Q.Find the general solution of $y^2\,dx+(x^2-xy+y^2)\,dy=0$.

The key idea is that this is a homogeneous differential equation — each term in $dx$ and $dy$ is of degree 2.

Step 1: Rewrite as

Step 2: Substitute $x = v y$, so $\frac{dx}{dy} = v + y\frac{dv}{dy}$. The equation becomes

Step 3: Simplify

$y\frac{dv}{dy} = -v^2 - 1 \quad\Rightarrow\quad \frac{dv}{v^2+1} = -\frac{dy}{y}$ …

The key is recognising the structure of the equation. When you see terms like $y^2\,dx$, $x^2\,dy$, and $xy\,dy$, the degrees of each term are the same — every term is of degree 2. That’s the hallmark of a homogeneous differential equation.

For a homogeneous equation, the standard trick is to substitute $x = vy$ (or $y = ux$). Why? Because it turns the equation into one where variables separate cleanly. Here, since the $dy$ term has more structure, substituting $x = vy$ works smoothly.

Let’s go through it step by step.

1. Rewrite the equation The given equation is:

We can write it as:

Or equivalently:

2. Substitute $x = vy$ Let $x = v y$, where $v$ is a function of $y$. Then:

Substitute into the equation:

3. Simplify the right-hand side Compute the numerator:

Dividing by $y^2$ gives:

So the equation becomes:

4. Separate variables Bring $v$ to the right:

So:

Now separate:

5. Integrate both sides The left integral is standard:

The right integral:

So:

6. Back-substitute $v = x/y$ Replace $v$: …

Method: Homogeneous DE solved by $x=vy$

Use this for a homogeneous equation where substituting $x=vy$ (rather than $y=vx$) is tidier — often when the $dx$ coefficient is the simpler one.

Steps

Step 1: Write $\dfrac{dx}{dy}$ as a function of $\dfrac{x}{y}$

Every term of degree two lets you divide by $y^2$ so only $\dfrac{x}{y}$ survives.

Step 2: Substitute $x=vy$

Then $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$; the equation reduces to $y\dfrac{dv}{dy}=(\text{function of }v)$. …

Common Mistakes

Mistake 1: Not recognising the equation is homogeneous (degree 2 terms)

Why it's wrong: $y^2\,dx+(x^2-xy+y^2)\,dy=0$ has all degree-2 terms, so $x=vy$ works. Correct approach: divide by $y^2$ and substitute.

Mistake 2: Choosing $y=vx$ when $x=vy$ is cleaner

Why it's wrong: because the $dx$ coefficient is simplest, $x=vy$ avoids messy algebra. Correct approach: substitute $x=vy$, $\frac{dx}{dy}=v+y\frac{dv}{dy}$. …