UPSC 2025 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

First-order higher-degree ODEs · ODEs · asked 5× in 13 yrs · Read the full method →

Question

Solve $\left(1 - y^2 + \dfrac{y^4}{x^2}\right)\left(\dfrac{dy}{dx}\right)^2 - 2\dfrac{y}{x}\dfrac{dy}{dx} + \dfrac{y^2}{x^2} = 0$ .

Technique

This is a first-order ODE of degree two. Clear $x^2$ to recognise it as a perfect square in $p=\dfrac{dy}{dx}$ , integrate the two resulting first-order ODEs, and read off the singular solution from the discriminant locus.

Solution

Write $p=\dfrac{dy}{dx}$ . Multiplying the equation by $x^2$ :

\bigl(x^2 - x^2y^2 + y^4\bigr)p^2 - 2xy\,p + y^2 = 0.

Key step — perfect square. Group the terms:

\underbrace{\bigl(x^2p^2 - 2xy\,p + y^2\bigr)}_{(px-y)^2} + \underbrace{\bigl(-x^2y^2p^2 + y^4p^2\bigr)}_{p^2y^2(y^2-x^2)} = 0,

(px - y)^2 = p^2 y^2 (x^2 - y^2).

Taking square roots,

px - y = \pm\,p\,y\sqrt{x^2 - y^2}\quad\Longrightarrow\quad p\bigl(x \mp y\sqrt{x^2-y^2}\bigr) = y.

Hence $\dfrac{dy}{dx}\bigl(x \mp y\sqrt{x^2-y^2}\bigr) = y$ , i.e.

x\,dy - y\,dx = \pm\, y\sqrt{x^2-y^2}\;dy.

Integrate. Divide by $y^2$ and recall $\dfrac{x\,dy - y\,dx}{y^2} = -\,d\!\left(\dfrac{x}{y}\right)$ . Put $u = \dfrac{x}{y}$ , so $\sqrt{x^2-y^2} = y\sqrt{u^2-1}$ (taking $y>0$ ):

-\,du = \pm\,\sqrt{u^2-1}\;dy \quad\Longrightarrow\quad \frac{du}{\sqrt{u^2-1}} = \mp\,dy.

Integrating gives $\cosh^{-1}u = \mp\,y + \text{const}$ , i.e.

\cosh^{-1}\!\left(\frac{x}{y}\right) = y + c \qquad\Longleftrightarrow\qquad x = y\cosh(y + c).

(The two signs collapse to one family on absorbing the sign into the arbitrary constant $c$ .)

Singular solution. The discriminant of the quadratic in $p$ is

B^2 - 4AC = \frac{4y^4(x^2-y^2)}{x^4},

which vanishes on $x^2 = y^2$ . The lines $y = x$ and $y = -x$ satisfy the original equation (with $y=\pm x$ , $p=\pm 1$ , the equation reduces to $0$ ), so they are singular solutions.

Answer

\boxed{\;x = y\cosh(y + c)\quad\text{(general solution)},\qquad y = \pm x\quad\text{(singular solutions)}.\;}