For the given differential equation find the general solution.

$\frac{d y}{d x} + 3 y = e^{- 2 x}$

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Solution

Given, $\frac{d y}{d x} + 3 y = e^{- 2 x}$
On comparing with the form $\frac{d y}{d x} + P y = Q$ we get P=3 and $Q = e^{- 2 x}$
$∴ I F = e^{\int P d x} \Rightarrow e^{\int 3 d x} \Rightarrow I F = e^{3 x}$
The solution of the given differential equation is given by
$y . I F = \int Q \times I F d x + C \Rightarrow y \times e^{3 x} = \int e^{- 2 x} . e^{3 x} d x + C \Rightarrow y e^{3 x} = \int e^{x} d x + C \Rightarrow y e^{3 x} = e^{x} + C \Rightarrow y = e^{- 2 x} + C e^{- 3 x}$

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