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Question
Solve the differential equation (1 + y2) tan–1xdx + 2y(1 + x2)dy = 0.
Solution
Given differential equation is (1 + y2) tan–1xdx + 2y(1 + x2)dy = 0
⇒ `2y(1 + x^2)"d"y = -(1 + y^2) . tan^-1x . "d"x`
⇒ `(2y)/(1 + y^2) "d"y = (tan^-1x)/(1 + x^2) . "d"x`
Integrating both sides, we get
`int (2y)/(1 + y^2) "d"y = -int (tan^-1x)/(1 + x^2) . "d"x`
⇒ `log|1 + y^2| = - 1/2(tan^-1x)^2 + "c"`
⇒ `1/2 (tan^-1x)^2 + log|1 + y^2|` = c
Which is the required solution.
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