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{\bf Question}

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Find the general solution of the differential equation
$$\frac{dx}{dt} + 3x = e^{-2t}$$

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{\bf Answer}

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$$\frac{dx}{dt} + 3x = e^{-2t}$$

Compare with the model equation $\displaystyle \frac{dx}{dt} +
p(t)x = q(t)$

Multiply by integrating factor $\displaystyle g(t) = e^{\int p(t)
\, dt}$ so that $\frac{d}{dt}[x g(t)] = g(t)q(t)$

Here the integrating factor is $\displaystyle g(t) = e^{\int3 \,
dt} = e^{3t}$

\begin{eqnarray*}
{\rm Thus} \hspace{.5in} e^{3t}\frac{dx}{dt} + 3e^{3t}x & = &
e^{3t}e^{-2t} = e^{3t-2t} = e^t \\ {\rm Hence} \hspace{.5in}
\frac{d}{dt}(xe^{3t}) & = & e^t \\\\ {\rm Finally} \hspace{.5in}
xe^{3t} & = & \int e^t \, dt = e^t + C \\ \Rightarrow x & = &
e^{-2t} + Ce^{-3t} \end{eqnarray*} Where C is a constant.



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