Question

Let $y$ be the solution of the differential equation $x\frac{dy}{dx}=\frac{y^2}{1-y\log x}$ satisfying $y\left(1\right)=1$. Then $y$ satisfies

• A. $y=x^{y-1}$
• B. $y=x^y$
• C. $y=x^{y+1}$
• D. $y=x^{y+2}$

Answer 1

The correct option is B $y=x^y$

Let $y=x^y\Rightarrow \log y=y\log x$
Differentiating both sides w.r.t. $x$, we get
$\frac{1}{y}\frac{dy}{dx}=y\times \frac{1}{x}+\log x\frac{dy}{dx}$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}-\log x\frac{dy}{dx}=\frac{y}{x}$

$\Rightarrow \frac{dy}{dx}(\frac{1}{y}-\log x)=\frac{y}{x}$

$\Rightarrow \frac{dy}{dx}\left(\frac{1-y\log x}{y}\right)=\frac{y}{x}$

$\Rightarrow \frac{xdy}{dx}=\frac{y^2}{(1-y\log x)}$