JEE Main 2023 (Online) 6th April Evening ShiftQuestion 1
If the solution curve $f(x,y)=0$ of the differential equation $(1+\log_ex)\frac{dx}{dy}-x\log_ex=e^{y}$, $x>0$, passes through the points $(1,0)$ and $(\alpha,2)$, then $\alpha^{\alpha}$ is equal to
Differential Equation
Solution
Concept. $(1+\ln x)\frac{dx}{dy}$ is $\frac{d}{dy}(x\ln x)$; with $u=x\ln x$ the equation is linear. Formulas used. (1) $\frac{d}{dy}(x\ln x)=(1+\ln x)\dfrac{dx}{dy}$Step 1 - Substitute. By (1), $u'-u=e^y$. IF $=e^{-y}$: $$(ue^{-y})'=1\ \Longrightarrow\ u=(y+C)e^y.$$ Step 2 - Constant. At $(1,0)$: $u=0$, so $C=0$ and $x\ln x=ye^y$. Step 3 - At $y=2$. $\alpha\ln\alpha=2e^2$, so $$\alpha^\alpha=e^{\alpha\ln\alpha}=e^{2e^2}.$$ Answer: (D)