Lecture 12: Integrating Factors
LECTURE 12 Integrating Factors
Recall that a differential equation of the form (12.1) M (x,y) + N (x,y)y = 0 is said to be exact if
∂M = ∂N , ∂y ∂x
(12.2)
and that in such a case, we could always find an implicit solution of the form (12.3) ψ(x,y) = C with ∂ψ = M (x,y) ∂x (12.4) ∂ψ = N (x,y) . ∂y Even if (12.1) is not exact, it is sometimes possible multiply it by another function of x and/or y to obtain an equivalent equation which is exact. That is, one can sometimes find a function µ(x,y) such that (12.5)
µ(x,y)M (x,y) + µ(x,y)N (x,y)y = 0
is exact. Such a function µ(x,y) is called an ntegrating factor. If an integrating factor can be found, then the original differential equation (12.1) can be solved by simply constructing a solution to the equivalent exact differential equation (12.5).
Example 12.1. Consider the differential equation
dy = 0 . x y + x 1 + y dx 2
This equation is not exact; for (12.6) and so
∂M ∂y ∂N ∂x
= =
3
2
xy
= 3x = 1+y
∂ 2 3 ∂y ∂ 1+ 2 ∂x
x
y
2
2
.
∂M = ∂N . ∂y ∂x
However, if we multiply both sides of the differential equation by µ(x,y) = 1 we get
xy
3
dy = 0 x + 1 +y y dx 2
3
which is not only exact, it is also separable. The general solution is thus obtained by calculating 1
12. INTEGRATING FACTORS
H ( x) 1
H2 (y)
and then demanding that
y
is related to
= =
xdx = 12 x2 1+ y 2 1 y 3 dy = 2y 2
2
+ ln |y|
x by H1 (x) + H2 (y) = C
or 1 2 x
2
−
1
2y2
+ ln |y| = C
.
Now, in general, the problem of finding an integrating factor µ(x, y) for a given differential equation is very difficult. In certain cases, it is rather easy to find an integrating factor.
0.1. Equations with Integrating Factors that depend only on x. Consider a general first order differential equation dy = 0 . (12.7) M (x,y) + N (x,y) dx We shall suppose that there exists an integrating factor for this equation that depends only on x: (12.8) µ = µ ( x) . If µ is to really be an integrating factor, then dy (12.9) µ(x)M (x,y) + µ(x)N (x,y) dx must be exact; i.e., ∂ ∂ (12.10) ∂y (µ(x)M (x,y)) = ∂x (µ(x)N (x,y)) .
Carrying out the differentiations (using the product rule, and the fact that µ(x) depends only on x), we get
or (12.11)
dµ N + µ ∂N µ ∂M = ∂y dx ∂x
dµ = 1 ∂M − ∂N µ . dx N ∂y ∂x Now if µ is depends only on x (and not on y), then necessarily dµ dx depends only on x.
Thus, the selfconsistency of equations (12.8) and (12.11) requires the right hand side of (12.11) to be a function of x alone. We presume this to be the case and set p(x) = − 1 ∂M − ∂N
so that we can rewrite (12.11) as
N ∂y ∂x
dµ + p(x)µ = 0 . (12.12) dx This is a first order linear differential equation for µ hat we can solve! According to the formula developed in Section 2.1, the general solution of (12.12) is ∂M ∂N (12.13) µ(x) = A exp −p(x)dx = A exp N1 ∂y − ∂y dx . The formula (12.13) thus gives us an integrating factor for (12.7) so long as 1 ∂M − ∂N depends only on x.
N ∂y ∂x
12. INTEGRATING FACTORS
3
y
0.2. Equations with Integrating Factors that depend only on .
first order differential equation
Consider again the general
dy = 0 . (12.14) M (x,y) + N (x,y) dx We shall suppose that there exists an integrating factor for this equation that depends only on y: (12.15) µ = µ( y ) . If µ is to really be an integrating factor, then
(12.16) must be exact; i.e., (12.17)
dy µ(y)M (x,y) + µ(y)N (x,y) dx
∂ ∂ ∂y (µ(y)M (x,y)) = ∂x (µ(y)N (x,y)) .
