JEE Main 2019 — Mathematics Calculus

Given differential equation is

$${y}^{2}dx=(\frac{1}{y}-x)dy$$y2dx=(y1​−x)dy

$$\Rightarrow {y}^{2}\frac{dx}{dy}+x=\frac{1}{y}$$⇒y2dydx​+x=y1​

$$\Rightarrow \frac{dx}{dy}+(\frac{1}{{y}^{2}})x=\frac{1}{{y}^{3}}$$⇒dydx​+(y21​)x=y31​

It is a linear differential equation whose integrating factor $\displaystyle I.F. ={e}^{\int \frac{dy}{{y}^{2}}}={e}^{-\frac{1}{y}}$I.F.=e∫y2dy​=e−y1​

Solution of a given differential equation can be written as

$$x{e}^{-\frac{1}{y}}=\int {e}^{-\frac{1}{y}}\frac{1}{{y}^{3}}dy=I$$xe−y1​=∫e−y1​y31​dy=I

Let $\displaystyle \frac{-1}{y}=t\Rightarrow \frac{dy}{{y}^{2}}=dt\Rightarrow I=\int (-t){e}^{t} dt$y−1​=t⇒y2dy​=dt⇒I=∫(−t)etdt

$={e}^{t}(1-t)+C$=et(1−t)+C, (Integrating by parts)

$\Rightarrow$⇒ Solution of differential equation is $x{e}^{-\frac{1}{y}}={e}^{-\frac{1}{y}}(1+\frac{1}{y})+C$xe−y1​=e−y1​(1+y1​)+C

Since for $x=1,$x=1,we have $y=1$y=1,

$$\Rightarrow 1.{e}^{-1}={e}^{-1}(1+1)+C\Rightarrow C=\frac{-1}{e}$$⇒1.e−1=e−1(1+1)+C⇒C=e−1​

$\Rightarrow$⇒ Solution with given condition is $x{e}^{\frac{-1}{y}}={e}^{\frac{-1}{y}}(1+\frac{1}{y})-{e}^{-1}$xey−1​=ey−1​(1+y1​)−e−1

or $x=1+\frac{1}{y}-{e}^{(\frac{1}{y}-1)}$x=1+y1​−e(y1​−1)

So, ${x|}_{y=2}=1+\frac{1}{2}-{e}^{\frac{1}{2}-1}=\frac{3}{2}-\frac{1}{\sqrt{e}}$x∣y=2​=1+21​−e21​−1=23​−e​1​