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The solution of x \frac{d y}{d x}+y=e^{x} is

(a) y=\frac{e^{x}}{x}+\frac{k}{x}
(b) y=x e^{x}+c x
(c) y=x e^{x}+k
(d) x=\frac{e^{y}}{y}+\frac{k}{y}

Answers (1)

The answer is the option (a) y=\frac{e^{x}}{x}+\frac{k}{x}$

Explanation: -

\Rightarrow \quad \frac{d y}{d x}+\frac{y}{x}=\frac{e^{x}}{x}$
This is a linear differential equation. Dn comparing it with $\frac{d y}{d x}+P y=Q$, we get
$$ P=\frac{1}{x} \text { and } Q=\frac{e^{x}}{x} $$
\therefore \quad \mathrm{IF}_{0}=e^{\int \frac{1}{x} d x}=e^{(\log x)}=x$
So, the general solution is:
\\y \cdot x=\int \frac{e^{x}}{x} x d x$ \\$\Rightarrow \quad y \cdot x=\int e^{x} d x$ \\$\Rightarrow \quad \quad y \cdot x=e^{x}+k$ \\$\Rightarrow$ $y=\frac{e^{x}}{x}+\frac{k}{x}$

Posted by

infoexpert22

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