UPSC Maths 2025 Paper 1 Q5b — Solution

10 marks · Section B

Question

Form the differential equation of all ellipses whose axes coincide with coordinate axes.

Technique

Eliminate the arbitrary constants: the family has two essential parameters, so differentiate twice and eliminate them to get a second-order ODE.

Solution

An ellipse with centre at the origin and axes along the coordinate axes is $\frac{x^2}{A} + \frac{y^2}{B} = 1,$ where $A,B$ are two arbitrary (positive) constants. Two constants $\Rightarrow$ a second-order ODE.

First differentiation with respect to $x$ : $\frac{2x}{A} + \frac{2y}{B}\,y' = 0 \quad\Longrightarrow\quad \frac{x}{A} + \frac{y\,y'}{B} = 0. \tag{1}$

Second differentiation of (1): $\frac{1}{A} + \frac{(y')^2 + y\,y''}{B} = 0. \tag{2}$

Eliminate $A,B$ . From (1), $\dfrac{1}{A} = -\dfrac{y\,y'}{B\,x}$ . Substitute into (2): $-\frac{y\,y'}{B\,x} + \frac{(y')^2 + y\,y''}{B} = 0.$

Multiply by $Bx$ (and cancel the common $B$ ): $-y\,y' + x\bigl[(y')^2 + y\,y''\bigr] = 0,$ that is, $x\,y\,y'' + x\,(y')^2 - y\,y' = 0.$

Answer

$\boxed{\;x\,y\,\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\,\frac{dy}{dx} = 0.\;}$