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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
Solution
`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]`
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