Solve: (x² + y²)dx - 2xy dy = 0 — homogeneous differential equation

Question

Solve the differential equation:

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

(NCERT Class 12, Exercise 9.5)

Solution — Step by Step

\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}

Check: replace $x$ with $tx$ and $y$ with $ty$ in the RHS. Every term has degree 2 in numerator and denominator, so $t^2$ cancels. This confirms the equation is homogeneous.

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

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

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

Left side: let $u = 1 - v^2$ , then $du = -2v\,dv$ :

\int \frac{2v}{1 - v^2}\,dv = -\int \frac{du}{u} = -\ln|u| = -\ln|1 - v^2|

Right side: $\int \frac{dx}{x} = \ln|x|$

So: $-\ln|1 - v^2| = \ln|x| + \ln|C_1|$

\ln|1 - v^2| = -\ln|C_1 x|

|1 - v^2| = \frac{1}{|C_1 x|}

1 - \frac{y^2}{x^2} = \frac{C}{x} \quad \text{(absorbing signs into constant } C\text{)}

\frac{x^2 - y^2}{x^2} = \frac{C}{x}

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

Why This Works

Homogeneous DEs have the property that $\frac{dy}{dx}$ depends only on the ratio $y/x$ . The substitution $v = y/x$ converts this ratio-dependence into a separable equation in $v$ and $x$ . Once separated, both sides integrate to logarithms (or standard forms), and we substitute back.

The solution $x^2 - y^2 = Cx$ represents a family of circles. Rewriting: $(x - C/2)^2 - y^2 = C^2/4$ , which is actually a family of hyperbolas passing through the origin.

Alternative Method — Exact equation approach

Rewrite as $M\,dx + N\,dy = 0$ where $M = x^2 + y^2$ and $N = -2xy$ .

Check: $\frac{\partial M}{\partial y} = 2y$ and $\frac{\partial N}{\partial x} = -2y$ . Since these aren’t equal, the equation is not exact. But $\frac{M_y - N_x}{N} = \frac{2y + 2y}{-2xy} = \frac{-2}{x}$ , which depends only on $x$ . So an integrating factor $\mu = x^{-2}$ makes it exact.

For CBSE boards, the homogeneous substitution ( $y = vx$ ) is the expected method. Don’t use the exact equation approach unless the question specifically asks for it. The marking scheme maps to: identify type (1 mark), substitute (1 mark), separate (1 mark), integrate (1 mark), back-substitute (1 mark).

Common Mistake

When separating variables, students sometimes write $\frac{2v}{1 + v^2}$ instead of $\frac{2v}{1 - v^2}$ . This happens when the subtraction $\frac{1+v^2}{2v} - v$ is done carelessly. Always simplify: $\frac{1+v^2 - 2v^2}{2v} = \frac{1-v^2}{2v}$ . The sign difference between $1 + v^2$ and $1 - v^2$ changes the integral completely — one gives $\tan^{-1}$ , the other gives $\ln$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Exact equation approach

Common Mistake

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