Q.Solve: $x^2\frac{dy}{dx}=x^2+xy+y^2$.

The key idea is that this is a homogeneous differential equation — the right-hand side can be written as a function of $y/x$.

Step 1: Rewrite in standard form

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

The equation becomes:

Step 3: Separate variables and integrate …

Why this approach works

When you see an equation like $x^2\frac{dy}{dx} = x^2 + xy + y^2$, the first instinct is to check if it's homogeneous. A differential equation is homogeneous if every term on the right-hand side has the same total degree when you treat $x$ and $y$ as variables. Here, $x^2$, $xy$, and $y^2$ are all degree 2 — so the right-hand side is a homogeneous function of degree 2 divided by $x^2$ (also degree 2). That means the ratio $\frac{dy}{dx}$ depends only on $\frac{y}{x}$, not on $x$ and $y$ separately.

The substitution $y = vx$ (where $v = y/x$) exploits this property. It turns the equation into one where $v$ and $x$ separate cleanly — and that's always solvable by integration.

Step-by-step solution

1. Rewrite the equation in standard form

Start with:

Divide both sides by $x^2$ (assuming $x \neq 0$):

This confirms the right-hand side is a function of $v = y/x$ alone.

2. Substitute $y = vx$

Let $v = \frac{y}{x}$, so $y = vx$. Differentiate with respect to $x$:

The original equation becomes:

3. Simplify to separate variables

Cancel $v$ from both sides:

Now the variables are separable — $v$ on one side, $x$ on the other:

$\frac{dv}{1 + v^2} = \frac{dx}{x}$ …

Method: Solving a homogeneous DE by the substitution $y=vx$

Use this whenever $\dfrac{dy}{dx}$ can be written as a function of $\dfrac{y}{x}$ alone (both parts the same degree).

Steps

Step 1: Confirm homogeneity and substitute

Rewrite the right side in terms of $\dfrac{y}{x}$. Put $y=vx$, so

Step 2: Reduce to separable in $v$ and $x$ …

Common Mistakes

Mistake 1: Not dividing by $x^2$ to expose $\frac{y}{x}$

Why it's wrong: homogeneity only becomes visible once the right side is written as $1+\frac{y}{x}+(\frac{y}{x})^2$. Correct approach: divide first, then substitute $y=vx$.

Mistake 2: Forgetting $\frac{dy}{dx}=v+x\frac{dv}{dx}$ …