Carrying out the differentiations (using the product rule, and the fact that µ(y) depends only on y), we get
dµ M + µ ∂M = µ ∂N dy ∂y ∂x or dµ = 1 ∂N − ∂M µ . (12.18) dy M ∂x ∂y Now since µ is depends only on y (and not on x), then necessarily dµ dy depends only on y . Thus, the selfconsistency of equations (12.15) and (12.18) requires the right hand side of (12.11) to be a function of y alone. We presume this to be the case and set ∂M − p(y) = − M1 ∂N ∂x ∂y so that we can rewrite (12.11) as
(12.19)
dµ + p(y)µ = 0 . dy
According to the formula developed in Section 2.1, the general solution of (12.19) is 1 ∂N ∂M µ(y) = A exp −p(y)dx = A exp (12.20) − dx
M ∂x ∂y
The formula (12.20) thus gives us an integrating factor for (12.14) so long as 1 ∂N − ∂M depends only on y.
M ∂x ∂y
Suppose that M (x,y) + N (x,y)y = 0
0.3. Summary: Finding Integrating Factors.
(12.21) is not exact. A. If
(12.22) depends only on x then (12.23) will be an integrating factor for (12.21).
F
1
=
µ(x) = exp
∂M ∂y
− ∂N ∂x
N
F (x)dx 1
.
12. INTEGRATING FACTORS
B. If
F
(12.24) depends only on y then
M
µ(y) = exp
(12.25)
− ∂M ∂y
∂N ∂x
=
2
4
F (y)dy
2
will be an integrating factor for (12.21). C. If neither A nor B is true, then there is little hope of constructing an integrating factor. Example 12.2.
x y xy + y )dx + (x + y )dy = 0
(3 2 + 2
(12.26) Here
3
M (x,y) N (x,y)
Since
∂M = 3x ∂y
2
this equation is not exact. We seek to find a function
2
x y xy + y x y .
3 2 +2 2+ 2
= =
3
x y x ∂N ∂x
+2 +3 2 =2 =
µ such that µ(x,y)(3x y + 2xy + y )dx + µ(x,y)(x + y )dy = 0 2
is exact. Now
F ≡
∂M ∂y
F ≡
∂N ∂x
1
2
2
3
− ∂N ∂x
M −
N
∂M ∂y
2
x
2
x y − 2x = 3(x + y ) = 3 x +y x + y −3 x + y 2x − 3x − 2x − 3y = = 3x y + 2xy + y 3x y + 2xy + y
=
3 2 +2 +3 2 2
2
2
2
2
2
2
2
2
2
3
2
2
3
Since F depends on both x and y, we cannot construct an integrating factor depending only on y from F . However, since F does not depend on y, we can consistently construct an integrating factor that is a function of x alone. Applying formula (12.23) we get 2
2
1
µ(x) = exp
F1 (x)dx
= exp
3dx = e x . 3
We can now employ this µ(x) as an integrating factor to construct a general solution of e x (3x + 2x + 3y ) + e x (x + y )y = 0 which, by construction, must be exact. So we seek a function ψ such that ∂ψ = e x (3x y + 2xy + y ) ∂x (12.27) ∂ψ = e x (x + y ) . ∂y Integrating the first equation with respect to x and the second equation with respect to y yeilds ψ(x, y) = x ye x + y e x + h (y) ψ(x, y) = x ye x + y e x + h (x) . Comparing these expressions for ψ (x, y) we see that we msut take h (y) = h (x) = C , a constant. Thus, function ψ satisfying (12.27) must be of the form ψ(x, y) = e x x y + e x y + C . 3
2
2
3
3
2
2
3
3
2
3
3
2
2
2
2
1 3 1 3
3 3 3 3
1 2
1
3
2
3
3
2
12. INTEGRATING FACTORS
Therefore, the general solution of (12.20) is found by solving e xx y + e xy = C for y. 3
2
3
3
5
Recall that a differential equation of the form (12.1) M (x,y) + N (x,y)y = 0 is said to be exact if
∂M = ∂N , ∂y ∂x
(12.2)
and that in such a case, we could always find an implicit solution of the form (12.3) ψ(x,y) = C with ∂ψ = M (x,y) ∂x (12.4) ∂ψ = N (x,y) . ∂y Even if (12.1) is not exact, it is sometimes possible multiply it by another function of x and/or y to obtain an equivalent equation which is exact. That is, one can sometimes find a function µ(x,y) such that (12.5)
µ(x,y)M (x,y) + µ(x,y)N (x,y)y = 0
is exact. Such a function µ(x,y) is called an ntegrating factor. If an integrating factor can be found, then the original differential equation (12.1) can be solved by simply constructing a solution to the equivalent exact differential equation (12.5).
