The method of undetermined coefficients

This method is more limited in scope; it applies only to the special case of (2), where p(t) is a constant and g(t) has some special form. The advantage of the method is that it does not require any integrations and is therefore quick to use. The homogeneous equation

$$y'+\lambda y=0$$

has the solution

$$y_h=Ce^{-\lambda t}.$$

To solve the inhomogeneous equation

$$y'+\lambda y=g(t),$$

it suffices to find one particular solution y_p(t). If y_p(t) is any particular solution, then the general solution is

$$y(t)=y_p(t)+Ce^{-\lambda t}.$$

The idea behind the method of undetermined coefficients is to look for y_p(t) which is of a form like that of g(t). This is possible only for special functions g(t), but these special cases arise quite frequently in applications.

We start with the case where g(t) is an exponential:

$$g(t)=Ae^{\alpha t}.$$

We look for y(t) in a similar form

$$y(t)=ae^{\alpha t}.$$

This leads to

$$y'=a\alpha e^{\alpha t},\quad y'+\lambda y=(\alpha+\lambda)ae^{\lambda t}.$$

So the differential equation becomes

$$(\alpha+\lambda)ae^{\alpha t}=Ae^{\alpha t}.$$

We can solve this to find $a=A/(\alpha+\lambda)$. This leads to the particular solution

$$y_p(t)={A\over \alpha+\lambda}e^{\alpha t},$$

and the general solution  

$$y(t)={A\over \alpha+\lambda}e^{\alpha t}+Ce^{-\lambda t}.$$ (3)

Example: Find the general solution of the equation

y'+2y=e^t.

The solution of the homogeneous equation is $C\exp(-2t)$, and we look for a particular solution in the form y_p=ae^t. Setting y=ae^t in the equation, we find

ae^t+2ae^t=e^t,

leading to a=1/3. The general solution is

$$y={1\over 3}e^t+Ce^{-2t}.$$

Why did this work? The idea is simply that if y is an exponential, then so is y', and so if both y and g are exponentials, then all terms in the equation are exponentials and we can hope to obtain a solution by setting coefficients equal to each other.

There are some other classes of functions for which this works. For instance, if y is a polynomial of degree n, then y' is a polynomial of degree n-1. If g is a polynomial, we can therefore look for polynomial solutions. Consider

y'+2y=t².

The right hand side is a polynomial of degree 2, so we look for a solution in the same form y=at²+bt+c. This leads to y'=2at+b, and

y'+2y=2at²+(2a+2b)t+b+2c=t².

To satisfy this, we want to set

$$2a=1,\quad 2a+2b=0,\quad b+2c=0.$$

This leads to a=1/2, b=-1/2, c=1/4. So a particular solution is

$$y_p={t^2\over 2}-{t\over 2}+{1\over 4}.$$

The general solution is

$$y={t^2\over 2}-{t\over 2}+{1\over 4}+Ce^{-2t}.$$

We note that the solution (3) breaks down if $\alpha=-\lambda$, since it would involve a division by zero. More generally, if the equation reads

$$y'+\lambda y=g(t),$$

and $g(t)=\exp(\alpha t)P_n(t)$, with P_n(t) an nth degree polynomial, then we can find a particular solution $y_p(t)=\exp(\alpha t)Q_n(t)$, where Q_n(t) is some other nth degree polynomial as long as $\alpha\neq -\lambda$.In the two examples above, we had $\lambda=-2$ and $\alpha=1$, $\alpha=0$,respectively, so $\alpha\neq -\lambda$. If, on the other hand, $\alpha=-\lambda$, we have to modify the procedure. The modification is simply to include an extra factor t in the solution. That is, instead of setting $y_p=\exp(\alpha t)Q_n(t)$, you set $y_p=t\exp(\alpha t)Q_n(t)$.

Examples:

1.

y'+2y=te^-2t.

Here $\lambda=2$ and $\alpha=-2$, so $\alpha=-\lambda$. The right hand side is a first degree polynomial times e^-2t. So we look for a solution of the form

y=te^-2t(at+b)=e^-2t(at²+bt).

We find

y'=e^-2t(-2at²+(2a-2b)t+b),

so that

y'+2y=e^-2t(2at+b)=te^-2t.

We compare coefficients to find a=1/2, b=0. The general solution of the equation is

$$y={1\over 2}t^2e^{-2t}+Ce^{-2t}.$$

2.

y'+2y=te^t.

In this case $\lambda=2$ and $\alpha=1$, so $\alpha\neq -\lambda$, and we do not need the extra factor t. So we look for a solution of the form

y=e^t(at+b).

This leads to

y'+2y=e^t(3at+3b+a)=te^t,

so we need

$$3a=1,\quad 3b+a=0,$$

leading to a=1/3, b=-1/9. The general solution is

$$y=({t\over 3}-{1\over 9})e^t+Ce^{-2t}.$$

3.

y'=t.

In this case $\lambda=\alpha=0$, and the right hand side is a first degree polynomial, so we look for a particular solution of the form y=t(at+b)=at² +bt. We find

y'=2at+b=t,

leading to a=1/2, b=0. The general solution is

$$y={t^2\over 2}+C.$$