Question

If $x.\frac{dy}{dx}+y=x.\frac{f\left(xy\right)}{f^{\prime }\left(xy\right)}$, then $f\left(xy\right)$ is equal to

• A. $k.e^{\frac{x^2}{2}}$
• B. $k.e^{\frac{y^2}{2}}$
• C. $k.e^{x^2}$
• D. $k.e^{\frac{xy}{2}}$

Answer 1

The correct option is A $k.e^{\frac{x^2}{2}}$

$x.\frac{dy}{dx}+y=x.\frac{f\left(xy\right)}{f^{\prime }\left(xy\right)}$
i.e., $\frac{d}{dx}\left(xy\right)=x\frac{f\left(xy\right)}{f^{\prime }\left(xy\right)}$
$\Rightarrow \frac{f^{\prime }\left(xy\right)}{f\left(xy\right)}d\left(xy\right)=xdx$
$\Rightarrow \int \frac{f^{\prime }\left(xy\right)}{f\left(xy\right)}d\left(xy\right)=\int xdx$
$\Rightarrow \log \left[f\right(xy\left)\right]=\frac{x^2}{2}+C$
$\Rightarrow f\left(xy\right)=e^{(\frac{x^2}{2}+C)}$
$=e^{\frac{x^2}{2}}.e^C=k.e^{\frac{x^2}{2}}$ .... $[e^C=k]$

Hence, $f\left(xy\right)=k.e^{\frac{x^2}{2}}$.