Answer to Question #286397 in Differential Equations for Uma

Question

d^2y/dx^2+4dy/dx-4y=sin3x

Answer 1

$y''+4y'-4y=\sin3x$

solution find in form

$y=y_1+y_2$

then

$1) y''+4y'-4y=0\\ \lambda^2+4\lambda-4=0\\ \lambda_1=-2-2\sqrt{2}, \lambda_2=-2+2\sqrt{2}\\ y_1=c_1e^{(-2-2\sqrt{2})x}+c_2e^{(-2+2\sqrt{2})x}\\ 2) y_2=a\cos3x+b\sin3x\\ y_2'=-3a\sin3x+3b\cos3x\\ y_2''=-9a\cos3x-9b\sin3x\\$

Plug in the equation

$-9a\cos3x-9b\sin3x-12a\sin3x+\\ +12b\cos3x-4a\cos3x-4b\sin3x=\sin3x\\ \cos3x: -9a+12b-4a=0\\ \sin3x: -9b-12a-4b=1\\ a=\frac{12b}{13}\\ -13b-12a=1, -13b-\frac{144b}{13}=1, -313b=13, \\ b=-\frac{13}{313}, a=-\frac{12}{313}\\ y_2=-\frac{12}{313}\cos3x-\frac{13}{313}\sin3x\\$

Answer:

$y=c_1e^{(-2-2\sqrt{2})x}+c_2e^{(-2+2\sqrt{2})x}-\\ -\frac{12}{313}\cos3x-\frac{13}{313}\sin3x$