Answer to Question #149013 in Differential Equations for Nikhil

Question

Solve the differential equation
d^2y/dx^2-2dy/dx+y=x^2e^2x

Answer 1

For the given differential equation, A.E. is given by,

$D^2-2D+1=0$

Solving it, we get, $D=1,1$

Hence, $y=ae^{x}+bxe^{x}$ (1)

P. I. $\frac{1}{D^2-2D+1}x^2e^{2x}=\frac{1}{(D-1)^2}x^2e^{2x}= e^{2x}\frac{1}{(D+2-1)^2}x^2$

$=e^{2x}\frac{1}{(D+1)^2}x^2= e^{2x}(D+1)^{-2}x^2$

$e^{2x}(1-2D+3D^3+.....)x^2 = e^{2x}(x^2-4x+6)$ (2)

So total solution for the equation is,

$y=ae^{x}+bxe^{x}+e^{2x}(x^2-4x+6)$