Question

lf $y+x\frac{dy}{dx}=x\frac{\varphi \left(xy\right)}{{\varphi }^1\left(xy\right)}$ then $\varphi \left(xy\right)$ is equal to

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

Answer 1

Answer: (A)

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

$\Rightarrow y+x{y}_1=\frac{x\varphi \left(xy\right)}{{\varphi }^1\left(xy\right)}$

$\Rightarrow \frac{d}{dx}\left(xy\right)=\frac{x\varphi \left(xy\right)}{{\varphi }^1\left(xy\right)}$

$\Rightarrow \frac{{\varphi }^1\left(xy\right)}{\varphi \left(xy\right)}d\left(xy\right)=xdx$.........eq(i)

let log $\left(\varphi \right(xy\left)\right)=t$

differentiate the above equation and substitute the values into eq(i), we get

$\Rightarrow \int dt=\int xdx-logk$

$\Rightarrow t=\frac{x^2}{2}-logk$

substitute the value of t and taking log term to left side and simplify, we get

$\Rightarrow \varphi \left(xy\right)=k{e}^{\frac{x^2}{2}}$