JEE Main 2022 (Online) 28th July Evening ShiftQuestion 1
Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx}+\frac{1}{x^2-1}y=\left(\frac{x-1}{x+1}\right)^{1/2}$, $x>1$ passing through the point $\left(2,\sqrt{\frac13}\right)$. Then $\sqrt7\,y(8)$ is equal to :
Differential Equation
Solution
Concept. IF $=\sqrt{\frac{x-1}{x+1}}$; the right side times IF is $\frac{x-1}{x+1}$. Formulas used. (1) $\int\dfrac{dx}{x^2-1}=\dfrac12\ln\dfrac{x-1}{x+1}$ for $x>1$(2) $\dfrac{x-1}{x+1}=1-\dfrac{2}{x+1}$Step 1 - IF. By (1), IF $=\sqrt{\dfrac{x-1}{x+1}}$. Step 2 - Solve. By (2), $$y\sqrt{\frac{x-1}{x+1}}=\int\frac{x-1}{x+1}\,dx=x-2\ln(x+1)+C.$$ Step 3 - Constant. At $x=2$: $\frac{1}{\sqrt3}\cdot\frac{1}{\sqrt3}=\frac13=2-2\ln3+C$, so $C=-\frac53+2\ln3$. Step 4 - At $x=8$. IF $=\frac{\sqrt7}{3}$: $$\frac{\sqrt7}{3}\,y=8-4\ln3-\frac53+2\ln3=\frac{19}{3}-2\ln3\ \Longrightarrow\ \sqrt7\,y(8)=19-6\ln3.$$ Answer: (D)