method for solving y''-2y'-3y=x^2

Question

Hello all,

Is it correct method:

First: solve y''-2y'-3y=0

use m^2-2m-3=0 -> m=2 or -3

then y_h = c1*e^2x + c2*x*e^-3x

and now hw can I find y_p and general answer as y?

Thank you

Answer 1

The characteristic equation for y"-2y'-3y = x² is (D²-2D-3)y = x² or (D-3)(D+1) = x²

So the general solution is y = Ae^-x + Be^3x + y_p

The particular solution can be found in a number of ways including repeated integrating factor for 2 first order equations, using Laplace Transforms, or other short methods described in textbooks etc,

Keep in mind the particular solution is the only solution if y(0) = y'(0) = 0

I used Laplace Transforms but then you have to deduce partial fractions but it can be done. In this case the particular solution y_p = (-14/27) + (4/9)x + (-1/3)x² + (1/2)e^-x + (1/54)e^3x