First-order linear differential equations: the general solution We will ...
First-order linear differential equations: the general solution We will consider equations of the form: (1) y' + P(x)y = Q(x), where x is the independent variable y(x) is the unknown function of x we want to solve for y' is shorthand for the first derivative dy/dx P(x), Q(x) are specified functions of x (but not of y) Equation (3.1) is called a first-order, linear, nonhomogeneous differential equation. First-order – no derivative higher than the first derivative, y' Linear – no powers in y higher than 1 Nonhomogeneous – Q(x) is not zero To see how to solve (1) for y(x), let's first consider the simpler equation when Q(x)=0, which is called a homogeneous equation: (2)
y' + P(x)y = 0, or
dy/dx = -P(x)y
This forms allows us to separate all the y-terms on the left and the x-terms on the right: (3)
dy/y = -P(x) dx, which gives: , where c is the constant of integration. Thus:
, where C = ec .
(4)
So if the function P(x) can be integrated, you get an analytical solution. Let's simplify notation and define the integral I: (5)
.
(6)
Note that as an indefinite integral, I is a function of x, and:
.
Then we notice that the solution (4) to equ (2) can be written: (7) (8)
yeI = C,
and therefore:
(yeI)' = y'eI + yeI I' = eI (y'+P(x)y) = 0
The equation (8) in bold is nothing more than the original differential equ (2) with both sides multiplied by eI. That means that if we had had the insight originally to look at equ (2) and just multiply it by eI, then the left-hand side would have been integrable. In other words, the function eI is an "integrating factor" that allows us to solve (2).
We will now generalize this approach to the nonhomogeneous equation (1). Let's just see if perhaps eI is also an integrating factor for equ (1). So we just multiply both sides of (1) by eI: (9)
eI(y'+P(x)y) = eI Q(x) .
But look! The left hand side is just what it was above from (8) – (yeI)' , and the right-hand side is a function only of x. So as it stands, both sides of equ (9) can be integrated:
(10)
, and so:
(11)
,
or:
(12)
, where c is again the constant of integration.
Thus, the final expression for y(x) is:
(13) This solution to (1) is totally general. Try applying it to some simple cases. (What does it reduce to, for instance, if Q(x) = 0 and P(x) is a constant?) Then see Examples 1&2 in Boas, p. 348. You should realize that the constant c must be determined for each specific problem you encounter. It can always be found by getting a general solution and then applying an "initial condition.' That is, after getting the general solution containing the unknown constant c, all you have to do is ask what is the initial value of y, y(0). You just write down the general solution with the value of y(0) that actually applies to the problem at hand, and plug in t=0. This will always give you an equation that you can solve for c. Try it out!
First-order linear differential equations appear all the time in kinetics, and luckily they always have analytical solutions, which always contain exponential terms. The solution above, which is taken from Boas Ch. 8, is totally general and always works. But for most of the problems we deal with it's a bit of overkill, and you may not want to memorize the solution (but make sure that you can apply it to problems). In kinetics, both P(x) and Q(x) are often constants (i.e., not functions of x), so the general solution collapses to one that's much easier to remember. You should be able to prove that in such a case the equation can be written (with more familiar constants "k" and "A" for P and Q, and time t for x) and solved: (14)
dy/dt + ky = A,
(15)
.
Therefore, this type of equation will always solve to: (16)
,
where the final value of y, y(4), is given in terms of k and A as above, and the initial value of y, y(0), is set by the "initial conditions" of the experiment, or of the specific problem under consideration. This is a very useful form of the solution, because it says that a differential equation of this type (first-order, linear, nonhomogeneous, constant coefficients!!) always gives an exponential relaxation of "amplitude" [y(4)-y(0)] with a rate constant k, where y(4) is given in terms of the constant coefficients. This means that you can just look at the differential equation and write down the answer! The exponential "rises" if y(4) > y(0), and it "falls" if y(4) < y(0). (Don't just take my word for this – play with the solution by trying out some numbers.)
y' + P(x)y = 0, or
dy/dx = -P(x)y
This forms allows us to separate all the y-terms on the left and the x-terms on the right: (3)
dy/y = -P(x) dx, which gives: , where c is the constant of integration. Thus:
, where C = ec .
(4)
So if the function P(x) can be integrated, you get an analytical solution. Let's simplify notation and define the integral I: (5)
.
(6)
Note that as an indefinite integral, I is a function of x, and:
.
Then we notice that the solution (4) to equ (2) can be written: (7) (8)
yeI = C,
and therefore:
(yeI)' = y'eI + yeI I' = eI (y'+P(x)y) = 0
The equation (8) in bold is nothing more than the original differential equ (2) with both sides multiplied by eI. That means that if we had had the insight originally to look at equ (2) and just multiply it by eI, then the left-hand side would have been integrable. In other words, the function eI is an "integrating factor" that allows us to solve (2).
We will now generalize this approach to the nonhomogeneous equation (1). Let's just see if perhaps eI is also an integrating factor for equ (1). So we just multiply both sides of (1) by eI: (9)
eI(y'+P(x)y) = eI Q(x) .
But look! The left hand side is just what it was above from (8) – (yeI)' , and the right-hand side is a function only of x. So as it stands, both sides of equ (9) can be integrated:
(10)
, and so:
(11)
,
or:
(12)
, where c is again the constant of integration.
Thus, the final expression for y(x) is:
(13) This solution to (1) is totally general. Try applying it to some simple cases. (What does it reduce to, for instance, if Q(x) = 0 and P(x) is a constant?) Then see Examples 1&2 in Boas, p. 348. You should realize that the constant c must be determined for each specific problem you encounter. It can always be found by getting a general solution and then applying an "initial condition.' That is, after getting the general solution containing the unknown constant c, all you have to do is ask what is the initial value of y, y(0). You just write down the general solution with the value of y(0) that actually applies to the problem at hand, and plug in t=0. This will always give you an equation that you can solve for c. Try it out!
First-order linear differential equations appear all the time in kinetics, and luckily they always have analytical solutions, which always contain exponential terms. The solution above, which is taken from Boas Ch. 8, is totally general and always works. But for most of the problems we deal with it's a bit of overkill, and you may not want to memorize the solution (but make sure that you can apply it to problems). In kinetics, both P(x) and Q(x) are often constants (i.e., not functions of x), so the general solution collapses to one that's much easier to remember. You should be able to prove that in such a case the equation can be written (with more familiar constants "k" and "A" for P and Q, and time t for x) and solved: (14)
dy/dt + ky = A,
(15)
.
Therefore, this type of equation will always solve to: (16)
,
where the final value of y, y(4), is given in terms of k and A as above, and the initial value of y, y(0), is set by the "initial conditions" of the experiment, or of the specific problem under consideration. This is a very useful form of the solution, because it says that a differential equation of this type (first-order, linear, nonhomogeneous, constant coefficients!!) always gives an exponential relaxation of "amplitude" [y(4)-y(0)] with a rate constant k, where y(4) is given in terms of the constant coefficients. This means that you can just look at the differential equation and write down the answer! The exponential "rises" if y(4) > y(0), and it "falls" if y(4) < y(0). (Don't just take my word for this – play with the solution by trying out some numbers.)
