# non-homogeneous differential equations

• Aug 6th 2009, 07:06 AM
Jonny-123
non-homogeneous differential equations
i have the following differential equation:

y''(x)+4y'(x)+20y(x)=10e^(-2x)*cos(rx)

i know that the solution = general solution + particular solution.

i know how to find the homogeneous solution but i don't know how to find the particular solution. what is the form of the particular solution?
• Aug 6th 2009, 07:19 AM
Calculus26
I assume you mean you know how to find the homogeneous Solution?

Other wise The general solution is the sum of the homogeneous and particular solution.

If you have the general solution separate out the homogeneous solution and what's left is the particular solution.

Another way is if you use variation of parameters or reduction of order
omit the integration constants and you will obtain just the pariticular solution.

If you use undetermined coefficients you have just the particular solution.

in your problem the characeristic eqn is L^2 + 4L + 20 = 0

the solution is L = -2 + 4 i
the homogeneous solution consists of

y1 = e^(-2x)cos(4x) and y2 =e^(-2x)sin(4x)

Now you can use variation of parameters to find the general solution.
• Aug 6th 2009, 07:24 AM
Jonny-123
sorry my mistake, i meant to say homogeneous solution
• Aug 6th 2009, 07:35 AM
Calculus26
I'd use variation of parameters to find the particular solution on this one.
• Aug 6th 2009, 04:26 PM
HallsofIvy
I would NOT use "variation of parameters"- too difficult.

I would use "undetermined coefficients".

Your original post has "= $10e^{-2x}cos(rt)$. Was that "r" a mistype for 4 or actually in the problem.

If it is actually "r" (and perhaps the point is to show what happens when r becomes 4), try a solution of the form $y= e^{-2x}(A cos(rt)+ B sin(rt)$. Find the first and second derivatives of that (not integrals), put them into the equation and solve for A and B. Interesting question: what happens to that solution as r goes to 4?

If it was supposed to be 4 then you will need to try $y= xe^{-2x}(A cos(4t)+ B sin(4t))$.
• Aug 6th 2009, 05:35 PM
Calculus26
Don't really see the difficulty of using variation of parameters:

y= y1z1 + y2z2

W(y1,y2) = 4*e^(-4*x)

z1 ' = -e^(-2x)sin(4x)*10e^(-2*x)cos(rx)/4e^(-4x)

= -5sin(4x)cos(rx)/2

z2 ' = e^(-2x)cos(4x)*10e^(-2*x)cos(rx)/4e^(-4x)

= 5cos(4x)cos(rx)/2

to me this is easier than considering separate cases especially
when r = 4 and you have to deal with

x*e^(-2x)[Acos(4x) + Bsin(4x)]

But thats a matter of personal taste