JEE 2023 — Mathematics PYQ

$(1+\ln x)\frac{dx}{dy}-x{\log }_{e}x=e^y$

Let $x\ln x=t$

$\Rightarrow (1+\ln x)\frac{dx}{dy}=\frac{dt}{dy}$

$\Rightarrow \frac{dt}{dy}-t=e^y$

Here $p=-1,Q=e^y$

$I.F.=e^{\int Pdy}=e^{\int -1dy}=e^{-y}$

$t(I.F.)=\int Q.(I.F.)dy$

$t(e^{-y})=\int e^{-y}\cdot e^{y}dy$

$(x\ln x)e^{-y}=y+C$

passes through $(1,0)\Rightarrow C=0$

passes through $(\alpha ,2)$

${\alpha }^{\alpha }=e^{2e^2}$