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Math Help - variation problem

  1. #1
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    variation problem

    (1-x)y''+xy'-y=2(x-1)^2e^x
    how to solve it?

    i was told to show first that x and e^x are a solution of some homogeneous equation
    i got that it the homogeneous equation?

    but the partivular solution is tough
    y_p=(ax^2+bx+c)e^x

    its hard to differentiate every time
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    (1-x)y''+xy'-y=2(x-1)^2e^x
    how to solve it?

    i was told to show first that x and e^x are a solution of some homogeneous equation
    i got that it the homogeneous equation?

    but the partivular solution is tough
    y_p=(ax^2+bx+c)e^x

    its hard to differentiate every time
    Unfortunately you cannot use the method of undermined coefficients when the ODE does not have constant coefficients. You will need to use the variation of parameters.
    Variation of parameters - Wikipedia, the free encyclopedia

    since you are given the solutions to the homogeneous problem, we can solve for the particular solution.

    1st. Put equation is standard form y''+\frac{x}{1-x}y'-\frac{1}{1-x}y=-2(x-1)e^{x}

    2st. Assume solution of the form y_p=u_1(x)\cdot x+u_2(x)\cdot e^{x}

    3nd.Calculate the Wronskian

    W=\begin{vmatrix}x & e^x \\ 1 & e^{x}  \end{vmatrix}=(x-1)e^{x}

    Now


    W_1=\begin{vmatrix}0 & e^x \\ -2(x-1)e^{x} & e^{x}  \end{vmatrix}=2(x-1)e^{2x}

    and


    W_2=\begin{vmatrix}x & 0 \\ 1 & -2(x-1)e^{x} \end{vmatrix}=-2x(x-1)e^{x}

    Now \displaystyle u_1'(x)=\frac{W_1}{W}=2e^{x} \implies u_1(x)= 2e^{x}

    and

    Now \displaystyle u_2'(x)=\frac{W_2}{W}=-2x \implies u_2(x)=-x^2

    Putting all of this together gives

    \displaystyle y_p=2xe^{x}-x^2e^{x}
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