JEE Main 2023 (Online) 29th January Evening ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $x\log_ex\frac{dy}{dx}+y=x^{2}\log_ex$, $(x>1)$. If $y(2)=2$, then $y(e)$ is equal to
Differential Equation
Solution
Concept. Divide by $x\ln x$: linear in $y$ with IF $\ln x$. Formulas used. (1) $\text{IF}=e^{\int\frac{dx}{x\ln x}}=\ln x$(2) $\int x\ln x\,dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}$Step 1 - Solve. $y'+\frac{y}{x\ln x}=x$. By (1) and (2), $$y\ln x=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C.$$ Step 2 - Constant. $y(2)=2$: $2\ln2=2\ln2-1+C$, so $C=1$. Step 3 - At $x=e$. $y=\frac{e^2}{2}-\frac{e^2}{4}+1=\dfrac{4+e^2}{4}$. Answer: (A)