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Math Help - [SOLVED] Differential

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Differential

    Find the general solution of the second-order, linear, inhomogeneous
    differential equation:

    y′′ + 3y′ + 2y = \frac{1}{(1 + e^x)^2}

    Please help solve this equation so far I have done:

    z^2 +3z+2=0
    (z+2)(z+1)=0
    z=-2,-1
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    Find the general solution of the second-order, linear, inhomogeneous
    differential equation:

    y′′ + 3y′ + 2y = \frac{1}{(1 + e^x)^2}

    Please help solve this equation so far I have done:

    z^2 +3z+2=0
    (z+2)(z+1)=0
    z=-2,-1
    Looks good up to here.

    The homogeneous solution is y_c=c_1e^{-2x}+c_2e^{-x}

    Here, I would consider using Variation of Parameters.

    Let y_p=u_1\!\left(x\right)e^{-2x}+u_2\!\left(x\right)e^{-x}.

    Thus, we need to evaluate three different Wronskians.

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

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

    W=\begin{vmatrix}e^{-2x} & e^{-x} \\ -2e^{-2x} & -e^{-x}\end{vmatrix}=-e^{-3x}+2e^{-3x}=e^{-3x}

    Thus, u_1^{\prime}\!\left(x\right)=\frac{W_1}{W}=-\frac{e^{2x}}{\left(1+e^x\right)^2} and u_2^{\prime}\!\left(x\right)=\frac{W_2}{W}=\frac{e  ^x}{\left(1+e^x\right)^2}


    Thus,

    u_1\!\left(x\right)=-\int \frac{e^{2x}}{\left(1+e^x\right)^2}\xrightarrow{z=  1+e^x}{}-\int\frac{z-1}{z^2}\,dz=-\ln\left|z\right|-\frac{1}{z}\implies -\ln\left|1+e^x\right|-\frac{1}{1+e^{x}}

    and

    u_2\!\left(x\right)=\int\frac{e^x}{\left(1+e^x\rig  ht)^2}\,dx\xrightarrow{z=1+e^x}{}\int\frac{\,dz}{z  ^2}=-\frac{1}{z}\implies -\frac{1}{1+e^x}

    Thus, our solution is

    y=c_1e^{-2x}+c_2e^{-x}-\left(\ln\left|1+e^x\right|-\frac{1}{1+e^x}\right)e^{-2x}+\left(-\frac{1}{1+e^{-x}}\right)e^{-x}

    =c_1e^{-2x}+c_2e^{-x}-\frac{\left(\ln\left(1+e^x\right)e^{-2x}\right)\left(1+e^x\right)+e^{-2x}+e^{-x}}{1+e^x}

    =\color{red}\boxed{c_1e^{-2x}+c_2e^{-x}-\frac{e^{-2x}\ln\left(1+e^x\right)+e^{-x}\ln\left(1+e^x\right)+e^{-2x}+e^{-x}}{1+e^x}}

    Does this make sense?
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  3. #3
    Member ronaldo_07's Avatar
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    Could you explain Thus, we need to evaluate three different Wronskians the W W1 W2. How this is calculated please. Thanks.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    Could you explain Thus, we need to evaluate three different Wronskians the W W1 W2. How this is calculated please. Thanks.
    See my explanation here.
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