JEE Main 2020 — Differential Equations Question with Solution

$\frac{dx}{dy}+x=y^2$

This equation is a linear differential equation of the type $\frac{dx}{dy}+Px=Q,$ where $P=1$ and $Q=y^2.$
Integrating Factor I.F. $=e^{\int 1dy}=e^y$

Now solution of the differential equation is $x\left(I.F.\right)=\int P\left(I.F.\right)dy+C$
$x·e^y=\int y^2·e^y·dy=y^2·e^y-\int 2y·e^y·dy$

$\Rightarrow x·e^y=y^2·e^y-2y·e^y+2e^y+c$

$\Rightarrow x=y^2-2y+2+c·e^{-y}$

Put $x=0, y=1$

$\Rightarrow 0=1-2+2+\frac{c}{e}$

$\Rightarrow c=-e$

Hence $x=y^2-2y+2+\left(-e\right)·e^{-y}$

On putting $y=0$ we get $x=0-0+2+\left(-e\right)\left(e^{-0}\right)$

$\Rightarrow x=2-e$