# Math216 Introduction to Differential Equations

Deniz Bilman, Department of Mathematics, University of Michigan

## Lecture 19: Nonhomogeneous linear differential equations

So far we mainly dealt with linear sistem of 2 first order homogeneous differential equations or second order homogeneous differential equations. Our goal now is to construct general solutions of nonhomogeneous differential equations.

Consider the second order ordinary differential equation:

$$y'' + p(t)y' +q(t)y = g(t),$$

where the coefficient functions $p(t)$, $q(t)$, and the external force (the nonhomogeneous term) $g(t)$ are all continuous on an interval $I \subset \mathbb{R}$. We express this nonhomogeneous ODE in the operator form. We define the second order linear differential operator $\mathcal{L}$ by

$$\mathcal{L}[f]=f''+p(t)f' +q(t)f.$$

Recall that operators are functions that operate on functions: $\mathcal{L}$ takes a function $f$ and maps it to the function $f''+p(t)f' +q(t)f$. With this definition, the nonhomogenous ODE above is expressed as:

$$\mathcal{L}[y]=g(t).$$

Key Observation: Suppose we have two particular solutions $y_1(t)$ and $y_2(t)$ of the ODE $\mathcal{L}[y]=g(t)$. Then their difference $y_2(t)-y_1(t)$ solves the corresponding homogeneous equation $\mathcal{L}[y]=0$ ($\mathcal{L}$ is the same operator.) Indeed:

$$\mathcal{L}[y_2-y_1]=\mathcal{L}[y_2] - \mathcal{L}[y_1] =g(t)-g(t)=0.$$

This true only because $\mathcal{L}$ is a linear operator, of course.

Now suppose that we have two linearly independent solutions (i.e., a fundamental set of solutions) $\bar{y}_1$ and $\bar{y}_2$ of the corresponding homogenous equation $\mathcal{L}[y]=0$. Since the difference $y_1(t)-y_2(t)$ is a solution of $\mathcal{L}[y]=0$ as well, it can be expressed as:

$$y_2(t)-y_1(t) = c_1 \bar{y}_1(t) + c_2 \bar{y}_2(t),$$

where $c_1$ and $c_2$ are some constants. By rearranging this expression as:

$$y_2(t) = c_1 \bar{y}_1(t) + c_2 \bar{y}_2(t) + + y_1(t).$$

This tells us that, if we know just one solution $y_1(t)$ of the nonhomogeneous equation $\mathcal{L}[y]=g(t)$, any solution $y(t)$ of this nonhomogeneous equation can be expressed as

$$y(t) = c_1 \bar{y}_1(t) + c_2 \bar{y}_2(t) + + y_1(t).$$

by choosing the constants $c_1$ and $c_2$ appropriately. We have arrived at the following fact.

Fact: The moment we have a particular solution $y_p(t)$ of the nonhomogeneous equation $\mathcal{L}[y]=g(t)$ and a fundamental set of solutions $\lbrace\bar{y}_1, \bar{y}_2\rbrace$ of the corresponding homogeneous equation $\mathcal{L}[y]=0$, the general solution of the nonhomogeneous equation $\mathcal{L}[y]=g(t)$ is constructed as:

$$y(t) = c_1 \bar{y}_1(t) + c_2 \bar{y}_2(t) + + y_p(t).$$

Note that the general solution of the nonhomogeneous equation is of the form

$$\text{general solution of the homogeneous equation} + \text{a particular solution of the nonhomogeneous equation}$$

and we have already seen in previous lectures that finding the general solution of the homogeneous equation is a rather straightforward procedure. To construct the general solution of the nonhomogeneous equation, we need to be able to find just one particular solution of the nonhomogeneous equation.

Question: How do we find $y_p(t)$?

The first method that addresses this problem is called the Method of Undetermined Coefficients.

## Method of Undetermined Coefficients

This method is based on guessing the form of the particular solution of the nonhomogeneous equation $\mathcal{L}[y]=g(t)$ based on the structure of the nonhomogeneous term $g(t)$.