Example 12.1. Consider the differential equation
dy = 0 . x y + x 1 + y dx 2
This equation is not exact; for (12.6) and so
∂M ∂y ∂N ∂x
= =
3
2
xy
= 3x = 1+y
∂ 2 3 ∂y ∂ 1+ 2 ∂x
x
y
2
2
.
∂M = ∂N . ∂y ∂x
However, if we multiply both sides of the differential equation by µ(x,y) = 1 we get
xy
3
dy = 0 x + 1 +y y dx 2
3
which is not only exact, it is also separable. The general solution is thus obtained by calculating 1
12. INTEGRATING FACTORS
H ( x) 1
H2 (y)
and then demanding that
y
is related to
= =
xdx = 12 x2 1+ y 2 1 y 3 dy = 2y 2
2
+ ln |y|
x by H1 (x) + H2 (y) = C
or 1 2 x
2
−
1
2y2
+ ln |y| = C
.
Now, in general, the problem of finding an integrating factor µ(x, y) for a given differential equation is very difficult. In certain cases, it is rather easy to find an integrating factor.
0.1. Equations with Integrating Factors that depend only on x. Consider a general first order differential equation dy = 0 . (12.7) M (x,y) + N (x,y) dx We shall suppose that there exists an integrating factor for this equation that depends only on x: (12.8) µ = µ ( x) . If µ is to really be an integrating factor, then dy (12.9) µ(x)M (x,y) + µ(x)N (x,y) dx must be exact; i.e., ∂ ∂ (12.10) ∂y (µ(x)M (x,y)) = ∂x (µ(x)N (x,y)) .
Carrying out the differentiations (using the product rule, and the fact that µ(x) depends only on x), we get
or (12.11)
dµ N + µ ∂N µ ∂M = ∂y dx ∂x
dµ = 1 ∂M − ∂N µ . dx N ∂y ∂x Now if µ is depends only on x (and not on y), then necessarily dµ dx depends only on x.
Thus, the selfconsistency of equations (12.8) and (12.11) requires the right hand side of (12.11) to be a function of x alone. We presume this to be the case and set p(x) = − 1 ∂M − ∂N
so that we can rewrite (12.11) as
N ∂y ∂x
dµ + p(x)µ = 0 . (12.12) dx This is a first order linear differential equation for µ hat we can solve! According to the formula developed in Section 2.1, the general solution of (12.12) is ∂M ∂N (12.13) µ(x) = A exp −p(x)dx = A exp N1 ∂y − ∂y dx . The formula (12.13) thus gives us an integrating factor for (12.7) so long as 1 ∂M − ∂N depends only on x.
N ∂y ∂x
12. INTEGRATING FACTORS
3
y
0.2. Equations with Integrating Factors that depend only on .
first order differential equation
Consider again the general
dy = 0 . (12.14) M (x,y) + N (x,y) dx We shall suppose that there exists an integrating factor for this equation that depends only on y: (12.15) µ = µ( y ) . If µ is to really be an integrating factor, then
(12.16) must be exact; i.e., (12.17)
dy µ(y)M (x,y) + µ(y)N (x,y) dx
∂ ∂ ∂y (µ(y)M (x,y)) = ∂x (µ(y)N (x,y)) .
