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Question

Solve the initial-value problem

y+5y+4y=16x+20ex,y(0)=0,y(0)=3y''+5y'+4y=16x+20e^x,y(0)=0,y'(0)=3

Solution

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Answered 2 years ago
Answered 2 years ago
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Find the homogeneous solution.

y+5y+4y=0\begin{align*} y^{\prime\prime}+5y^{\prime}+4y&= 0 \end{align*}

    m2+5m+4=0(m+4)(m+1)=0m={4,1}\begin{align*} \implies\\ m^2+5m+4 &= 0\\ \left(m+4\right)\left(m+1\right)&= 0\\\\ m=\left\{-4,-1\right\} \end{align*}

    yc=C1e4x+C2ex\begin{align*} \implies\\ y_c &= C_1e^{-4x}+C_2e^{-x} \end{align*}

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