The order and degree of the differential equation ( d^3 y d x^3 )^ 1 2 -2 ( d y d x )^ 1 4 +x y=0 are respectively

Step-by-step solution

Given, y₃^ 1 2 -2 (y₁ )^ 1 4 +x y=0 (y₃^ 1 2 -x y )^4= [2 (y₁ )^ 1 4 ]^4 aligned & y₃^ 1 2 [ ^4 C₁ (y₃ ) x y+ ^4 C₃(x y)^3 ] & = [16 y₁- ^4 C₀ (y₃ )^2- ^4 C₂ (y₃ )(x y)^2- ^4 C₄(x y)^4 ] aligned Squaring both side we have =y₃ [ ^4 C₁ (y₃ ) x y+ ^4 C₃(x y)^3 ]^2= [16 y₁- ^4 C₀ (y₃ )^2- ^4 C₂ (y₃ ) . .(x y)^2- ^4 C₄(x y)^4 ]^2 which is now free from fractions in the power of derivatives (y₁, y₂, y₃ ) . Clearly, order =3 Since y₃ will be having power 4 in right side. Hence degree =4