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Chris L T521 · #6 May 24th, 2008, 09:24 PM

Second Order NON-Homogeneous Differential Equations

Ah...the dreadful second order non-homogeneous differential equation has the form:

$a\frac{{d^2 y}}{{dx^2 }} + b\frac{{dy}}{{dx}} + cy = f\left( x\right)$

I will go through three techniques on how to solve these nasties:

Method of Undetermined Coefficients
The "Annihilator" Method (Similar to #1)
Variation of Parameters

Technique 1 : Method of Undetermined Coefficients:

In the case we have a differential equation like $a\frac{{d^2 y}}{{dx^2 }} + b\frac{{dy}}{{dx}} + cy = f\left( x\right)$ , we can guess what the particular solution to a DE may be, depending on what $f(x)$ is. For example, let us say that $f(x)=3x+7$ . We would assume that a particular solution to the DE would be $y_p=Ax+B$ . To find the Undetermined Coefficients, plug $y_p$ back into the original DE.

If $f(x)=5e^{-3x}$ , we assume that the particular solution to the DE would have the form of $y_p=Ae^{-3x}$ . We too would substitute $Ae^{-3x}$ into the DE to find the unknown coefficient value.

If $f(x)=3\cos(x)$ , we assume that the particular solution to the DE would have the form of $y_p=A\cos(x)+B\sin(x)$ . Again, to find the unknown coefficients, substitute $y_p$ into the original DE.

Sometimes, the guess of $y_p$ isn't that obvious. Try to think outside the box when solving these problems!

Note that the solution to the non-homogeneous DE is a linear combination of the complimentary solution (solution to the homogeneous equation) and the particular solution (solution to the non-homogeneous equation)

Example 16:

Solve $y'' - 3y' + 2y = 3e^{ - x} - 10\cos \left( {3x} \right)$ ; $y(0)=1$ ; $y'(0)=2$ .

Solve the homogeneous equation $y'' - 3y' + 2y = 0$ first.

$y''-3y'+2y=0 \implies r^2-3r+2=0 \implies r=2 \ \text{or} \ r=1$ .

Thus, $y_c=c_1e^{x}+c_2e^{2x}$ .

Now solve the non-homogeneous equation.

$y''-3y'+2y=3e^{ - x} - 10\cos \left( {3x} \right)$ .

Using the method of Undetermined Coefficients, we guess and assume that the particular solution will have the form:

$y_p=Ae^{-x}+B\cos(3x)+C\sin(3x)$

$\therefore y_{p}^{/}=-Ae^{-x}-3B\sin(3x)+3C\cos(3x)$

$\therefore y_{p}^{//}=Ae^{-x}-9B\cos(3x)-9C\sin(3x)$ .

Substituting these values into the original DE, we get:

$(Ae^{-x}-9B\cos(3x)-9C\sin(3x))-3(-Ae^{-x}-3B\sin(3x)+3C\cos(3x))+2(Ae^{-x}+$
$B\cos(3x)+C\sin(3x))=3e^{ - x} - 10\cos \left( {3x} \right)$

$\implies 6Ae^{-x}+(-7B-9C)\cos(3x)+(9B-7C)\sin(3x)=3e^{-x}-10\cos(3x)$

Now compare the coefficients (like in Partial Fractions)

$\begin{aligned}6A&=3 \\7B-9C&=-10 \\9B-7C&=0\end{aligned}$

Solving for the constants, we get $A=\frac{1}{2}$ , $B=\frac{7}{13}$ and $C=\frac{9}{13}$ (verify).

Thus, $y_p=\frac{1}{2}e^{-x}+\frac{1}{13}\left(7\cos(3x)+9\sin(3x)\right)$ .

Therefore, the solution is

$y=c_1e^{x}+c_2e^{2x}+\frac{1}{2}e^{-x}+\frac{1}{13}\left(7\cos(3x)+9\sin(3x)\right)$ .

Find $y^{/}$ so we can apply the initial conditions:

$y^{/}=c_1e^{x}+2c_2e^{2x}-\frac{1}{2}e^{-x}+\frac{1}{13}\left(21\cos(3x)+27\sin(3x)\right)$ .

Apply the initial conditions.

$y(0)=1=c_1+c_2+\frac{1}{2}+\frac{7}{13}$
$y^{/}(0)=2=c_1+2c_2-\frac{1}{2}+\frac{21}{13}$ .

Solving for $c_1 \ \text{and} \ c_2$ , get $c_1=-\frac{1}{2} \ \text{and} \ c_2=\frac{6}{13}$ (verify)

Therefore, the solution to the initial value problem is:

$\color{red}\boxed{y=-\frac{1}{2}e^{x}+\frac{6}{13}e^{2x}+\frac{1}{2}e^{-x}+\frac{1}{13}\left(7\cos(3x)+9\sin(3x)\right)}$ .

I will discuss the next two sometime later...

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