UPSC 2025 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Formulation of differential equations · ODEs · asked 2× in 14 yrs · Read the full method →

Question

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

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

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.

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