Question

Find the solution for the given conditions: y''+3y'+2y = cosx. When x=0, y=0 and y'=0. How do you find out the particular integral?

Answer 1

Is this a differential equations class?

Answer 2

use the method of annihilators:
You need a 0 on the right side, therefore you need to multiply both sides by something that will give you 0 on the left.

Answer 3

So the characteristic equation is:
r^2 + 3r + 2 = Cosx,
How do you annihilate cosine.. it's the same as the solution or whatever you want to call it as {e^(ix), e^(-ix)}
or r^2 + 1 = 0
So now multiply both sides by (r^2+1 )

Answer 4

(r^2+1)(r^2+3r+2)=0
we already now the solutions to r^2+1 = 0, r1 = i, r2 = -i, now solve r^2 + 3r + 2 =0, you will get: r = -2 and r =-1
So your solutions so far are:
r = i, r=-i, r = -2, r = -1
{e^(ix), e^(-ix), e^(-2x), e^(-x)}
{Cosx, Sinx, e^(-2x), e^(-x)}
y = ACosx + BSinx + Ce^(-2x)+De^(-x).
Find y' and, plug in x=0, y=0, y'=0 and solve for constants.