dt
0
e−st cos at dt by parts:
R Z 1 a R −st e−st cos at dt = − e−st cos at − e sin at dt s s 0 0 1 1 = − e−sR cos aR s s ! R Z a 1 −st a R −st − − e sin at + e cos at dt s s s 0 0 Z 1 a a2 R −st 1 e cos at dt = − e−sR cos aR + 2 e−sR sin aR − 2 s s s s 0
Thus,
a2 1+ 2 s
Z
R
−st
e
R
Z
1 a 1 − e−sR cos aR + 2 e−sR sin aR s s s 1 1 a − e−sR cos aR + 2 e−sR sin aR s s s
e−st cos at dt =
0
cos at dt =
0
1 1+
a2 s2
so Z F (s) = lim
R→∞
=
R
e−st cos at dt
0
1 s
if s > 0
2
1 + as2 s = 2 s + a2
if s > 0
and diverges otherwise.
58
2. Use the properties of the Laplace transform and the table at the front of the book to determine the Laplace Transform of the following functions. (a) 5 + 11t. L {5 + 11t} = 5L {1} + 11L {t} =
5 11 + 2. s s
(b) t4 e−2t . L t4 e−2t =
4! 24 = . 5 (s + 2) (s + 2)5
(c) t3 sin 3t. d3 L t3 sin 3t = (−1)3 3 L {sin 3t} ds 3 d3 = (−1)3 3 2 ds s + 32 72s(s2 − 9) = (s2 + 9)4 (d) e5t sin 4t. L e5t sin 4t =
4 4 = 2 . 2 2 (s − 5) + 4 s − 10s + 41
(e) t2 cos 2t + t3 e−4t . L t2 cos 2t + t3 e−4t = L t2 cos 2t + L t3 e−4t d2 3! L {cos 2t} + ds2 (s + 4)4 d2 s 6 = 2 2 + ds s + 22 (s + 4)4 2s(s2 − 12) 6 = + (s2 + 4)3 (s + 4)4 = (−1)2
59
3. Find the Laplace transform of the following step functions. ( 0 t