Example: Consider the following ODE

$$y''-3y'-4y=3e^{2t}.$$

The external force term (the term that introduces nonhomogeneity) is $g(t)=3e^{2t}$. The differential operator on the left hand side is

$$\mathcal{L}[y]=y''-3y'-4y.$$

We are looking for a solution $y_p(t)$ so that $\mathcal{l}[y_p]=3e^{2t}$. Since $\mathcal{L}[y]$ consists of the second derivative, the first derivative, and the function $y$ itself with constant coefficients, it is reasonable to assume that $y_p(t)$ is of the form

$$y_p(t) = A e^{2t}$$

where $A$ is a coefficient to be determined (hence the name of the method). This ansatz makes sense because the derivatives of the exponential function will produce the function itself. We plug this into the nonhomogeneous ODE and determine the constant coefficient $A$. We compute that

$$y_p'(t)=2Ae^{2t},\quad y_p''(t)=4 Ae^{2t},$$

and plugging in gives

$$(4A-6A-4A)e^{2t} = 3e^{2t},$$

and hence $A=\frac{-1}{2}$. Thus a particular solution of the nonhomogeneous equation is found to be:

$$y_p(t)=-\frac{1}{2}e^{2t}.$$

Example: Now consider the same ODE with a different nonhomogeneity $g(t)=2\sin t$:

$$y''-3y'-4y=2\sin t.$$

Now, note that the operator $\mathcal{L}$ involves taking the first derivative of a function it acts on. So if we assume $y_p(t)=A\sin t$, $\mathcal{L}$ will produce a term involving $\cos t$, but that term is absent on the right hand side. Therefore we need to ''seed'' a cosine term in the solution as well to offset the cosine produced by the operator. So we consider the following ansatz:

$$y_p(t)=A\sin t + B\cos t.$$

Plugging this in the nonhomogeneous ODE gives:

$$(-A + 3B -4A)\sin t + (-B-3A-4B)\cos t = 2\sin t.$$

Thus we have

$$-A + 3B -4A = 2\quad\text{and}\quad -B-3A-4B=0.$$

solving this algebraic system of equations for $A$ and $B$ gives:

$$A= -\frac{5}{17}\quad\text{and}\quad B=\frac{3}{17},$$

and a particular solution of the nonhomogeneous equation is found:

$$y_p(t)=-\frac{5}{17}\sin t +\frac{3}{17}\cos t.$$

What if the right hand side is has a more complicated structure?

Example: Now consider the same ODE with the nonhomogeneity $g(t)=-8e^t\cos(2t)$:

$$y''-3y'-4y=-8e^t\cos(2t).$$

The structure of the differential operator dictates us the ansatz

$$y_p(t)=Ae^t\cos(2t) + Be^t\sin(2t),$$

because the product rule for differentiation will produce linear combinations of such terms. Plugging this in the ODE gives:

$$(-3A+4B -3A+6B-4A)e^t\cos(2t)+(-4A -3B -3B+6A-4B)e^t\sin(2t)=-8e^t\cos(2t),$$

which results in the algebraic system for $A$ and $B$:

$$-3A+4B -3A+6B-4A=-8\quad\text{and}\quad -4A -3B -3B+6A-4B=0.$$

The unique solution to this system is:

$$A=\frac{10}{3}\quad\text{and}\quad B=\frac{2}{13},$$

and hence a particular solution of the nonhomogeneous ODE is found:

$$y_p(t)=\frac{10}{3} e^t\cos(2t) + \frac{2}{13}e^t\sin(2t).$$

Question: What if $g(t)$ is even more complicated, involving several terms?

Example: Find a particular solution of $y''-3y'-4y=2\sin t+3e^{2t} + 4t^2$.

We can use a divide and conquer approach using the following principle.

Superposition Principle for Nonhomogeneous Equations. Suppose we would like to find a solution of a nonhomogeneous equation of the form $\mathcal{L}[y]=g_1(t)+g_2(t)$. Let $y_1(t)$ be a solution of $\mathcal{L}[y]=g_1(t)$ and $y_2(t)$ be a solution of $\mathcal{L}[y]=g_2(t)$. Then the sum $y_1(t)+y_2(t)$ is a solution of $\mathcal{L}[y]=g_1(t)+g_2(t)$.

