# Solving second order non homogeneous differential equation

1. Apr 22, 2012

### the0

1. The problem statement, all variables and given/known data

The problem is to solve:

$y''-2y'+5y = e^{x}(cos^{2}(x)+x^{2})$

2. Relevant equations

3. The attempt at a solution

I (think I) have solved the associated homogeneous equation:

$y''-2y'+5y = 0$

giving the solution as:

$y_{h} = e^{x}(C_{1}cos(2x)+C_{2}sin(2x))$
(This could of course be rearranged using the trig identities.... would that help?)

And now I don't know how to go about finding a particular solution, could somebody please point me in the right direction as to which method I could use?
Thanks a lot!

2. Apr 22, 2012

### LCKurtz

I would suggest the method of finding a particular solution with the method of Undetermined Coefficients, more formally known as the method of annihilators. Have you studied that?

3. Apr 22, 2012

### the0

Yes I have studied this method (a little) however I am struggling to choose the right form for the particular solution:

Should $y_{p} = Asin(2x)+Bcos(2x)+e^{x}(C+Dx+Ex^{2})$ work?

4. Apr 22, 2012

### LCKurtz

That $\cos^2x$ on the right side of your DE is $\frac{1+\cos(2x)} 2$ so you basically have a term $e^x(\frac{1+\cos(2x)} 2)$ in the NH term. So there's no reason to expect $\sin(2x)$ and $\cos(2x)$ to work. You would want to multiply them by $e^x$. But that is in the complementary solution. So how do you fix that?

5. Apr 22, 2012

### the0

would simply multiplying by $x$ solve this?
i.e.
$y_{p} = e^{x}(Axsin(2x)+Bxcos(2x)+C+Dx+Ex^{2})$

6. Apr 22, 2012

### LCKurtz

Well, that's the standard fix for that situation. I think that is a good trial for the particular solution. Only way to be sure you haven't overlooked something is to go ahead and try it

7. Apr 22, 2012

### sharks

How about the method of inverted operator techniques? (which i personally find easier). Although some people use the Wronskian method, i think.

The D operator is simply $\frac{d}{dx}$.
Expressing first, the L.H.S. in terms of D: $L(D)=D^2-2D+5$, where L(D) is a function of D.
$$L(D)y_p=e^x(cos^{2}(x)+x^{2})$$
So, $$y_p=\frac{1}{L(D)}(e^xcos^{2}(x)+e^xx^{2}) \\y_p=\frac{1}{L(D)}e^xcos^{2}(x)+\frac{1}{L(D)}e^xx^{2}$$
Here, $y_p=y_1+y_2$,
$$y_1=\frac{1}{L(D)}e^xcos^{2}(x)$$
$$y_2=\frac{1}{L(D)}e^xx^{2}$$
Now, you have to use the Shift theorem to evaluate both $y_1$ and $y_2$.
The Shift theorem simply shifts D to (D+a) and simultaneously separates $e^x$ from $(cos^2x+x^2)$.

To find $y_1$:
$$y_1=\frac{1}{L(D)}e^xcos^{2}(x)=\frac{1}{(D^2-2D+5)}e^xcos^{2}(x)$$
Applying the Shift theorem:
$$y_1=e^x\frac{1}{((D+1)^2-2(D+1)+5)}cos^{2}(x) \\=e^x\frac{1}{(D^2+5)}cos^{2}(x) \\=e^x\frac{1}{(D^2+5)}(\frac{1}{2}+\frac{1}{2} \cos 2x) \\=e^x\frac{1}{(D^2+5)}(\frac{1}{2})+e^x\frac{1}{(D^2+5)}(\frac{1}{2} \cos 2x) \\=e^x(\frac{1}{10})+e^x(\frac{1}{2} \cos 2x) \\=e^x(\frac{1}{10}+\frac{1}{2} \cos 2x)$$
Now, find $y_2$. Applying the Shift theorem again, you'll get:
$$y_2=e^x\frac{1}{(D^2+5)}x^2 \\=e^x(\frac{1}{5}-\frac{D^2}{25})x^2 \\=e^x(\frac{1}{5}x^2-\frac{2}{25})$$
Therefore, $$y_p=e^x(\frac{1}{10}+\frac{1}{2} \cos 2x)+e^x(\frac{1}{5}x^2-\frac{2}{25}) \\=\frac{1}{50}e^x+\frac{1}{5}e^xx^2+ \frac {1}{2} e^x \cos 2x$$
The final answer (the general solution) is: $y_c+y_p$ where $y_c$ is the complementary function that you've already obtained in your first post.

Last edited: Apr 22, 2012