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Question
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
x + y = tan–1y : y2 y′ + y2 + 1 = 0
Solution
x + y = tan-1y
1 = y’ = `1/(1 + y^2)` (y’)
⇒ (1 + y') (1 + y2) = y’
⇒ 1 + y2 + y' + y2y' = y'
⇒ 1 + y2 + y2y' = 0
Hence, the given function x + y = tan-1y is a solution to the given differential equation.
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