JEE Main 2020 — Mathematics Calculus

$$\frac{dx}{dy}+x={y}^{2}$$dydx​+x=y2

This equation is a linear differential equation of the type $\frac{dx}{dy}+Px=Q,$dydx​+Px=Q, where $P=1$P=1 and $Q={y}^{2}.$Q=y2.

Integrating Factor I.F. $\displaystyle ={e}^{\int 1dy}={e}^{y}$=e∫1dy=ey

Now solution of the differential equation is $\displaystyle x(I.F.)=\int P(I.F.)dy+C$x(I.F.)=∫P(I.F.)dy+C

$$x\cdot {e}^{y}=\int {y}^{2}\cdot {e}^{y}\cdot dy={y}^{2}\cdot {e}^{y}-\int 2y\cdot {e}^{y}\cdot dy$$x⋅ey=∫y2⋅ey⋅dy=y2⋅ey−∫2y⋅ey⋅dy

$$\Rightarrow x\cdot {e}^{y}={y}^{2}\cdot {e}^{y}-2y\cdot {e}^{y}+2{e}^{y}+c$$⇒x⋅ey=y2⋅ey−2y⋅ey+2ey+c

$$\Rightarrow x={y}^{2}-2y+2+c\cdot {e}^{-y}$$⇒x=y2−2y+2+c⋅e−y

Put $x=0,y=1$x=0,y=1

$$\Rightarrow 0=1-2+2+\frac{c}{e}$$⇒0=1−2+2+ec​

$$\Rightarrow c=-e$$⇒c=−e

Hence $x={y}^{2}-2y+2+(-e)\cdot {e}^{-y}$x=y2−2y+2+(−e)⋅e−y

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

$$\Rightarrow x=2-e$$⇒x=2−e