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Let $y$ be the solution of the differential equation
$x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x}$ satisfying $y(1)=1 .$ Then, $y$
satisfies
Options:
$x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x}$ satisfying $y(1)=1 .$ Then, $y$
satisfies
Solution:
1278 Upvotes
Verified Answer
The correct answer is:
$y=x^{y}$
Let $y=x^{y} \Rightarrow \log y=y \log x$
Differentiating both sides w.r.t. $x$, we get $\frac{1}{y} \frac{d y}{d x}=y \times \frac{1}{x}+\log x \frac{d y}{d x}$
$\Rightarrow \quad \frac{1}{y} \frac{d y}{d x}-\log x \frac{d y}{d x}=\frac{y}{x}$
$\Rightarrow \quad \frac{d y}{d x}\left(\frac{1}{y}-\log x\right)=\frac{y}{x}$
$\Rightarrow \quad \frac{d y}{d x}\left(\frac{1-y \log x}{y}\right)=\frac{y}{x}$
$\Rightarrow \quad \frac{x d y}{d x}=\frac{y^{2}}{(1-y \log x)}$
Differentiating both sides w.r.t. $x$, we get $\frac{1}{y} \frac{d y}{d x}=y \times \frac{1}{x}+\log x \frac{d y}{d x}$
$\Rightarrow \quad \frac{1}{y} \frac{d y}{d x}-\log x \frac{d y}{d x}=\frac{y}{x}$
$\Rightarrow \quad \frac{d y}{d x}\left(\frac{1}{y}-\log x\right)=\frac{y}{x}$
$\Rightarrow \quad \frac{d y}{d x}\left(\frac{1-y \log x}{y}\right)=\frac{y}{x}$
$\Rightarrow \quad \frac{x d y}{d x}=\frac{y^{2}}{(1-y \log x)}$
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