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Question
The integrating factor of the differential equation.
`(1 - y^2) dx/dy + yx = ay(-1 < y < 1)` is ______.
Options
`1/(y^2 - 1)`
`1/sqrt(y^2 - 1)`
`1/(1 - y^2)`
`1/sqrt(1 - y^2)`
Solution
The integrating factor of the differential equation.
`(1 - y^2) dx/dy + yx = ay(-1 < y < 1)` is `underline(1/sqrt(1 - y^2)).`
Explanation:
The differential equation is,
`(1 – y^2)dy/dx + yx = ay`
or `dx/dy + y/(1 - y^2) x = y/(1 - y^2)`
Comparing with `dx/dy + Px = Q`,
`P = y/(1 - y^2), Q = y/(1 - y^2)`
`int P dx = int y/(1 - y^2) dy`
`= e^(- 1/2 int (- 2y)/(1 - y^2) dy)`
Let `= - 1/2 int (- 2y)/(1 - y^2) dy`
`1 - y^2` = t
∴ - 2y dy = dt
`= - 1/2 int dt/t = - 1/2 log t`
`= - 1/2 log (1 - y^2)`
`= log 1/sqrt(1 - y^2)`
`I.F. = e^(int P dx) = e^(log 1 sqrt(1 - y^2))`
`= 1/sqrt(1 - y^2)`
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