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Q.
The solution of $\frac{dy}{dx}+y=e^{-x}$, $y\left(0\right)=0$ is
Differential Equations
Solution:
$\frac{dy}{dx}+y=e^{-x}$
It is linear differential equation with
$I.F.=e^{\int dx}=e^{x}$
Now, solution is $y\cdot e^{x}=\int e^{x}\,e^{-x}\,dx+c$
$\Rightarrow ye^{x} = x + c$
$\Rightarrow y = \left(x + c\right)e^{-x}$
Now, $y\left(0\right)=0$
$\Rightarrow c=0$
Hence, $y = xe^{-x}$