Solution of equation `(xy^4 + y) dx – xdy = 0 `is

Options

A. `4x^(4)y^(3)+3x^(3)=Cy^(3)`
B. `3x^(3)y^(4)+4y^(3)=Cx^(3)`
C. `3x^(4)y^(3)+4x^(3)=Cy^(3)`
D. none of these

Answer

C

Explanation

We have,
`(xy^(4)+y)dx-xdy=0`
`rArr" "(dy)/(dx)=(xy^(4)+y)/(x)`
`rArr" "(dy)/(dx)-(y)/(x)=y^(4)`
`rArr" "(1)/(y^(4))(dy)/(dx)+((-1)/(y^(3)))(1)/(x)=1`
Let `-y^(-3)=v." Then, "3y^(-4)(dy)/(dx)=(dv)/(dx)`
`therefore" "(1)/(3)(dv)/(dx)+(v)/(x)=1rArr (dv)/(dx)+(3)/(x)v=3" ...(i)"`
This is a linear differential equaiton with integrating factor `x^(3)`.
Multiplying both sides of (i) by `x^(3)` and integrating, we get
`vx^(3)=(3x^(4))/(4)+C`
`rArr" "(-x^(3))/(y^(3))=(3x^(4))/(4)+C`
`rArr" "-4x^(3)=3x^(4)y^(3)+4y^(3)C`
`rArr" "3x^(4)y^(3)+4x^(3)=-4Cy^(3)rArr 3x^(4)y^(3)+4x^(3)=lambday^(3)`