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Home / Differential Equations / Second Order DE's / Nonhomogeneous Differential Equations
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Section 3-8 : Nonhomogeneous Differential Equations

It’s now time to start thinking about how to solve nonhomogeneous differential equations. A second order, linear nonhomogeneous differential equation is

$$$y'' + p\left( t \right)y' + q\left( t \right)y = g\left( t \right)\label{eq:eq1}$$$

where $$g(t)$$ is a non-zero function. Note that we didn’t go with constant coefficients here because everything that we’re going to do in this section doesn’t require it. Also, we’re using a coefficient of 1 on the second derivative just to make some of the work a little easier to write down. It is not required to be a 1.

Before talking about how to solve one of these we need to get some basics out of the way, which is the point of this section.

First, we will call

$$$y'' + p\left( t \right)y' + q\left( t \right)y = 0\label{eq:eq2}$$$

the associated homogeneous differential equation to $$\eqref{eq:eq1}$$.

Now, let’s take a look at the following theorem.

Theorem

Suppose that $$Y_{1}(t)$$ and $$Y_{2}(t)$$ are two solutions to $$\eqref{eq:eq1}$$ and that $$y_{1}(t)$$ and $$y_{2}(t)$$ are a fundamental set of solutions to the associated homogeneous differential equation $$\eqref{eq:eq2}$$ then,

${Y_1}\left( t \right) - {Y_2}\left( t \right)$

is a solution to $$\eqref{eq:eq2}$$ and it can be written as

${Y_1}\left( t \right) - {Y_2}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)$

Note the notation used here. Capital letters referred to solutions to $$\eqref{eq:eq1}$$ while lower case letters referred to solutions to $$\eqref{eq:eq2}$$. This is a fairly common convention when dealing with nonhomogeneous differential equations.

This theorem is easy enough to prove so let’s do that. To prove that $$Y_{1}(t) - Y_{2}(t)$$ is a solution to $$\eqref{eq:eq2}$$ all we need to do is plug this into the differential equation and check it.

\begin{align*}{\left( {{Y_1} - {Y_2}} \right)^{\prime \prime }} + p\left( t \right){\left( {{Y_1} - {Y_2}} \right)^\prime } + q\left( t \right)\left( {{Y_1} - {Y_2}} \right) & = 0\\ {Y_1}^{\prime \prime } + p\left( t \right){Y_1}^\prime + q\left( t \right){Y_1} - \left( {{Y_2}^{\prime \prime } + p\left( t \right){Y_2}^\prime + q\left( t \right){Y_2}} \right) & = 0\\ g\left( t \right) - g\left( t \right) & = 0\\ 0 & = 0\end{align*}

We used the fact that $$Y_{1}(t)$$ and $$Y_{2}(t)$$ are two solutions to $$\eqref{eq:eq1}$$ in the third step. Because they are solutions to $$\eqref{eq:eq1}$$ we know that

\begin{align*}{Y_1}^{\prime \prime } + p\left( t \right){Y_1}^\prime + q\left( t \right){Y_1} & = g\left( t \right)\\ {Y_2}^{\prime \prime } + p\left( t \right){Y_2}^\prime + q\left( t \right){Y_2} & = g\left( t \right)\end{align*}

So, we were able to prove that the difference of the two solutions is a solution to $$\eqref{eq:eq2}$$.

Proving that

${Y_1}\left( t \right) - {Y_2}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)$

is even easier. Since $$y_{1}(t)$$ and $$y_{2}(t)$$ are a fundamental set of solutions to $$\eqref{eq:eq2}$$ we know that they form a general solution and so any solution to $$\eqref{eq:eq2}$$ can be written in the form

$y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)$

Well, $$Y_{1}(t) - Y_{2}(t)$$ is a solution to $$\eqref{eq:eq2}$$, as we’ve shown above, therefore it can be written as

${Y_1}\left( t \right) - {Y_2}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)$

So, what does this theorem do for us? We can use this theorem to write down the form of the general solution to $$\eqref{eq:eq1}$$. Let’s suppose that $$y(t)$$ is the general solution to $$\eqref{eq:eq1}$$ and that $$Y_{P}(t)$$ is any solution to $$\eqref{eq:eq1}$$ that we can get our hands on. Then using the second part of our theorem we know that

$y\left( t \right) - {Y_P}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)$

where $$y_{1}(t)$$ and $$y_{2}(t)$$ are a fundamental set of solutions for $$\eqref{eq:eq2}$$. Solving for $$y(t)$$ gives,

$y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + {Y_P}\left( t \right)$

We will call

${y_c}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)$

the complementary solution and $$Y_{P}(t)$$ a particular solution. The general solution to a differential equation can then be written as.

$y\left( t \right) = {y_c}\left( t \right) + {Y_P}\left( t \right)$

So, to solve a nonhomogeneous differential equation, we will need to solve the homogeneous differential equation, $$\eqref{eq:eq2}$$, which for constant coefficient differential equations is pretty easy to do, and we’ll need a solution to $$\eqref{eq:eq1}$$.

This seems to be a circular argument. In order to write down a solution to $$\eqref{eq:eq1}$$ we need a solution. However, this isn’t the problem that it seems to be. There are ways to find a solution to $$\eqref{eq:eq1}$$. They just won’t, in general, be the general solution. In fact, the next two sections are devoted to exactly that, finding a particular solution to a nonhomogeneous differential equation.

There are two common methods for finding particular solutions : Undetermined Coefficients and Variation of Parameters. Both have their advantages and disadvantages as you will see in the next couple of sections.