Solution of ` x (dy)/(dx) + y = x^(2) y^(4)` is

Options

A. `x^(2)y^(2)(3+Cx)=1`
B. `x^(2)y^(3)(3+Cx)=1`
C. `x^(3)y^(3)(3+Cx)=1`
D. `x^(2) y^(3) (3- Cx)=1`

Answer

b

Explanation

` (dy)/(dx) +1/x y = xy^(4)`
` rArr y^(-4) (dy)/(dx) =1/x y^(-3 = x `
Let `z = y^(-3)`
` rArr (dz)/ (dx) = -3y ^(-4) (dy)/(dx) `
` rArr -1/3 (dz)/(dx) +1/(x)z = x`
Above equation is linear differential equation.
` :. IF = e^(-int3/xdx) = e ^(-3logx) = 1/(x^(3))`
` :. ` Solution is given by
` z * 1/(x^(3)) = int 1/(x^(3)) xx (-3x) dx+C`
` rArr Z/(x^(3)) = -3 1/(x^(2)) dx+ C = -3 ((-1)/x)+C`
` rArr Z/(x^(3)) = -3 int 1/(x^(2)) dx + C = -3 ((-1)/3)+C`
` rArr 1/(x^(3)y^(3))=3/x+C " " [ :. z = y^(-3)]`
` rArr 1= x^(2) y^(3) (3+Cx)`