Skip to content
Miscellaneous Exercise 6(II) · Q113

Q.Determine the order and degree: 1+(dydx)23=d2ydx2\sqrt[3]{1+\left(\dfrac{dy}{dx}\right)^2}=\dfrac{d^2y}{dx^2}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
64% · 113/177 Questions
✓ Free question

1+(dydx)23=d2ydx2\sqrt[3]{1+\left(\dfrac{dy}{dx}\right)^2}=\dfrac{d^2y}{dx^2}. Cubing: 1+(dydx)2=(d2ydx2)31+\left(\dfrac{dy}{dx}\right)^2=\left(\dfrac{d^2y}{dx^2}\right)^3. Order 22 (from d2ydx2\dfrac{d^2y}{dx^2}), degree 33.

✓Final answer

order 2, degree 3

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.