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Question
For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0
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Solution
`(x - y^2x)dx - (y + x^2y)dy` = 0, when x = 2, y = 0
∴ x(1 – y2)dx – y(1 + x2)dy = 0
∴ `x/(1 + x^2)dx - y/(1 - y^2)dy` = 0
∴ `(2x)/(1 + x^2)dx - (2y)/(1 - y^2)dy` = 0
Integrating both sides, we get
`int (2x)/(1 + x^2)dx + int(-2y)/(1 - y^2)dy = c_1`
Each of these integrals is of the type
`int (f^'(x))/(f(x))dx = log |f(x)| + c`
∴ The general solution is log |1 + x2| + log |1 - y2| = log c, where c1 = log c
∴ log |(1 + x2)(1 – y2)| = log c
∴ (1 + x2)(1 – y2) = c
When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ The particular solution is (1 + x2)(1 – y2) = 5.
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