Skip to main contentSkip to solution

Match the LIST-I with LIST-II

LIST-I Differential equationLIST-II Order and degree
A. y′′=[1+(y′)2]3/2y'' = [1+(y')^2]^{3/2}I. 3, 4
B. y′′′+y′+y=2xy''' + y' + y = 2xII. 2, 2
C. 2(y′′)3−(y′)4+y=02(y'')^3 - (y')^4 + y = 0III. 3, 1
D. (y′′′)4+(y′′)3+y′+y=0(y''')^4 + (y'')^3 + y' + y = 0IV. 2, 3

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

Order is the highest derivative present; degree is its power after removing radicals.

A: squaring gives (y′′)2=[1+(y′)2]3(y'')^2 = [1+(y')^2]^3, so order 2, degree 2 (II).

B: order 3, degree 1 (III).

C: order 2, degree 3 (IV).

D: order 3, degree 4 (I).

Keyboard Shortcuts

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