JEE Main 2022 — Differential Equations Question with Solution

Given,

$$(1+e^{2x})\frac{dy}{dx}+2(1+y^2)e^x=0$$

$$\Rightarrow \frac{dy}{dx}=\frac{-2(1+y^2)e^x}{1+e^{2x}}$$

$$\Rightarrow \int \frac{dy}{1+y^2}=\int \frac{-2e^{x}dx}{(1+e^{2x})}$$

$$\Rightarrow {\tan }^{-1}(y)=\int \frac{-2e^{x}dx}{(1+e^{2x})}$$

Now,

Let $$e^x=t$$

$$\Rightarrow e^{x}dx=dt$$

$$\Rightarrow {\tan }^{-1}(y)=\int \frac{-2dt}{1+t^2}$$

$$\Rightarrow {\tan }^{-1}(y)=-2{\tan }^{-1}(t)+C$$

$$\Rightarrow {\tan }^{-1}(y)=-2{\tan }^{-1}(e^x)+C$$ [Putting value of t]

Given, y(0) = 0 means when x = 0 the y = 0

$$\therefore$$ $${\tan }^{-1}(0)=-2{\tan }^{-1}(e^o)+C$$

$$\Rightarrow 0=-2\times \frac{\pi }{4}+C$$

$$\Rightarrow C=\frac{\pi }{2}$$

$$\therefore$$ $${\tan }^{-1}(y)=2{\tan }^{-1}(e^x)+\frac{\pi }{2}$$

$$\Rightarrow y=\tan \left(-2{\tan }^{-1}(e^x)+\frac{\pi }{2}\right)$$

Differentiating both sides, we get

$$y^{\prime }={\sec }^2\left(-2{\tan }^{-1}(e^x)+\frac{\pi }{2}\right)-2.\frac{e^x}{1+e^{2x}}$$

$$={\sec }^2\left(-2{\tan }^{-1}(e^x)+\frac{\pi }{2}\right)\times \frac{-2e^x}{1+e^{2x}}$$

$$\therefore$$ $$y^{\prime }(0)={\sec }^2\left(-2{\tan }^{-1}(e^o)+\frac{\pi }{2}\right)\times \frac{-2e^o}{1+e^o}$$

$$={\sec }^2\left(-2\times \frac{\pi }{4}+\frac{\pi }{2}\right)\times \frac{-2}{2}$$

$$={\sec }^2(0)x-1$$

$$=1\times 1$$

$$=-1$$

And

$$y\left({\log }_e^{\sqrt{3}}\right)=\tan \left(-2{\tan }^{-1}\left(e^{{\log }_e^{\sqrt{3}}}\right)+\frac{\pi }{2}\right)$$

$$=\tan \left(-2{\tan }^{-1}(\sqrt{3})+\frac{\pi }{2}\right)$$

$$=\tan \left(-2\times \frac{\pi }{3}+\frac{\pi }{2}\right)$$

$$=\tan \left(-\frac{\pi }{6}\right)$$

$$=-\tan \left(\frac{\pi }{6}\right)$$

$$=-\frac{1}{\sqrt{3}}$$

$$\therefore$$ $$6\left(y^{\prime }(0)+{\left(y({\log }_e^{\sqrt{3}})\right)}^2\right)$$

$$=6\left(-1+{\left(\frac{-1}{\sqrt{3}}\right)}^2\right)$$

$$=6\left(-1+\frac{1}{3}\right)=6\times \frac{-2}{3}=-4$$