Animated Solution for Mathematics - Differential Equations: Let $\alpha x=\exp (x^{\beta }{y}^{\gamma })$ be the solution of the differential equation $2x^{2}ydy-(1-x{y}^2)dx=0,x>0,y(2)=\sqrt{{\log }_{e}2}$. Then $\alpha +\beta -\gamma$ equals :

Imagine you are standing on the edge of a mathematical cliff, staring at the differential equation $2x^{2}y\frac{dy}{dx}-(1-x{y}^2)=0$. At first glance, it looks intimidating, perhaps even impossible.

But in the world of JEE Advanced, every equation is a puzzle waiting for the right key. Our journey begins by rearranging the terms. We move the $dx$ component to the right, and after a bit of algebraic maneuvering, we find ourselves with:

Now, look closer. The equation is not linear in $y$, but it has a hidden symmetry. If we cross-multiply and bring the $y$ terms to one side, we get $2x^{2}y\frac{dy}{dx}+x{y}^2=1$.

To isolate the derivative part, we divide the entire equation by $x^2$. This gives us:

Notice the structure here? It is starting to look familiar. We have a $y^2$ term and a $2y\frac{dy}{dx}$ term.

This is the 'Aha!' moment. We can make this linear using a clever substitution. Let $t=y^2$. Differentiating with respect to $x$, we get:

This is perfect because $2y\frac{dy}{dx}$ is exactly the first term in our rearranged equation.

With our substitution $t=y^2$ and $\frac{dt}{dx}=2y\frac{dy}{dx}$, the equation transforms into:

This is a standard linear differential equation of the form $\frac{dt}{dx}+P(x)t=Q(x)$, where $P(x)=\frac{1}{x}$ and $Q(x)=\frac{1}{x^2}$.

To solve this, we need the Integrating Factor (I.F.). The formula is:

The integrating factor is simply $x$. Now, the general solution is $t\cdot (I.F.)=\int Q(x)\cdot (I.F.)dx+C$.

Substituting our values, we get $t\cdot x=\int \frac{1}{x^2}\cdot xdx+C$, which simplifies to:

Integrating $\frac{1}{x}$ gives us $\ln x$, so $t\cdot x=\ln x+C$. Replacing $t$ with $y^2$, we get our general solution:

We are given the initial condition $y(2)=\sqrt{\ln 2}$. Substituting $x=2$ and $y=\sqrt{\ln 2}$ into our general solution, we get:

This simplifies to $2\ln 2=\ln 2+C$. Thus, $C=\ln 2$.

Substituting $C$ back, we get $y^{2}x=\ln x+\ln 2=\ln (2x)$. Taking the exponential on both sides, we get:

Rearranging this to match the form $\alpha x=\exp (x^{\beta }{y}^{\gamma })$, we get $2x=\exp (x^1{y}^2)$. Comparing this with the target expression, we find $\alpha =2$, $\beta =1$, and $\gamma =2$.

The question asks for $\alpha +\beta -\gamma$, which is $2+1-2=1$. And there it is—the elegance of the final cancellation. You have successfully navigated the complexity and arrived at the answer 1.