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Question

# Solve the initial-value problem$y''+5y'+4y=16x+20e^x,y(0)=0,y'(0)=3$

Solution

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Step 1
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Find the homogeneous solution.

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

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

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

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