The solution of the differential equation `x(x^(2)+1)(dy//dx)=y(1-x^(2))+x^(3)logx` is

Options

A. `y(x^(2)+1)//x=1/4x^(2)logx+1/2x^(2)+c`
B. `y^(2)(x^(2)-1)//x=1/2x^(2)logx-1/4x^(2)+c`
C. `y(x^(2)+1)x=1/2x^(2)logx-1/4x^(2)+c`
D. None of these

Answer

C

Explanation

The given equation can be rewritten as
`(dy)/(dx) +(x^(2)-1)/(x(x^(2)-1)y)=(x^(2)logx)/(x^(2)+1)`…………..(1)
which is linear, Also,
`P=(x^(2)-1)/(x(x^(2)+1))` and `Q=(x^(2)logx)/(x^(2)+1)`
`intPdx=int[(2x)/(x^(2)+1)-1/x]dx` [resolving into partial fractions]
`log(x^(2)+1)-logx`
`therefore` I.F. `=[e^(log[x^(2)+1)//x)]=(x^(2)+1)/(x)`
Hence, the required solution of equation (1) is
`y(x^(2)+1)/(x) = int((x^(2)+1)/(x)(x^(2)logx)/(x^(2)+1))dx+c`
`=intxlogxdx+c`
`=1/2x^(2)logx-int1/x^(2)/2dx+c`
`therefore y(x^(2)+1)//x=1/2x^(2)logx-1/4x^(2)+c`