What is the solution for y^'=2(x+y)^2 ?

The solution for y^'=2(x+y)^2 is y=(tan(sqrt(2)(x+c_{1})))/(sqrt(2))-x

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y^'=2(x+y)^2

y^{^{'}} = 2 (x + y)^{2}

y = \frac{tan ( 2 ( x + c _{1} ) )}{2} - x

Solution steps

y^{'} (x) = 2 (x + y)^{2}

Solve using substitution:

y = \frac{tan ( 2 ( x + c _{1} ) )}{2} - x

y = \frac{tan ( 2 ( x + c _{1} ) )}{2} - x

y = \frac{4}{x ^{2}}

\frac{\partial}{\partial x} (x sin (x + y))

y^{^{'}} + e^{x} y = x^{2} y^{2}

y = x^{2}, y = 17 x

\int_{0}^{1} \frac{e ^{x}}{e ^{x} + 1} d x

What is the solution for y^'=2(x+y)^2 ?
The solution for y^'=2(x+y)^2 is y=(tan(sqrt(2)(x+c_{1})))/(sqrt(2))-x