JEE Advanced 2022 Paper 2 · Q02 · Differential Equations

Question

If $y(x)$ is the solution of the differential equation $x\,dy - (y^2 - 4y)\,dx = 0$ for $x > 0$, $y(1) = 2$, and the slope of the curve $y = y(x)$ is never zero, then the value of $10\, y(\sqrt{2})$ is _____.

SolveKar AI · Accepted Answer

Separate variables: $\dfrac{dy}{y^2-4y}=\dfrac{dx}{x}$. Partial fractions: $\dfrac{1}{y(y-4)}=\dfrac{1}{4}\left(\dfrac{1}{y-4}-\dfrac{1}{y}ight)$. Integrating: $\dfrac{1}{4}\ln\left|\dfrac{y-4}{y}ight|=\ln|x|+C$. At $x=1, y=2$: $\dfrac{1}{4}\ln 1 = 0$, so $C=0$, giving $\left|\dfrac{y-4}{y}ight|=x^4$. Since slope $
e 0$, the curve stays with $0

JEE Advanced 2022 Paper 2 · Q02 · Differential Equations

Solution