Question

If $x\frac{dy}{dx}+y=x.\frac{f(x.y)}{f^{{}^{\prime }}(x.y)}$ then $f(x.y)$ is equal to ($k$ being an arbitrary constant):

• A. $k{e}^{x^2/2}$
• B. $k{e}^{y^2/2}$
• C. $k{e}^{xy/2}$
• D. none of these

Answer 1

Answer: (A)

$k{e}^{x^2/2}$

$x\frac{dy}{dx}+y=x\frac{f\left(xy\right)}{f\left(xy\right)}\Rightarrow \frac{d\left(xy\right)}{dx}=x\frac{f\left(xy\right)}{f^{\prime }\left(xy\right)}$
$\Rightarrow \int \frac{f^{\prime }\left(xy\right)}{f\left(xy\right)}d\left(xy\right)=\int xdx$
$\Rightarrow \left(f\left(xy\right)\right)=\frac{1}{2}{x}^2+c\Rightarrow f\left(xy\right)=e^{\frac{x^2}{2}}{e}^c$
$\Rightarrow f\left(xy\right)=k{e}^{\frac{x^2}{2}}$