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  • The solution of \frac{d y}{d x}+y=e^{-x}, y(0)=0 is (a) y=e^{x}(x-1)$

The solution of \frac{d y}{d x}+y=e^{-x}, y(0)=0 is
(a) y=e^{x}(x-1)
(b) y=x e^{-x}
(c) y=x e^{-x}+1
(d) y=(x+1) e^{-x}

Answers (1)

The answer is the option (b) y=x e^{-x}$

Explanation: -

This is a linear differential equation.
On comparing it with \frac{d y}{d x}+P y=Q$, we get
$$ P=1, Q=e^{-x} $$

\mathrm{IF}=e^{[P d x} e^{\int d x}=e^{x}

So, the general solution is:
\\y \cdot e^{x}=\int e^{-x} e^{x} d x+C$ \\$\Rightarrow \quad y \cdot e^{x}=\int d x+C$ \\$\Rightarrow \quad y \cdot e^{x}=x+C$
Given that when x=0 and y=0
\\\Rightarrow$ $0=0+C$ \\$\Rightarrow$ $C=0$
Eq. (i) becomes y \cdot e^{x}=x$
\Rightarrow \quad y=x e^{-x}$

Posted by

infoexpert22

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