Does anyone know how to solve$\displaystyle \frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=10exp(-x)sinx$ thnx

Printable View

- May 30th 2009, 07:34 AMoxrigbyinhomogenous eqn
Does anyone know how to solve$\displaystyle \frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=10exp(-x)sinx$ thnx

- May 30th 2009, 08:47 AMTheEmptySet

First solve the homogenious equation. The auxillary equation is

$\displaystyle m^2+2m+2=0 \iff m^2+2m+1=-1 \iff (m+1)^2=-1 \iff m= -1 \pm i$

So the complimentry solution is

$\displaystyle y_c=c_1e^{-x}\cos(x)+c_2e^{-x}\sin(x) $

Since $\displaystyle e^{-x}\sin(x)$ appears in the original equation the form of your particular solution is

$\displaystyle y_p=Axe^{-x}\cos(x)+Be^{-x}\sin(x)$

now take two derivatives to get

$\displaystyle y_p'=(A-Ax+Bx)e^{-x}\cos(x)+(-Ax+B-Bx)e^{-x}\sin(x)$

$\displaystyle y_p''=(-2Ax+2B-2Bx)e^{-x}\cos(x)+(-2A+2Ax-2B)e^{-x}\sin(x)$

Now plug all of this into the ODE and simplify to get

$\displaystyle -2Ae^{-x}\sin(x)+2Be^{-x}\cos(x)=10e^{-x}\sin(x)$

So we can see that $\displaystyle B=0,A=-5$

So $\displaystyle y_p=-5xe^{-x}\cos(x)$

$\displaystyle y=y_c+y_p$