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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

\[\begin{equation}y'' + p\left( t \right)y' + q\left( t \right)y = g\left( t \right)\label{eq:eq1}\end{equation}\]

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

\[\begin{equation}y'' + p\left( t \right)y' + q\left( t \right)y = 0\label{eq:eq2}\end{equation}\]

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.