Someone solve this for me

[Rob]

Please! I have this inhomogenous second order differential equation and I keep mucking it up. You mathemagicians are such clever people... a hint would be useful!

d^2y/dx^2 + 4y = 2sin(2x)

Thank you.

[Ben]

The solution of the homogeneous solution is:
y=Asin(2x)+Bcos(2x) === A*cos(2x+B) ((although be careful to note that they're not the same A and B))
then the particular solution is done by taking y=C*x*sin(2x)+D*x*cos(2x) and plugging it into the ODE and equating LHS and RHS.
dy/dx = C(sin(2x)+2x*cos(2x))+D(cos(2x)-2x*sin(2x))
d(dy/dx)/dx = C(cos(2x)+2(cos(2x)-2x*sin(2x))-D(sin(2x)+2(sin(2x)+2x*cos(2x)))

trying to solve that horrific algebraic mess isn't fun. However,
(Dy)^2=C*cos(2x)+2*C*cos(2x)-4*C*x*sin(2x)-D*sin(2x)-2*D*sin(2x)-4*D*x*cos(2x)
(Dy)^2=Cos(2x)[C+2C-4Dx]-sin(2x)[D+2D-4Cx]

So then we have to add in the y term, giving us

LHS = Cos(2x)[3C-4Dx]-Sin(2x)[3D-4Cx]+Cxsin(2x)+Dxcos(2x)
which simplifies a bit to
Cos(2x)[3C-3Dx]-Sin(2x)[3D-5Cx]

So we have:
Cos(2x)[3C-3Dx]-Sin(2x)[3D-5Cx] = 2Sin(2x)

Then i forget what to do This is probably all wrong anyway. Arse...

[Rob]

Yeah, I got something similar... the problem is, that when equating the coefficients C comes out as a number AND zero and so does D. I hate my fucking maths lecture person. He likes to test us on how we deal with the longest line of simple algebra as possible. The only joy he has brought to me is that his name sounds exactly the same as the noise Mario and Wario make on Mario Party...

GABOOOOORRR!

[Ben]

haha, That's not his first name is it? because we have a lecturer called Gabor Megyesi. Is your lecturer bald, wears glasses, russian, and NASAL?

[oasis2000]

As I explained to you I can't solve this, but in this website: www.sosmath.com/CBB/index.php?sid=2b1a6584a2345baeadfb952620849417 there is people that I'm sure they can

[Ben]

Oh, i've solved it, using a maths package, but I just wanted to know how to get the answer, i suppose.

[Rob]

For those interested (i.e. Ben), I think this is the answer. As ever, it was silly maths mistakes that put me into the wrong to start with...

y'' + 4y = 2sin(2x)

Complementary function:

y = Acos(2x) + Bsin(2x)

Particular integral:

y = Cxcos(2x) + Dxsin(2x)

y'' = -2Csin(2x) - 2Csin(2x) - 4Cxcos(2x) + 2Dcos(2x) + 2Dcos(2x) -4Dxsin(2x)

Now, substituting these back into the original equation and equating the coefficents, we get

C = -0.5 and D = 0

So the particular integral is

y = -0.5xcos(2x)

And the general solution is this plus the complementary function

y = -0.5cos(2x) + Acos(2x) + Bsin(2x)

Gabor makes life so boring. I had to take a 4x4 matrix to the power of three by hand too, and lots more boring, long things...

[Ben]

haha. the method i was about to suggest for making life simpler involved just as many calculations, so never mind. But if you decomposed A=PDP^T, where P was orthogonal and D was diagonal, then A³=PD³P^T, which is pretty cool.

Hmm, i must have made a mistake as well then. Buggery. Still, never mind, i knew the technique and that's all that I needed!