JEE Advanced 2022 · Paper 2, Mathematics question 2

Separate.
$$\frac{dy}{y(y-4)}=\frac{dx}{x}\implies\frac14\ln\left|\frac{y-4}{y}\right|=\ln x+C.$$
At $x=1$, $y=2$: $\left|\frac{-2}{2}\right|=1$, so $C=0$ and $\left|\frac{y-4}{y}\right|=x^4$.

Choose the branch. The slope $\frac{y^2-4y}{x}$ is never zero, so $y$ never reaches 0 or 4; starting at 2 it stays in $(0,4)$. There $\frac{y-4}{y}<0$, so
$$\frac{4-y}{y}=x^4\implies y=\frac{4}{1+x^4}.$$

Answer. $y(\sqrt2)=\frac45$, so $10\,y(\sqrt2)=8$.