Answer to Question #187920 in Differential Equations for Uttam

Question

Find the general solution of y"-y'-2y=4x²

Answer 1

Solution

To get general solution of homogeneous equation y₀"-y₀'-2y₀=0 let’s solve characteristic equation r²-r-2=0.

r1,2 = (1±√(1+4*2))/2 = (1±3)/2 => r₁ = -1, r₂ = 2

Therefore y₀(x) = C₁e^-x+C₂e^2x    

Partial solution y₁(x) may be find in the form of polynomial y₁(x) = Ax²+Bx+C. Substituting into equation we will get

2A-2Ax-B-2Ax²-2Bx-2C = 4x²  =>  

2A-B-2C = 0; -2A-2B = 0; -2A = 4  => 

A = -2; B = 2; C = -3

So y₁(x) = -2x²+2x-3

General solution of the given equation is

y(x) = y₀(x)+ y₁(x) =  C₁e^-x+C₂e^2x -2x²+2x-3

Answer

y(x) = C₁e^-x+C₂e^2x -2x²+2x-3