JEE Main 2023 (Online) 29th January Morning ShiftQuestion 1
Let $y=f(x)$ be the solution of the differential equation $y(x+1)\,dx-x^{2}dy=0$, $y(1)=e$. Then $\lim_{x\to0^{+}}f(x)$ is equal to
Differential Equation
Solution
Concept. Separate the variables; then study $x\,e^{2-1/x}$ as $x\to0^+$. Formulas used. (1) $\int\left(\frac1x+\frac1{x^2}\right)dx=\ln x-\frac1x$Step 1 - Separate. $\dfrac{dy}{y}=\dfrac{x+1}{x^2}\,dx$, so by (1), $\ln y=\ln x-\frac1x+c$. Step 2 - Constant. $y(1)=e$: $1=0-1+c$, so $c=2$ and $y=x\,e^{2-1/x}$. Step 3 - Limit. As $x\to0^+$, $x\to0$ and $e^{-1/x}\to0$, so $f(x)\to0$. Answer: (A)