Solving second order non homogeneous differential equation

[the0]

Homework Statement

The problem is to solve:

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

Homework Equations

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!

 

[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?

 

[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?

 

[LCKurtz]

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?

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?

 

[the0]

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

 

[LCKurtz]

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

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

 

[DryRun]

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!

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})<br /> \\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)<br /> \\=e^x\frac{1}{(D^2+5)}cos^{2}(x)<br /> \\=e^x\frac{1}{(D^2+5)}(\frac{1}{2}+\frac{1}{2} \cos 2x)<br /> \\=e^x\frac{1}{(D^2+5)}(\frac{1}{2})+e^x\frac{1}{(D^2+5)}(\frac{1}{2} \cos 2x)<br /> \\=e^x(\frac{1}{10})+e^x(\frac{1}{2} \cos 2x)<br /> \\=e^x(\frac{1}{10}+\frac{1}{2} \cos 2x)<br />
Now, find y_2. Applying the Shift theorem again, you'll get:
y_2=e^x\frac{1}{(D^2+5)}x^2<br /> \\=e^x(\frac{1}{5}-\frac{D^2}{25})x^2<br /> \\=e^x(\frac{1}{5}x^2-\frac{2}{25})<br />
Therefore, y_p=e^x(\frac{1}{10}+\frac{1}{2} \cos 2x)+e^x(\frac{1}{5}x^2-\frac{2}{25})<br /> \\=\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.