Q.Solve the following differential equation: $(x^2 - y^2) dx + 2xy dy = 0$

Concept: Homogeneous Differential Equation — the equation is of the form $M dx + N dy = 0$ where $M$ and $N$ are homogeneous functions of the same degree.

Step 1: Check homogeneity

$M = x^2 - y^2$, $N = 2xy$ — both are degree 2. So substitute $y = vx$, $dy = v dx + x dv$.

Step 2: Substitute and simplify

$(x^2 - v^2 x^2) dx + 2x(vx)(v dx + x dv) = 0$

$x^2(1 - v^2) dx + 2v x^2 (v dx + x dv) = 0$

$x^2[(1 - v^2 + 2v^2) dx + 2v x dv] = 0$

$x^2[(1 + v^2) dx + 2v x dv] = 0$

Step 3: Separate variables …

1. Recognising the type

Look at the equation:

Every term in $dx$ and $dy$ is a polynomial of degree 2. That’s the hallmark of a homogeneous differential equation: the coefficients of $dx$ and $dy$ are homogeneous functions of the same degree.

When you see that, the standard trick is to set $y = vx$ (or $x = vy$). Why? Because then every term becomes a function of $v$ times a power of $x$, and the $x$ factors cancel beautifully, leaving a separable equation in $v$ and $x$.

2. Substituting $y = vx$

Let $y = vx$. Then $dy = v\,dx + x\,dv$.

Plug into the equation:

Simplify each piece:

• $x^2 - v^2 x^2 = x^2(1 - v^2)$
• $2x(vx) = 2vx^2$

So the equation becomes:

Factor $x^2$ out of both terms:

Since $x^2 \neq 0$ (we can handle $x=0$ separately later), we divide through by $x^2$:

Combine the $dx$ terms:

3. Separating variables

Now we have a separable equation:

Divide both sides by $x(1+v^2)$ (assuming $x \neq 0$):

4. Integrating both sides

Integrate:

The left side gives $\log|x|$. For the right side, use the substitution $u = 1+v^2$, $du = 2v\,dv$:

(Since $1+v^2 > 0$, we can drop the absolute value.)

So we have:

where $\log|C|$ is the constant of integration (written as $\log|C|$ for convenience).

Combine the logs:

Exponentiate both sides:

$|x| = \frac{|C|}{1+v^2}$ …

Method: Homogeneous equation via $y=vx$ (giving $x^2+y^2=Cx$)

Use this for a homogeneous first-order equation like $(x^2-y^2)\,dx+2xy\,dy=0$, where every term has the same degree so $\dfrac{dy}{dx}$ depends only on $\dfrac{y}{x}$.

Steps

Step 1: Write as $\dfrac{dy}{dx}=F(y/x)$ and substitute $y=vx$

Solve for the derivative, then set $y=vx$, $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.

Step 2: Separate $v$ from $x$

After substituting, isolate $x\dfrac{dv}{dx}$ and separate; here the $v$-side becomes $\dfrac{2v}{1+v^2}\,dv$, a logarithmic form. …

Common Mistakes

Mistake 1: Sign error solving for $\dfrac{dy}{dx}$

Why it's wrong: from $(x^2-y^2)dx+2xy\,dy=0$, $\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$; a dropped sign changes the whole integral. Correct approach: isolate the derivative carefully before substituting $y=vx$.

Mistake 2: Not recognising $\dfrac{2v}{1+v^2}$ as a log integrand …