Answer to Question #106418 in Calculus for Suraj Singh

Question

solve by method of undetermined coefficient : y'' + 2(y') + 5y = x(e^-x)cos(2x)

Answer 1

$y''+2y'+5y=xe^{-x}\cos(2x)$

General solution will be sum of the complementary solution and particular solution.

Find the complementary solution by solving $y''+2y'+5y=0$

Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$

Substitute $y(x)=e^{\lambda x}$ into the differential equation

$(\lambda^2+2\lambda+5)e^{\lambda x}=0$

solve for $\lambda$ : $\lambda = -1+2i$ or $\lambda=-1-2i$

These roots give this general solution

$y_g(x)=c_1e^{-x}\cos{2x}+c_2e^{-x}\sin{2x}$

To find partial solution, we put into equation this form of solution

$y_p(x)=axe^{-x}\cos(2x)+bxe^{-x}\sin(2x)+cx^2e^{-x}\cos(2x)+dx^2e^{-x}\sin(2x)$

and by computing first and second derivative and putting everything into equation, we determine all the coefficients

Partial solution:

$y_p(x)=1/16xe^{-x}\cos{2x}+1/8x^2e^{-x}\sin{2x}$

Then $y(x)=y_g(x)+y_p(x)=c_1e^{-x}\cos{2x}+c_2e^{-x}\sin{2x}+1/16xe^{-x}\cos{2x}+1/8x^2e^{-x}\sin{2x}$