JEE Main 2023 — Mathematics Calculus

Given,

$$(1-{x}^{2}{y}^{2})dx=ydx+xdy$$(1−x2y2)dx=ydx+xdy

$$\Rightarrow dx=\frac{ydx+xdy}{1-{(xy)}^{2}}$$⇒dx=1−(xy)2ydx+xdy​

$$\Rightarrow dx=\frac{d(yx)}{1-{(xy)}^{2}}$$⇒dx=1−(xy)2d(yx)​

$$\Rightarrow dx=\frac{1}{2}(\frac{d(xy)}{1-xy}+\frac{d(xy)}{1+xy})$$⇒dx=21​(1−xyd(xy)​+1+xyd(xy)​)

Now integrating both side we get,

$$\Rightarrow 2x+c=\mathrm{ln}|\frac{1+xy}{1-xy}|$$⇒2x+c=ln∣1−xy1+xy​∣

$$\Rightarrow |\frac{xy+1}{xy-1}|={e}^{c}{e}^{2x}$$⇒∣xy−1xy+1​∣=ece2x

Now given, $y(1)=2$y(1)=2 so putting the value in above equation we get,

$$3={e}^{c}{e}^{2}$$3=ece2

$$\Rightarrow {e}^{c}=\frac{3}{{e}^{2}}$$⇒ec=e23​

Hence, the equation becomes $|\frac{xy+1}{xy-1}|=3{e}^{2x-2}$∣xy−1xy+1​∣=3e2x−2

Now finding $y(2)$y(2) so putting $x=2$x=2 in above equation we get,

$$|\frac{2y+1}{2y-1}|=3{e}^{2}$$∣2y−12y+1​∣=3e2

$$\Rightarrow 2y+1=2\cdot 3{e}^{2}y-3{e}^{2}$$⇒2y+1=2⋅3e2y−3e2

$$\Rightarrow 1+3{e}^{2}=2y(3{e}^{2}-1)$$⇒1+3e2=2y(3e2−1)

$$\Rightarrow y(2)=\frac{1+3{e}^{2}}{2(3{e}^{2}-1)}$$⇒y(2)=2(3e2−1)1+3e2​