In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

x + y = tan^{–1}y : y^{2}y′ + y^{2} + 1 = 0

It is given that x + y = tan^{-1}y

Now, differentiating both sides w.r.t. x, we get,

Now, Substituting the values of y’ in the given differential equations, we get,

LHS = y^{2}y’ + y^{2} + 1 =

= -1 – y^{2} + y^{2} + 1

= 0 = RHS

Therefore, the given function is the solution of the corresponding differential equation.

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