Answer to Question #199275 in Differential Equations for Didues

Question

(D² - 4) y= xe² , xe^x is y_p = (ax + b) e^x

Answer 1

$k^2-4=0$

$k_{1}=-2,k_2=2$

$y=c_1e^{-2x}+c_2e^{2x}$

$y_p=(ax^2+bx)e^{2x}$

$y'_p=(2ax+b)e^{2x}+2(ax^2+bx)e^{2x}=(2ax^2+(2a+2b)x+b)e^{2x}$

$y''_p=(4ax+2a+2b)e^{2x}+2(2ax^2+(2a+2b)x+b)e^{2x}$

$4ax^2+8ax+4bx+2a+4b-4ax^2-4bx=x$

$a=1/8,b=-1/16$

$y=c_1e^{-2x}+c_2e^{2x}+(x^2/8-x/16)e^{2x}$