Animated Solution for Mathematics - Differential Equations: The curve amongst the family of curves, represented by the differential equation, $(x^2-y^2)dx+2xydy=0$ which passes through $(1,1)$ is :

Unlocking the Homogeneous Equation

Welcome, future engineers. Today, we are not just solving a differential equation; we are embarking on a journey of pattern recognition.

When you look at the equation $(x^2-y^2)dx+2xydy=0$, what do you see? Do you see a wall of algebra, or do you see a hidden symmetry?

In the world of JEE Advanced, the ability to spot the 'soul' of an equation is what separates the good from the great. Let us peel back the layers of this problem together.

The Recognition

First, let us analyze the structure. We have terms like $x^2$, $y^2$, and $2xy$.

Notice the degree of each term. $x^2$ is degree 2, $y^2$ is degree 2, and $2xy$ is degree 2 because $1+1=2$.

This is the defining characteristic of a first-order homogeneous differential equation. When every term in your equation shares the same degree, the universe is telling you that there is a specific, elegant path to the solution.

The Master Key Substitution

Whenever you see this homogeneity, the substitution $y=vx$ is your best friend. It transforms the equation into a separable one, effectively changing your coordinate system to align with the flow of the curve.

Since $y=vx$, we must also find the derivative. Using the product rule, we differentiate with respect to $x$:

Now, let us substitute this into our rearranged equation. We started with $\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$. Substituting $y=vx$, we get:

Look at the magic that happens next. We factor out $x^2$ from the numerator and denominator:

The $x^2$ terms cancel out completely. We are left with $v+x\frac{dv}{dx}=\frac{v^2-1}{2v}$. The variable $x$ is no longer tangled with $y$; we have successfully decoupled the system.

The Calculus Journey

Now, we isolate the variables. Moving $v$ to the right side gives us:

Separating the variables, we bring all $v$ terms to the left and all $x$ terms to the right:

This is a beautiful integral. On the left, the numerator $2v$ is exactly the derivative of the denominator $v^2+1$. This is the classic $\int \frac{f^{\prime }(v)}{f(v)}dv=\ln |f(v)|$ form.

Integrating both sides, we get $\ln (v^2+1)=-\ln |x|+\ln C$. Using the properties of logarithms, this simplifies to $\ln (v^2+1)=\ln (\frac{C}{x})$, which implies:

The Geometric Revelation

We are almost home. Substituting $v=\frac{y}{x}$ back into our equation, we get $(\frac{y}{x}{)}^2+1=\frac{C}{x}$. Multiplying by $x^2$, we arrive at $y^2+x^2=Cx$, or $x^2+y^2-Cx=0$.

We are told the curve passes through $(1,1)$. Plugging these coordinates in, we find $1^2+1^2=C(1)$, so $C=2$. Our final equation is:

To see the geometry, we rearrange: $x^2-2x+y^2=0$. Completing the square for $x$, we add $1$ to both sides:

This is a circle with center $(1,0)$ and radius $1$.

Look at the center $(1,0)$. The $y$-coordinate is zero, meaning the center lies perfectly on the $x$-axis. We have solved the mystery! The curve is a circle with its center on the $x$-axis.