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Differential Equation of a Hyperbola | JEE

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Question
The equation dydx=x2+y2+12xy\dfrac{dy}{dx} = \dfrac{x^2 + y^2 + 1}{2xy}, with y(1)=1y(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=y2u = y^2 (so dudx=2ydydx\dfrac{du}{dx} = 2y\dfrac{dy}{dx}):
2xydydx−y2=x2+1⇒xdudx−u=x2+1.2xy\frac{dy}{dx} - y^2 = x^2 + 1 \Rightarrow x\frac{du}{dx} - u = x^2 + 1.
 Step 2: Linear form and IF:
dudx−1xu=x+1x,IF=1x.\frac{du}{dx} - \frac{1}{x}u = x + \frac{1}{x}, \qquad \text{IF} = \frac{1}{x}.
ux=∫(1+1x2)dx=x−1x+c⇒y2=x2−1+cx.\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)=1y(1)=1: 1=1−1+c⇒c=11 = 1 - 1 + c \Rightarrow c = 1, so y2=x2+x−1y^2 = x^2 + x - 1, i.e.
(x+12)2−y2=54,\left(x + \tfrac{1}{2}\right)^2 - y^2 = \tfrac{5}{4},
 which is a hyperbola. Correct answer: (3)
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