Carrying out the differentiations (using the product rule, and the fact that µ(y) depends only on y), we get
dµ M + µ ∂M = µ ∂N dy ∂y ∂x or dµ = 1 ∂N − ∂M µ . (12.18) dy M ∂x ∂y Now since µ is depends only on y (and not on x), then necessarily dµ dy depends only on y . Thus, the selfconsistency of equations (12.15) and (12.18) requires the right hand side of (12.11) to be a function of y alone. We presume this to be the case and set ∂M − p(y) = − M1 ∂N ∂x ∂y so that we can rewrite (12.11) as
(12.19)
dµ + p(y)µ = 0 . dy
According to the formula developed in Section 2.1, the general solution of (12.19) is 1 ∂N ∂M µ(y) = A exp −p(y)dx = A exp (12.20) − dx
M ∂x ∂y
The formula (12.20) thus gives us an integrating factor for (12.14) so long as 1 ∂N − ∂M depends only on y.
M ∂x ∂y
Suppose that M (x,y) + N (x,y)y = 0
0.3. Summary: Finding Integrating Factors.
(12.21) is not exact. A. If
(12.22) depends only on x then (12.23) will be an integrating factor for (12.21).
F
1
=
µ(x) = exp
∂M ∂y
− ∂N ∂x
N
F (x)dx 1
.
12. INTEGRATING FACTORS
B. If
F
(12.24) depends only on y then
M
µ(y) = exp
(12.25)
− ∂M ∂y
∂N ∂x
=
2
4
F (y)dy
2
will be an integrating factor for (12.21). C. If neither A nor B is true, then there is little hope of constructing an integrating factor. Example 12.2.
x y xy + y )dx + (x + y )dy = 0
(3 2 + 2
(12.26) Here
3
M (x,y) N (x,y)
Since
∂M = 3x ∂y
2
this equation is not exact. We seek to find a function
2
x y xy + y x y .
3 2 +2 2+ 2
= =
3
x y x ∂N ∂x
+2 +3 2 =2 =
µ such that µ(x,y)(3x y + 2xy + y )dx + µ(x,y)(x + y )dy = 0 2
is exact. Now
F ≡
∂M ∂y
F ≡
∂N ∂x
1
2
2
3
− ∂N ∂x
M −
N
∂M ∂y
2
x
2
x y − 2x = 3(x + y ) = 3 x +y x + y −3 x + y 2x − 3x − 2x − 3y = = 3x y + 2xy + y 3x y + 2xy + y
=
3 2 +2 +3 2 2
2
2
2
2
2
2
2
2
2
3
2
2
3
Since F depends on both x and y, we cannot construct an integrating factor depending only on y from F . However, since F does not depend on y, we can consistently construct an integrating factor that is a function of x alone. Applying formula (12.23) we get 2
2
1
µ(x) = exp
F1 (x)dx
= exp
3dx = e x . 3
We can now employ this µ(x) as an integrating factor to construct a general solution of e x (3x + 2x + 3y ) + e x (x + y )y = 0 which, by construction, must be exact. So we seek a function ψ such that ∂ψ = e x (3x y + 2xy + y ) ∂x (12.27) ∂ψ = e x (x + y ) . ∂y Integrating the first equation with respect to x and the second equation with respect to y yeilds ψ(x, y) = x ye x + y e x + h (y) ψ(x, y) = x ye x + y e x + h (x) . Comparing these expressions for ψ (x, y) we see that we msut take h (y) = h (x) = C , a constant. Thus, function ψ satisfying (12.27) must be of the form ψ(x, y) = e x x y + e x y + C . 3
2
2
3
3
2
2
3
3
2
3
3
2
2
2
2
1 3 1 3
3 3 3 3
1 2
1
3
2
3
3
2
12. INTEGRATING FACTORS
Therefore, the general solution of (12.20) is found by solving e xx y + e xy = C for y. 3
2
3
3
5
