JEE Advanced 2022 Paper 2Question 1
If $y(x)$ is the solution of the differential equation $$x\,dy-(y^{2}-4y)\,dx=0 \text{ for } x>0,\quad y(1)=2,$$ and the slope of the curve $y=y(x)$ is never zero, then the value of $10\,y(\sqrt{2})$ is ___________.
Differential Equation
Answer$8$
Solution
Concept. Separate with partial fractions. Slope never zero means $y$ never reaches $0$ or $4$, so $y$ stays in $(0,4)$. Formulas used. (1) $\dfrac{1}{y(y-4)}=\dfrac14\left(\dfrac{1}{y-4}-\dfrac1y\right)$Step 1 - Integrate. By (1), $$\frac14\ln\left|\frac{y-4}{y}\right|=\ln x+c.$$ At $x=1$, $y=2$: $\left|\frac{-2}{2}\right|=1$, so $c=0$ and $\left|\frac{y-4}{y}\right|=x^4$. Step 2 - Branch. $y'=\frac{y(y-4)}{x}\ne0$, so $y$ never equals $0$ or $4$; starting at $2$ it stays in $(0,4)$, where $\frac{y-4}{y}<0$: $$\frac{y-4}{y}=-x^4\ \Longrightarrow\ y=\frac{4}{1+x^4}.$$ Step 3 - Value. $y(\sqrt2)=\frac45$, so $10\,y(\sqrt2)=8$. Answer: 8