Now back to the example. Note that $g(t)=2\sin t+3e^{2t} + 4t^2$, set $\mathcal{L}[y]:=y''-3y'-4y$, and recall that we found a particular solution of

$$\mathcal{L}[y]=2\sin t$$

to be $y_{p,1}(t)=-\frac{5}{17}\sin t +\frac{3}{17}\cos t$. We have also found a particular solution of

$$\mathcal{L}[y]=3e^{2t}$$

to be $y_{p,2}(t)=-\frac{1}{2}e^{2t}$. We will now find a particular solution of

$$\mathcal{L}[y]=4t^2.$$

Since the right hand side is a second degree polynomial, we assume that the solution is of the form

$$y_{p,3}(t)=At^2 + Bt + C.$$

Plugging this into $\mathcal{L}[y]=4t^2$ gives: $A=-1$, $B=\frac{3}{2}$, and $C=-\frac{3}{8}$, hence

$$y_{p,3}(t)=-t^2 + \frac{3}{2}t + \frac{3}{8}.$$

Using the superposition principle, a particular solution of the equation

$$\mathcal{L}[y]=2\sin t+3e^{2t} + 4t^2$$

is given by

$$y_p(t)=y_{p,1}(t)+y_{p,2}(t)+y_{p,3}(t) = -\frac{5}{17}\sin t +\frac{3}{17}\cos t -\frac{1}{2}e^{2t}-t^2 + \frac{3}{2}t + \frac{3}{8}.$$

Let's now proceed and find the general solution of the nonhomogeneous equation $\mathcal{L}[y]=2\sin t+3e^{2t} + 4t^2$. As we have a particular solution already, all we need to do is to find a general solution of the homogeneous equation $\mathcal{L}[y]=0$.

Solving the homogeneous equation: $\mathcal{L}[y]=0$ has the characteristic equation

$$\lambda^2 - 3\lambda -4 =0,$$

which has two distinct real roots $\lambda_1 = 4$ and $\lambda_2=-1$. Then

$$\bar{y}_1(t)=e^{4t}\quad\text{and}\quad \bar{y}_2(t)=e^{-t}$$

form a fundamental set of solutions of $\mathcal{L}[y]=0$.

General solution of the nonhomogeneous equation: General solution of the nonhomogeneous equation $\mathcal{L}[y]=2\sin t+3e^{2t} + 4t^2$ is therefore given by

$$y(t) = c_1 e^{4t} + c_2 e^{-t} -\frac{5}{17}\sin t +\frac{3}{17}\cos t -\frac{1}{2}e^{2t}-t^2 + \frac{3}{2}t + \frac{3}{8},$$

where $c_1$ and $c_2$ are to be determined by initial data given in an initial value problem.

The remarkable thing about the Method of Undetermined Coefficients (MOUC) is the fact that finding a solution of a differential equation boils down to solving a linear system of algebraic equations. This is straightforward to solve, especially when the differential equation has constant coefficients.

Therefore, given a nonhomogeneous second order linear differential equation, MOUC seems to be the first thing to try, but it has major shortcomings:

• MOUC only works when the external force term ($g(t)$, the nonhomogeneity) has a special form so that a second order linear differential operator can produce $g(t)$ with an appropriately chosen ansatz for the particular solution. For example, if the right handside is a rational function, the simple approach we tried in this lecture wouldn't work, e.g.,
$$y''-3y'-4y=\frac{t^2+1}{3t^2+1},$$
• But even in the case of 'simple' nonhomogeneity $g(t)$, MOUC might fail. Consider the following example:

Example: Find a particular solution of $y''-3y'-4y=2e^{-t}$.

As before, we take the ansatz $y_p(t)=A e^{-t}$ and plug this in the nonhomogeneous ODE. Doing so yields

$$Ae^{-t}+ 3Ae^{-t}-4Ae^{-t}=2e^{-t}$$

which enforces

$$(A+3A-4A)=2.$$

This equation obviously has no solution. Therefore, there is no value of $A$ such that $y_p(t)=A e^{-t}$ solves the nonhomogeneous equation. But why would that be?

This happens because the nonhomogeneity $g(t)=2e^{-t}$ is proportional to a solution of the homogeneous equation we found earlier $y_2(t)=e^{-t}$. Any function $y(t)=Ae^{-t}$ is a solution of the homogeneous equation, therefore it cannot possibly be a solution of the nonhomogeneous equation.

As we have seen today, MOUC applies in certain special cases, and when it applies, it is a very straightforward way to obtain a solution of a second order linear nonhomogenous differential equation. Please see Table 4.5.1 in the book for a summary including the forms of the solutions assumed given the structure of the nonhomogeneity.

We will soon cover the method called Variation of Parameters, which works in the cases MOUC fails.



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