Skip to content
Miscellaneous Exercise 6(I) · Q108

Q.x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 is a solution of... (A) d2ydx2+yx+(dydx)2=0\dfrac{d^2y}{dx^2}+yx+\left(\dfrac{dy}{dx}\right)^2=0 (B) xyd2ydx2+2(dydx)2−ydydx=0xy\dfrac{d^2y}{dx^2}+2\left(\dfrac{dy}{dx}\right)^2-y\dfrac{dy}{dx}=0 (C) yd2ydx2+2(dydx)2+y=0y\dfrac{d^2y}{dx^2}+2\left(\dfrac{dy}{dx}\right)^2+y=0 (D) xydydx+yd2ydx2=0xy\dfrac{dy}{dx}+y\dfrac{d^2y}{dx^2}=0

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
61% · 108/177 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Eliminating the two arbitrary constants a,ba,b from x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 by differentiating twice and combining the resulting relations gives xyd2ydx2+x(dydx)2−ydydx=0xy\dfrac{d^2y}{dx^2}+x\left(\dfrac{dy}{dx}\right)^2-y\dfrac{dy}{dx}=0 — the same structural family as option (B), which is the intended match (the printed coefficient of the middle …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.