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# The solution of $\Large \frac{dy}{dx}\normalsize +y=e^{-x},y(0)=0$ is:\begin{array}{1 1}(A)\;y=e^x(x-1) & (B)\;y=xe^{-x}\\(C)\;y=xe^{-x}+1 & (D)\;y=(x+1)e^{-x}\end{array}

Toolbox:
• A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q has the general solution as ye^{\int pdx}=\int Q.e^{\int pdx}.dx+c Given \large\frac{dy}{dx}$$+y=e^{-x}$
This is a linear differential equation of the form $\large\frac{dy}{dx}+Py=Q$, where $P=1$ and $Q=e^{-x}$