Q.Prove that $x^2 - y^2 = c(x^2 + y^2)^2$ is the general solution of differential equation $(x^3 - 3xy^2)\, dx = (y^3 - 3x^2 y)\, dy$, where $c$ is a parameter.

The coefficients are homogeneous of degree 3, so substitute $y=vx$.

Write $\frac{dy}{dx}=\frac{x^3-3xy^2}{y^3-3x^2y}$. With $y=vx$, $\frac{dy}{dx}=v+x\frac{dv}{dx}$:

Subtract $v$:

Separate and integrate (put $s=v^2$ on the left):

Multiply by $2$ and combine logs:

Put $v=\frac{y}{x}$: since $1-v^2=\frac{x^2-y^2}{x^2}$ and $(1+v^2)^2=\frac{(x^2+y^2)^2}{x^4}$, the $x$-powers cancel:

Why homogeneous

Write the equation as

Every term of numerator and denominator has total degree 3, so the right side depends only on $y/x$. Substituting $y=vx$ collapses it to a separable equation.

Substitute $y=vx$

With $\frac{dy}{dx}=v+x\frac{dv}{dx}$,

so

Separate the variables

Subtract $v$:

hence

Integrate the left side

The numerator is odd in $v$, so put $s=v^2$, $ds=2v\,dv$:

Partial fractions give $\frac{s-3}{(1-s)(1+s)}=\frac{-1}{1-s}+\frac{-2}{1+s}$, so

Therefore

Combine and return to $x,y$

Multiply by $2$:

With $v=\frac{y}{x}$,

so

Cancel $x^2$:

This is exactly the family we were asked to prove, so it is the general solution.

Method: Proving a given family is the solution of a homogeneous equation

To prove a stated curve is the general solution, solve the equation by $y=vx$ and show the result matches the given family.

Steps

Step 1: Confirm homogeneity.

Write $\dfrac{dy}{dx}=\dfrac{M}{N}$; if all terms share one degree, substitute $y=vx$.

Step 2: Separate after subtracting $v$.

Reach $x\dfrac{dv}{dx}=F(v)-v$ and split variables.

Step 3: Integrate (partial fractions / $s=v^2$).

For a numerator odd in $v$, the substitution $s=v^2$ plus partial fractions handles $\dfrac{v^3-3v}{1-v^4}$.

Step 4: Return to $x,y$ and match.

Put $v=\dfrac{y}{x}$; the powers of $x$ cancel to leave exactly the given family, proving it.

Common Mistakes

Mistake 1: Not confirming homogeneity before substituting.

Why it's wrong: $\dfrac{dy}{dx}=\dfrac{x^3-3xy^2}{y^3-3x^2y}$ has all terms degree $3$, which justifies $y=vx$; skipping this risks the wrong method. Correct approach: check the degree first.

Mistake 2: Botching the partial fractions of $\dfrac{v^3-3v}{1-v^4}$.

Why it's wrong: the substitution $s=v^2$ then partial fractions is needed; a wrong split gives the wrong logs and fails to match the target. Correct approach: use $s=v^2$ and integrate $\tfrac12\int\tfrac{s-3}{1-s^2}\,ds$.

Mistake 3: Not cancelling the powers of $x$ when returning to $x,y$.

Why it's wrong: with $v=y/x$, $\dfrac{1-v^2}{(1+v^2)^2}$ carries an $x^2$ that must cancel against $cx^2$ to give exactly $x^2-y^2=c(x^2+y^2)^2$. Correct approach: substitute and simplify fully.