← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q5a — Solution

10 marks · Section B

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

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

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