Question

The solution of $x \frac{dy}{dx}+y=e^{x}$ is

• A. $y=\frac{e^{x}}{x}+\frac{k}{x}$
• B. $y = xe^x+ cx$
• C. $y = xe^x + k$
• D. $x=\frac{e^{y}}{y}+\frac{k}{y}$

Answer 1

Answer: (A)

$x \frac{dy}{dx}+y=e^{x}$
$\frac{dy}{dx}+\frac{1}{x}y=\frac{e^{x}}{x}$
It is a linear differential equation with
$I.F.=e^{\int \frac{1}{x}dx}=e^{logx}=x$
Now, solution is $y\cdot x=\int \frac{e^{x}}{x}\cdot x\,dx+k$
$\Rightarrow yx=e^{x}+k$
$\Rightarrow y=\frac{e^{x}}{x}+\frac{k}{x}$