How Can I Solve This Second-Order Differential Equation?

[Exulus]

Could anyone help me with this? I need to solve this equation:

$\frac{d^2y}{dx^2} + 4y = \sin{2x}$

Everything I seem to try for y ends up canceling itself out, as the second diffferential is always the negative of what you begin with, and the 2x part of the sin means you get a factor of 4 in the second differential, which then cancels with the 4y. Eg:

Trying $y = c \sin{2x}$

$\frac{dy}{dx} = 2c \cos{2x}$
$\frac{d^2y}{dx^2} = -4c \sin{2x}$

Substituting these into the LHS of my question gives:

$-4c \sin{2x} + 4c \sin{2x} = 0$

The only thing i can think of now is that its a complex solution..but we haven't been taught that yet!

Any help appreciated :(

(ps no idea why the itex stuff is coming out so small..sorry about that).

 

[estalniath]

You'll have to find the homogeneous solution first before applying the method of undetermined coefficients. Let (D^2+4)y=r^2+4=0. hence r=2i,-2i
Thus y(homogenous)=Acos2x+Bsin2x
Therefore, let the trial function be y=AxSin2x+BxCos2x
You should get the answer from here. =)

 

[estalniath]

*Ammendments:Trial solution=CxSin2x+DxCos2x

 

[Exulus]

Hi,

I tried your trial solution but it still comes out as 0 :(

$y = Cx\sin{2x} + Dx\cos{2x}$

$\frac{dy}{dx} = 2Cx\cos{2x} + C\sin{2x} - 2Dx\sin{2x} + D\cos{2x}$

$\frac{d^2y}{dx^2} = -4Cx\sin{2x} + 2C\cos{2x} - 2C\cos{2x} - 4Dx\cos{2x} - 2D\sin{2x} - 2D\sin{2x} =$
$= -4x(C\sin{2x} + D\cos{2x}) - 4D\sin{2x}$

Subbing back into question:

$-4x(C\sin{2x} + D\cos{2x}) + 4Cx\sin{2x} + 4Dx\cos{2x} = 0$

:(

 

[dextercioby]

Try Lagrange's method (of constant variation)

Assume the inhomogenous solution

$$y_{p}=C(x)\sin 2x+D(x)\cos 2x$$

And solve for the unknown functions,by plugging it in the original ODE.

Daniel.

 

[Exulus]

doh, it looks like i did my sums wrong i think. I've now got $y = -\frac{1}{4}\cos{2x}$ which looks like the right answer (haven't tested yet).

Cheers for the help guys! :)

 

[Palindrom]

If you multiplicate it by x I'll believe you...

 

[dextercioby]

Anyways,there's an alternative to 'trial solutions'.Lagrange's method is suitable for these problems.

Daniel.

 

[estalniath]

After subbing the values of d^2y/dx^2,dy/dx and y into the differential equation, the RHS of the equation should be sin2x, then you'll have to do comparing of the coefficients to determine the value of the constants =)

 

[Exulus]

If you multiplicate it by x I'll believe you...

Sorry yeah i typed it wrong, whoops. I did actually have it with the x written down..honest ;)

Thanks for the other method as well, dextercioby :)