Animated Solution for Mathematics - Differential Equations: If $x=f(y)$ is the solution of the differential equation $(1+y^2)+(x-2e^{{\tan }^{-1}y})\frac{dy}{dx}=0,$ $y\in (-\frac{\pi }{2},\frac{\pi }{2})$ with $f(0)=1$, then $f(\frac{1}{\sqrt{3}})$ is equal to:

Imagine you are standing in front of a complex puzzle. The differential equation $(1+y^2)+(x-2e^{{\tan }^{-1}y})\frac{dy}{dx}=0$ is exactly that—a puzzle designed to test your intuition.

At first glance, it looks like a chaotic mess of $y^2$ and exponential functions. If you try to solve this as a linear differential equation in $y$, you will quickly find yourself hitting a wall.

But here is the secret of the JEE Advanced masters: sometimes, the solution isn't found by pushing forward, but by shifting your perspective. What if we treat $x$ as the dependent variable and $y$ as the independent variable?

By multiplying the entire equation by $\frac{dx}{dy}$, we transform the equation into:

Suddenly, the structure begins to emerge.

Now that we have $\frac{dx}{dy}$ in the mix, our goal is to force this equation into the standard linear form: $\frac{dx}{dy}+P(y)x=Q(y)$.

To do this, we divide the entire equation by $(1+y^2)$. This gives us:

Now, look at that! It is a perfectly linear equation in $x$. We can clearly identify our components:

Identifying these correctly is the most critical step; a single sign error here would derail the entire process.

With our standard form established, we reach for our most powerful tool: the Integrating Factor (IF). For an equation linear in $x$, the IF is defined as:

Substituting our $P(y)$, we get $IF=e^{\int \frac{1}{1+y^2}dy}$. Recalling our fundamental calculus, we know that the integral of $\frac{1}{1+y^2}$ is simply ${\tan }^{-1}y$.

Thus, our Integrating Factor simplifies beautifully to $e^{{\tan }^{-1}y}$. This factor is the key that unlocks the differential equation, turning the left-hand side into the derivative of the product $x\cdot e^{{\tan }^{-1}y}$.

We now write the general solution:

Substituting our $Q(y)$, we have:

Simplifying the integrand, we get:

This looks intimidating, but let's use a simple substitution: let $u={\tan }^{-1}y$. Then $du=\frac{1}{1+y^2}dy$.

The integral transforms into $\int 2e^{2u}du$, which is simply $e^{2u}$. Substituting back, we get:

We are almost there! We use the initial condition $f(0)=1$, which means when $y=0$, $x=1$.

Plugging these into our equation, we get $1\cdot e^0=e^0+C$, which simplifies to $1=1+C$, meaning $C=0$. Our specific solution is $x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}$, or $f(y)=e^{{\tan }^{-1}y}$.

Finally, to find $f(\frac{1}{\sqrt{3}})$, we substitute $y=\frac{1}{\sqrt{3}}$. Since ${\tan }^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi }{6}$, our final answer is:

You have successfully navigated the complexity and arrived at the elegant truth!