Animated Solution for Mathematics - Differential Equations: The solution of the differential equation $\frac{dy}{dx}=-\left(\frac{x^2+3y^2}{3x^2+y^2}\right),y(1)=0$ is

Imagine you are standing before a complex differential equation:

At first glance, it looks like a tangled mess of variables. But look closer at the symmetry; every term in the numerator and denominator has a degree of 2.

This is not a coincidence; it is the hallmark of a Homogeneous Differential Equation. This structure is a gift, and our strategy is to exploit it.

To break the deadlock, we use the substitution $y=vx$. This is our bridge from the world of $x$ and $y$ to the world of $v$ and $x$.

By differentiating $y=vx$ with respect to $x$, we get:

Now, we substitute this into our original equation. The left side becomes $v+x\frac{dv}{dx}$, and the right side transforms into:

Notice how the $x^2$ terms factor out and vanish. This is the magic of homogeneity, leaving us with:

Now, we move the $v$ to the right side:

Combining these terms requires a steady hand. The numerator becomes $-(1+3v^2+3v+v^3)$, which is the expansion of $-(v+1{)}^3$.

Our equation is now beautifully separable:

Integrating the right side is trivial: $-\ln |x|+C$. But the left side is where the artistry lies.

Instead of a long, tedious partial fraction decomposition, we rewrite the numerator $3+v^2$ as $(v+1-1{)}^2+3$. This expands to $(v+1{)}^2-2(v+1)+4$.

Dividing this by $(v+1{)}^3$ gives us three simple integrals:

These are standard power rule integrals. We obtain:

We are almost there. We substitute $v=\frac{y}{x}$ back into our expression.

The term $\ln |v+1|$ becomes $\ln |\frac{y}{x}+1|$, which is $\ln |\frac{x+y}{x}|$. Using the property $\ln |\frac{x+y}{x}|=\ln |x+y|-\ln |x|$, we see the $\ln |x|$ terms on both sides cancel out.

With the initial condition $y(1)=0$, we find $C=0$.

The final result is:

This is a testament to the power of symmetry and substitution. You have conquered the equation!