1. ## Nonhomogenious Equation

I am a bit confused on how to solve the following using the method of undetermined coefficients:

equation: y'' + y - 2y = x + sin2x

So what i have done so far is:
roots are 1, -2 so
Yc = Ae^(x) + Be^(-2x)

But from the homogenious equation above i dont know how to go on. can someone help me solve this?

Thanks

2. Originally Posted by taurus
I am a bit confused on how to solve the following using the method of undetermined coefficients:

equation: y'' + y - 2y = x + sin2x

So what i have done so far is:
roots are 1, -2 so
Yc = Ae^(x) + Be^(-2x)

But from the homogenious equation above i dont know how to go on. can someone help me solve this?

Thanks
the particular solution will be of the form
$y_p(x)=Ax^2+Bx+C+D\sin(2x)+E\cos(2x)$

$y'_p=2Ax+B+2D\cos(2x)-2E\sin(2x)$

$y''_p=2A-4D\sin(2x)-4E\cos(2x)$

just sub these back into the non homogeonious equation and solve for the constants.

I hope this helps.

3. Dont quite get it, which equation to you sub back?
and is A=1, B, 1, C=-2 So whats D and E?

thanks

4. Originally Posted by taurus
Dont quite get it, which equation to you sub back?
and is A=1, B, 1, C=-2 So whats D and E?

thanks
y'' + y - 2y = x + sin2x

$y_p=D\sin(2x)+E\cos(2x)$

$y_p=2D\cos(2x)-2E\sin(2x)$

$y_p=-4D\sin(2x)-4E\cos(2x)$

$=-4D\sin(2x)-4E\cos(2x)+2D\cos(2x)-2E\sin(2x)-2[D\sin(2x)+E\cos(2x)]$
$=\sin(2x)$

$[-6D-2E]\sin(2x)+[-6E+2D]\cos(2x)=\sin(2x)$

so we get

$-6D-2E=1$
$2D-6E=0$

Solve this system gives $E=-\frac{1}{20},D=-\frac{3}{20}$

I hope this clears it up.

5. Wait so now i put what back into the nonhomogenious equation?

so it becomes:
y = e^x + e^(-2x) + x^2 + -2 + (-3/20)sin(2x) + (-1/20)cos(2x) ?
Is this correct?

6. Actually I think its this:

y = Ae^x + Be^(-2x) + x^2 + -2 + (-3/20)sin(2x) + (-1/20)cos(2x)

and then i can solve for A and B using y(0)=1 and y'(0)=0

is this correct?