Substitute y = vx to convert homogeneous ODE to separable form.

Homogeneous Differential Equation — (x² + y²)dx - 2xy dy = 0

Question

Solve the differential equation:

(x^2 + y^2)\,dx - 2xy\,dy = 0

This appeared in JEE Main 2023. The key signal that unlocks this problem is recognising it as a homogeneous equation — both terms have the same degree (2), so the substitution y = vx will reduce it to a separable form.

Solution — Step by Step

Rewrite as $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy}$ .

The right-hand side has degree 2 in both numerator and denominator, so it depends only on the ratio $y/x$ . That confirms homogeneity — we substitute y = vx.

Put $y = vx$ , so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ .

The right-hand side becomes:

\frac{x^2 + v^2x^2}{2x \cdot vx} = \frac{1 + v^2}{2v}

Our equation is now:

v + x\frac{dv}{dx} = \frac{1 + v^2}{2v}

x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}

Now separate variables — all $v$ terms on one side, all $x$ terms on the other:

\frac{2v}{1 - v^2}\,dv = \frac{dx}{x}

The left side is a standard log form. Notice that $\dfrac{d}{dv}(1 - v^2) = -2v$ , so:

\int \frac{2v}{1 - v^2}\,dv = -\ln|1 - v^2| + C_1

The right side gives $\ln|x|$ . Combining:

-\ln|1 - v^2| = \ln|x| + C

\ln|1 - v^2| + \ln|x| = -C = \ln|A| \quad \text{(say)}

x(1 - v^2) = A

Replace $v = y/x$ :

x\!\left(1 - \frac{y^2}{x^2}\right) = A

x \cdot \frac{x^2 - y^2}{x^2} = A

\boxed{x^2 - y^2 = Ax}

This is the general solution.

Why This Works

A homogeneous ODE has the structure $\dfrac{dy}{dx} = f\!\left(\dfrac{y}{x}\right)$ . When we substitute $y = vx$ , the function of $y/x$ becomes a function of $v$ alone — the $x$ dependence disappears from the right side. This converts a two-variable ODE into one separable in $v$ and $x$ .

The trick on the left integral — spotting that the numerator is (up to a sign) the derivative of the denominator — is a pattern you’ll see repeatedly in integration. Train yourself to check this before reaching for partial fractions.

The final curve $x^2 - y^2 = Ax$ is actually a family of rectangular hyperbolas passing through the origin. This makes geometric sense: the original equation has a neat orthogonal-trajectory interpretation.

Alternative Method — Exact Equation Check

Rewrite as $(x^2 + y^2)\,dx - 2xy\,dy = 0$ . Check exactness:

M = x^2 + y^2, \quad N = -2xy

\frac{\partial M}{\partial y} = 2y, \quad \frac{\partial N}{\partial x} = -2y

Not exact. But compute the integrating factor using $\dfrac{M_y - N_x}{N}$ :

\frac{2y - (-2y)}{-2xy} = \frac{4y}{-2xy} = \frac{-2}{x}

This depends on $x$ alone, so $\mu = e^{\int -2/x\,dx} = x^{-2}$ .

Multiply through by $\mu = 1/x^2$ :

\frac{x^2 + y^2}{x^2}\,dx - \frac{2y}{x}\,dy = 0

Now verify exactness and integrate — you’ll arrive at the same answer $x^2 - y^2 = Ax$ . This method is cleaner if you’re comfortable with integrating factors, though most students find the $y = vx$ route faster under exam conditions.

In JEE, both homogeneous substitution and the integrating factor approach are valid. The substitution $y = vx$ is almost always faster. Use the exact-equation route only if the problem specifically asks you to find an integrating factor.

Common Mistake

The most common slip: students write $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy}$ and then substitute correctly, but mess up the algebra when subtracting $v$ from the right side. They forget to find a common denominator:

\frac{1+v^2}{2v} - v \neq \frac{1+v^2 - v}{2v}

The correct step is $v = \dfrac{2v^2}{2v}$ , so the numerator becomes $1 + v^2 - 2v^2 = 1 - v^2$ . This error flips the entire integral and gives a wrong answer that still “looks” like a valid family of curves — so you won’t catch it by sanity-checking the form.

Question

Solution — Step by Step

Why This Works

Alternative Method — Exact Equation Check

Common Mistake

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