Skip to main contentSkip to solution

If pp and qq are degree and order of the differential equation (d2ydx2)3−2(dydx)4=6\left(\dfrac{d^2y}{dx^2}\right)^3 - 2\left(\dfrac{dy}{dx}\right)^4 = 6 respectively. Then the value of 2p+3q2p+3q is

Solution

✅ Correct Option: 2

The highest derivative is d2ydx2\frac{d^2y}{dx^2}, so the order is q=2q=2, and its highest power is 33, so the degree is p=3p=3. Hence 2p+3q=6+6=122p+3q=6+6=12.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question