JEE Main 2022 — Mathematics Calculus

$$({\mathrm{tan}}^{-1}y-x)dy=(1+{y}^{2})dx$$(tan−1y−x)dy=(1+y2)dx

$$\Rightarrow \frac{dx}{dy}+(\frac{1}{1+{y}^{2}})x=\frac{{\mathrm{tan}}^{-1}y}{1+{y}^{2}}$$⇒dydx​+(1+y21​)x=1+y2tan−1y​

I.F. $\displaystyle ={e}^{\int \frac{1}{1+{y}^{2}}dy}={e}^{{\mathrm{tan}}^{-1}y}$=e∫1+y21​dy=etan−1y

General solution will be

$$x({e}^{{\mathrm{tan}}^{-1}y})=\int {e}^{{\mathrm{tan}}^{-1}y}\times \frac{{\mathrm{tan}}^{-1}y}{1+{y}^{2}}dy$$x(etan−1y)=∫etan−1y×1+y2tan−1y​dy

$$\Rightarrow x({e}^{{\mathrm{tan}}^{-1}y})={\mathrm{tan}}^{-1}y({e}^{{\mathrm{tan}}^{-1}y})-{e}^{{\mathrm{tan}}^{-1}y}+C$$⇒x(etan−1y)=tan−1y(etan−1y)−etan−1y+C

Since the curve passes through $(1,0)$(1,0) $\Rightarrow C=2$⇒C=2

So, the curve reduces to $x({e}^{{\mathrm{tan}}^{-1}y})={\mathrm{tan}}^{-1}y({e}^{{\mathrm{tan}}^{-1}y})-{e}^{{\mathrm{tan}}^{-1}y}+2$x(etan−1y)=tan−1y(etan−1y)−etan−1y+2

when $y=\mathrm{tan}1,x(e)=1(e)-e+2$y=tan1,x(e)=1(e)−e+2

$$\Rightarrow x=\frac{2}{e}$$⇒x=e2​