Question

Using the method of undetermined coefficients, write the trial solution of the equation

d^2y/dx^2 + 2dy/dx +5y= xe^-x cos2x


and hence solve it.

Answer 1

$\frac{d^2y}{dx^2}+2\frac{dy}{dx}+5y=xe^{-x} \cos2x$

Solution of equation can be found in the form

$y=y_1+y_2$

Solution $y_1$ can be found from equation

$y''+2y'+5y=0$

$k^2+2k+5=0\\ D=4-20=-16=16i^2\\ k_1=\frac{-2-4i}{2}=-1-2i\\ k_2=\frac{-2+4i}{2}=-1+2i\\ y_1=e^{-x}(C_1\cos2x+C_2\sin2x)$

Solution $y_2$ can be found in form

$y_2=xe^{-x}((ax+b)\cos2x+(cx+m)\sin2x)=\\ =e^{-x}((ax^2+bx)\cos2x+(cx^2+mx)\sin2x))\\ y_2' =e^{-x}\cos2x(x^2(-a+2c)+\\+x(2a-b+2m)+b)+\\ +e^{-x}\sin2x(x^2(-2a-c)+\\+x(-2b+2c-m)+m)\\ y_2'' =e^{-x}\cos2x(x^2(-3a-4c)+\\+x(-4a-3b-4m+8c)-2b+2a+4m)+\\ +e^{-x}\sin2x(x^2(4a-3c)+\\+x(-8a+4b-3m-4c)-4b-2m+2c)\\$

Plug $y_2, y_2', y_2''$ into the equation, then

$e^{-x}\cos2x(x^2(-3a-4c)+\\+x(-4a-3b-4m+8c)-2b+2a+4m)+\\ +e^{-x}\sin2x(x^2(4a-3c)+\\+x(-8a+4b-3m-4c)-4b-2m+2c)+\\ +2e^{-x}\cos2x(x^2(-a+2c)+\\+x(2a-b+2m)+b)+\\ +2e^{-x}\sin2x(x^2(-2a-c)+\\+x(-2b+2c-m)+m)+\\ +5e^{-x}((ax^2+bx)\cos2x+(cx^2+mx)\sin2x))=\\ =xe^{-x}\cos2x\\ \cos2x:\\ x^2:-3a-4c-2a+4c+5a=0\\\implies0=0\\ x:-4a-3b-4m+8c+4a-2b+4m+5b=1\\ \implies8c=1, c=\frac{1}{8}\\ x^0:-2b+2a+4m+2b=0\\ \implies2a+4m=0\\ \sin2x:\\ x^2:4a-3c-4a-2c+5c=0\\ \implies0=0\\ x:-8a+4b-3m-4c-4b+\\+4c-2m+5m=0\\ \implies a=0\\ x^0:-4b-2m+2c+2m=0\\ \implies-4b+2c=0\\ b=\frac{1}{16}\\ m=0$

Then

$y_2=e^{-x}(\frac{1}{16}x\cos2x+\frac{1}{8}x^2\sin2x)$

The solution of the equation is

$y=e^{-x}(C_1\cos2x+C_2\sin2x)+\\ +e^{-x}(\frac{1}{16}x\cos2x+\frac{1}{8}x^2\sin2x)$