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Thread: inhomogenous eqn

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    inhomogenous eqn

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

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

    $\displaystyle y=y_c+y_p$
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