JEE Main 2015 (Offline)Question 1
Let $y(x)$ be the solution of the differential equation $(x\log x)\frac{dy}{dx}+y=2x\log x$, $(x\ge1)$. Then $y(e)$ is equal to :
Differential Equation
Solution
Concept. Divide by $x\ln x$: linear with IF $\ln x$. The constant comes from $y$ being finite at $x=1$. Formulas used. (1) $e^{\int\frac{dx}{x\ln x}}=\ln x$(2) $\int\ln x\,dx=x\ln x-x$Step 1 - Linear form. $y'+\dfrac{y}{x\ln x}=2$, IF $=\ln x$ by (1). Step 2 - Integrate. By (2), $y\ln x=2(x\ln x-x)+C$. Step 3 - The constant. At $x=1$ the left side is $0$ (as $y(1)$ is finite), so $0=-2+C$ and $C=2$. Step 4 - At $x=e$. $y=2(e-e)+2=2$. Answer: (A)