Question
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-2\text{x}}$

Answer

Here, $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-2\text{x}}$ This is a linear differential equation, comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=1,\text{Q}=\text{e}^{-2\text{x}}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int2\text{dx}}$ $=\text{e}^{\text{x}}$Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\times\text{e}^{\text{x}}=\int\text{e}^{-2\text{x}}\times\text{e}^{\text{x}}\text{dx + C}$ $=\int\text{e}^{-\text{x}}+\text{C}$ $\text{y}\text{e}^{\text{x}}=\frac{\text{e}^{-\text{x}}}{-1}+\text{C}$ $\text{y}=-\text{e}^{-2\text{x}}+\text{C}\text{e}^{-\text{x}}$

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