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Differential Equation of a Hyperbola | JEE

JEE Maths question with a full step-by-step solution.

Question

The equation

\dfrac{dy}{dx} = \dfrac{x^2 + y^2 + 1}{2xy}

, with

y(1) = 1

, is the differential equation of:

Aa system of parabolas

Ba system of circles

Ca hyperbolacorrect

Da pair of lines

Solution

Step 1: Clear the fraction and substitute

u = y^2

(so

\dfrac{du}{dx} = 2y\dfrac{dy}{dx}

2xy\frac{dy}{dx} - y^2 = x^2 + 1 \Rightarrow x\frac{du}{dx} - u = x^2 + 1.

Step 2: Linear form and IF:

\frac{du}{dx} - \frac{1}{x}u = x + \frac{1}{x}, \qquad \text{IF} = \frac{1}{x}.

\frac{u}{x} = \int\left(1 + \frac{1}{x^2}\right)dx = x - \frac{1}{x} + c \Rightarrow y^2 = x^2 - 1 + cx.

Step 3: Apply

y(1)=1

1 = 1 - 1 + c \Rightarrow c = 1

, so

y^2 = x^2 + x - 1

, i.e.

\left(x + \tfrac{1}{2}\right)^2 - y^2 = \tfrac{5}{4},

which is a hyperbola. Correct answer: (3)

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