Question

For the differential equation find a particular solution satisfying the given condition:

`x(x^2 - 1) dy/dx = 1` , y = 0  when x = 2

Answer 1

`x (x^2 - 1) dy/dx = 1`

`=> dy = 1/(x (x^2 - 1)) dx`

`dy = dx/(x(x + 1)(x - 1))`

`=> dy = [- 1/x + 1/2 (1/(x + 1) + 1/(x - 1))] dx`

Integrating by Partial Fractions

`int 1. dy = - int 1/x  dx + 1/2 int 1/(x + 1)  dx + 1/2 int 1/(x - 1)  dx`

`y = - log x + 1/2 log (x + 1) + 1/2  log (x - 1) + log C`

`y = 1/2 [-2  log x + log (x - 1)] + C`

`y = 1/2 [-2 log x + log (x^2 - 1)] + C`

`y = 1/2 [- log x^2 + log (x^2 - 1)] + C`

`y = 1/2 [log (x^2 - 1)/x^2] + C`    .... (i)

Given y = 0 and x = 2

`therefore 0 = 1/2 [log (4 - 1)/4] + C`

`0 = 1/2 [log  3/4] + C`

`=> C = 1/2  log (3/4)`

Putting this value of C in equation (i)

`y = 1/2  log ((x^2 - 1)/x^2) - 1/2  log  3/4`

`y = 1/2  [log  ((x^2 - 1)/x^2) - "log"  3/4]`