Question
Verify that the function $x^{2}=2 y^{2} \log y$ (implicit or explicit) is a solution of the differential equation $\left(x^{2}+y^{2}\right) \frac{d y}{d x}-x y=0$ 

Answer

It is given that x2 = 2y2 log y
Now, differentiating both sides w.r.t. x, we get,
$2 x=2 \cdot \frac{d}{d x}\left(y^{2} \log y\right)$ 
$\Rightarrow \mathrm{x}=\left[2 \mathrm{y} \cdot \log \mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{2} \cdot \frac{1}{\mathrm{y}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right]$ 
$\Rightarrow \mathrm{x}=\frac{\mathrm{dy}}{\mathrm{dx}}(2 \mathrm{ylogy}+\mathrm{y})$ 
$\Rightarrow \frac{d y}{d x}=\frac{x}{y(1+2 \log y)}$ 
Now, substituting the value of $\frac{d y}{d x}$ in the LHS of the given differential equation, we get,
$\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{xy}=\left(2 \mathrm{y}^{2} \log \mathrm{y}+\mathrm{y}^{2}\right) \cdot \frac{\mathrm{x}}{\mathrm{y}(1+2 \log \mathrm{y})}-\mathrm{xy}$ 
= $\mathrm{y}^{2}(1+2 \log \mathrm{y}) \cdot \frac{\mathrm{x}}{\mathrm{y}(1+2 \log \mathrm{y})}-\mathrm{xy}$ 
= xy - xy
= 0
Therefore, the given function is the solution of the corresponding differential equation.

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