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    Difficult DiffyQ

    y'' - 4y = xe^(x)

    Use the method of vartiation of parameters to find the general solution of the given differential equation.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bearej50 View Post
    y'' - 4y = xe^(x)

    Use the method of vartiation of parameters to find the general solution of the given differential equation.
    First solve the homogeneous equation y^{\prime\prime}-4y=0.

    The characteristic equation is r^2-4=0\implies r=\pm 2.

    Thus, the complimentary solution is y=c_1e^{-2x}+c_2e^{2x}.

    Now, let y_1\!\left(x\right)=e^{-2x} and y_2\!\left(x\right)=e^{2x}.

    We now need to find u_1^{\prime}\!\left(x\right) and u_2^{\prime}\!\left(x\right)

    u_1^{\prime}\!\left(x\right)=\frac{\displaystyle\b  egin{vmatrix}0 & y_2\\f\!\left(x\right)&y_2^{\prime}\end{vmatrix}}{  \displaystyle\begin{vmatrix}y_1 & y_2\\y_1^{\prime} & y_2^{\prime}\end{vmatrix}}=\frac{\displaystyle\beg  in{vmatrix}0 & e^{2x}\\xe^{x} & 2e^{2x}\end{vmatrix}}{\displaystyle\begin{vmatrix}  e^{-2x} & e^{2x}\\-2e^{-2x} & 2e^{2x}\end{vmatrix}}=\tfrac{1}{4}xe^{3x}

    Thus, u_1\!\left(x\right)=\tfrac{1}{4}\int xe^{3x}\,dx

    I leave it for you to find u_1\!\left(x\right) (See spoiler to check your answer).

    Spoiler:
    u_1\!\left(x\right)=\tfrac{1}{36}e^{3x}\left(3x-1\right)


    u_2^{\prime}\!\left(x\right)=\frac{\displaystyle\b  egin{vmatrix}y_1 & 0\\ y_1^{\prime} & f\!\left(x\right)\end{vmatrix}}{\displaystyle\begi  n{vmatrix}y_1 & y_2\\y_1^{\prime} & y_2^{\prime}\end{vmatrix}}=\frac{\displaystyle\beg  in{vmatrix}e^{-2x} &0\\-2e^{-2x} & xe^{x}\end{vmatrix}}{\displaystyle\begin{vmatrix}e  ^{-2x} & e^{2x}\\-2e^{-2x} & 2e^{2x}\end{vmatrix}}=\tfrac{1}{4}xe^{-x}

    Thus, u_2\!\left(x\right)=\tfrac{1}{4}\int xe^{-x}\,dx

    I leave it for you to find u_2\!\left(x\right) (See spoiler to check your answer).

    Spoiler:
    u_2\!\left(x\right)=-\tfrac{1}{4}e^{-x}\left(x+1\right)


    Once you find u_1\!\left(x\right) and u_2\!\left(x\right), the solution to your differential equation would be y=u_1\!\left(x\right)e^{-2x}+u_2\!\left(x\right)e^{2x}=\dots (See spoiler to check your answer)

    Spoiler:
    y=\tfrac{1}{36}e^{x}\left(3x-1\right)-\tfrac{1}{4}e^{x}\left(x+1\right)=\boxed{-\tfrac{1}{18}e^x\left(3x+5\right)}
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