Animated Solution for Mathematics - Differential Equations: If $y=y(x)$ is the solution of the differential equation $(1+e^{2x})\frac{dy}{dx}+2(1+y^2)e^x=0$ and $y(0)=0$, then $6[y^{\prime }(0)+(y({\log }_e\sqrt{3}){)}^2]$ is equal to:

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a problem that might look like a tangled mess of exponentials and derivatives, but it is a beautifully choreographed dance.

When you first look at the equation $(1+e^{2x})\frac{dy}{dx}+2(1+y^2)e^x=0$, it is natural to feel a moment of hesitation. Remember, in the world of JEE Advanced, complexity is often just a mask for elegance.

Our first mission is to bring order to chaos by isolating our variables. By rearranging the terms, we arrive at the following separation:

Look at the left side—it is the classic derivative of ${\tan }^{-1}(y)$. On the right side, we observe $e^x$ in the numerator and $(e^x{)}^2$ in the denominator.

This is a perfect setup for a substitution. If we let $u=e^x$, then $du=e^{x}dx$. The integral becomes a standard form, and the path forward clears.

Integrating both sides, we obtain:

This $C$ is our constant of integration, the signature of our specific curve. We are given the initial condition $y(0)=0$.

Plugging these values in, we find $0=-2{\tan }^{-1}(1)+C$. Since ${\tan }^{-1}(1)=\frac{\pi }{4}$, we find that $C=\frac{\pi }{2}$. Our particular solution is now locked in:

We need to evaluate $6[y^{\prime }(0)+(y({\log }_e\sqrt{3}){)}^2]$. We do not need to differentiate the entire function to find $y^{\prime }(0)$.

Simply return to the original differential equation and substitute $x=0$ and $y=0$. This yields $2y^{\prime }(0)+2=0$, which gives us $y^{\prime }(0)=-1$.

Next, we tackle $y({\log }_e\sqrt{3})$. Substituting $x={\log }_e\sqrt{3}$ into our particular solution, we get $e^x=\sqrt{3}$.

Thus, ${\tan }^{-1}y=\frac{\pi }{2}-2{\tan }^{-1}(\sqrt{3})$. Since ${\tan }^{-1}(\sqrt{3})=\frac{\pi }{3}$, we have ${\tan }^{-1}y=\frac{\pi }{2}-\frac{2\pi }{3}=-\frac{\pi }{6}$. Therefore, $y=\tan (-\frac{\pi }{6})=-\frac{1}{\sqrt{3}}$.

We are at the finish line. We have $y^{\prime }(0)=-1$ and $y=-\frac{1}{\sqrt{3}}$. Plugging these into our expression:

The final result is -4. We started with a daunting differential equation, navigated through integration, applied initial conditions, and arrived at a clean, integer result. This is the beauty of mathematics—the logic holds, the steps align, and the truth reveals itself.