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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 |
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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)2 = (dy/dx - d3y/dx3)3 ∴ Order = 3, Degree = 3 ii. (2 - y")2 = y"2 + 2y' 4 - 4y" + (y")2 = (y")2 + 2y' ⇒ 4 - 4y" = 2y' ∴ Order = 2, Degree = 1 iii. (dy/dx)3 + y = x - dx/dy multiplting the equation by dy/dx (dy/dx)4 + y(dy/dx) = x(dy/dx) - 1 ∴ Order = 1, Degree = 4 |
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