1.

Find the order and degree of the following differential equations.i. d2y/dx2 + y + (dy/dx - d3y/dx3)3/2 = 0ii. (2 - y")2 = y"2 + 2y'iii. (dy/dx)3 + y = x - dx/dy

Answer»

i. d2y/dx2 + y + (dy/dx - d3y/dx3)3/2 = 0

Here we elimenate the radical sign

For this write the equation as  d2y/dx2 + y = -(dy/dx - d3y/dx3)3/2

Squaring both the sides, we get (d2y/dx2 + y)= (dy/dx - d3y/dx3)3

∴ Order = 3, Degree = 3

ii. (2 - y")2 = y"2 + 2y'

4 - 4y" + (y")= (y")+ 2y'

⇒ 4 - 4y" = 2y'

∴ Order = 2, Degree = 1

iii. (dy/dx)+ y = x - dx/dy

multiplting the equation by dy/dx

(dy/dx)+ y(dy/dx) = x(dy/dx) - 1

∴ Order = 1,     Degree = 4



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