Find the particular solution of the differential equation `x(x^(2)-1)(dy)/(dx) =1,` it being given that `y = 0` when `x = 2`.

Options

A. `y = (1)/(2)log|(4(x^(2)-1))/(x^(2))|,`
B. `y = (1)/(2)log|(7(x^(2)-1))/(3x^(2))|,`
C. `y = (1)/(2)log|(4(x^(2)-1))/(3x^(2))|,`
D. `y = log|(4(x^(2)-1))/(3x^(2))|,`

Answer

C

Explanation

We have
`x(x^(2)-1)(dy)/(dx)=1`
`rArr dy=(1)/(x(x^(2)-1))dx`
`rArr int dy=int(dx)/(x(x-1)(x+1)) " " ` ...(i)
Let ` (1)/(x(x-1)(x+1))=(A)/(x)+(B)/((x-1)) +(C)/((x+1)).`
Then, `A(x-1)(x+1)+Bx(x+1)+Cx(x-1)-=1." " `...(ii)
Putting x = 0 in (ii), we get A = -1.
Putting x = 1 in (ii), we get B = `(1)/(2)`.
Putting x = -1 in (ii), we get C = `(1)/(2).`
`therefore (1)/(x(x-1)(x+1))=(-1)/(x)+(1)/(2(x-1))+(1)/(2(x+1))`
Puttiing this value in (i), we get
`int dy = int (-dx)/(x)+(1)/(2)int (dx)/((x-1)) +(1)/(2)int (dx)/((x+1))+C_(1), " where " C_(1)` is an arbitrary constant
`rArr y = -log|x|+(1)/(2)log|x-1|+(1)/(2)log|x+1|+C_(1). " " ` ...(iii)
We are given y = 0 when x = 2.
Putting x = 2 and y = 0 in (iii), we get
`-log2 + (1)/(2)log 3 + C_(1) =0`
`rArr C_(1)=log2 -(1)/(2)log 3 =(1)/(2) log 4-(1)/(2)log 3 = (1)/(2)log ((4)/(3)).`
Putting `C_(1) = (1)/(2)log((4)/(3))` in (iii) we get
`y=-log |x|+(1)/(2)log|x-1|+(1)/(2)log|x+1|+(1)/(2)"log"(4)/(3)`
`rArr y = -(1)/(2)logx^(2)+(1)/(2)log|(x-1)(x+1) |+(1)/(2) "log"(4)/(3)`
`rArr y = (1)/(2)log|(4(x^(2)-1))/(3x^(2))|,` which is the required solution .