The solution of frac{dy}{dx} + y = e^{-x}, y( | vedclass

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(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = e^{-x}$.First,find t

Vedclass · Accepted Answer

$y = xe^{-x}$

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(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = e^{-x}$.First,find the Integrating Factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int 1 dx} = e^x$.The general solution is given by $y 	imes (I.F.) = \int (Q 	imes I.F.) dx + c$.Substituting the values:$y 	imes e^x = \int (e^{-x} 	imes e^x) dx + c$$y e^x = \int 1 dx + c$$y e^x = x + c$.Using the initial condition $y(0) = 0$:$0 	imes e^0 = 0 + c \implies c = 0$.Thus,the particular solution is $y e^x = x$,which simplifies to $y = x e^{-x}$.

The solution of $\frac{dy}{dx} + y = e^{-x}, y(0) = 0$ is

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