Differential Equations - Second Order: Non-homogeneous

t’s now time to start thinking about how to solve non-homogeneous differential equations. A second order, linear non-homogeneous differential equation is

\begin{matrix}  \end{matrix}

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{matrix}  \end{matrix}

the associated homogeneous differential equation to

(1)

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

Theorem

Suppose that

Y_{1} (t)

and

Y_{2} (t)

are two solutions to

(1)

and that

y_{1} (t)

and

y_{2} (t)

are a fundamental set of solutions to the associated homogeneous differential equation

(2)

then,

Y_{1} (t) - Y_{2} (t)

is a solution to

(2)

and it can be written as

Y_{1} (t) - Y_{2} (t) = c_{1} y_{1} (t) + c_{2} y_{2} (t)

Note the notation used here. Capital letters referred to solutions to

(1)

while lower case letters referred to solutions to

(2)

. This is a fairly common convention when dealing with non-homogeneous 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

(2)

all we need to do is plug this into the differential equation and check it.

\begin{aligned} {(Y_{1} - Y_{2})}^{''} + p (t) {(Y_{1} - Y_{2})}^{'} + q (t) (Y_{1} - Y_{2}) & = 0 \\ {Y_{1}}^{''} + p (t) {Y_{1}}^{'} + q (t) Y_{1} - ({Y_{2}}^{''} + p (t) {Y_{2}}^{'} + q (t) Y_{2}) & = 0 \\ g (t) - g (t) & = 0 \\ 0 & = 0 \end{aligned}

We used the fact that

Y_{1} (t)

and

Y_{2} (t)

are two solutions to

(1)

in the third step. Because they are solutions to

(1)

we know that

\begin{aligned} {Y_{1}}^{''} + p (t) {Y_{1}}^{'} + q (t) Y_{1} & = g (t) \\ {Y_{2}}^{''} + p (t) {Y_{2}}^{'} + q (t) Y_{2} & = g (t) \end{aligned}

So, we were able to prove that the difference of the two solutions is a solution to

(2)

Proving that

Y_{1} (t) - Y_{2} (t) = c_{1} y_{1} (t) + c_{2} y_{2} (t)

is even easier. Since

y_{1} (t)

and

y_{2} (t)

are a fundamental set of solutions to

(2)

we know that they form a general solution and so any solution to

(2)

can be written in the form

y (t) = c_{1} y_{1} (t) + c_{2} y_{2} (t)

Well,

Y_{1} (t) - Y_{2} (t)

is a solution to

(2)

, as we’ve shown above, therefore it can be written as

Y_{1} (t) - Y_{2} (t) = c_{1} y_{1} (t) + c_{2} y_{2} (t)

So, what does this theorem do for us? We can use this theorem to write down the form of the general solution to

(1)

. Let’s suppose that

y (t)

is the general solution to

(1)

and that

Y_{P} (t)

is any solution to

(1)

that we can get our hands on. Then using the second part of our theorem we know that

y (t) - Y_{P} (t) = c_{1} y_{1} (t) + c_{2} y_{2} (t)

where

y_{1} (t)

and

y_{2} (t)

are a fundamental set of solutions for

(2)

. Solving for

y (t)

gives,

y (t) = c_{1} y_{1} (t) + c_{2} y_{2} (t) + Y_{P} (t)

We will call

y_{c} (t) = c_{1} y_{1} (t) + c_{2} y_{2} (t)

the complementary solution and

Y_{P} (t)

a particular solution. The general solution to a differential equation can then be written as.

y (t) = y_{c} (t) + Y_{P} (t)

So, to solve a non-homogeneous differential equation, we will need to solve the homogeneous differential equation,

(2)

, which for constant coefficient differential equations is pretty easy to do, and we’ll need a solution to

(1)

This seems to be a circular argument. In order to write down a solution to

(1)

we need a solution. However, this isn’t the problem that it seems to be. There are ways to find a solution to

(1)

. 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 non-homogeneous 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.

Comments

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Calculus III - 3-Dimensional Space: Equations of Lines