What is the solution for y^'=(x+y)^2,y(0)=1 ?

The solution for y^'=(x+y)^2,y(0)=1 is y=tan(x+pi/4+pin)-x

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y^'=(x+y)^2,y(0)=1

y^{^{'}} = (x + y)^{2}, y (0) = 1

y = tan (x + \frac{π}{4} + πn) - x

Solution steps

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

Solve using substitution:

y = tan (x + \frac{π}{4} + πn) - x

y = tan (x + \frac{π}{4} + πn) - x

sin (x)

\frac{\partial}{\partial x} (2 + y ln (4 + x (y)^{2}))

\frac{d}{d x} (\frac{x y sin ( x - y )}{1 + x ^{2} + y ^{2}})

f (x) = 3^{s i n (3 x)}

\int \frac{2 + w ^{4}}{w ^{3} + 9 w} d w

What is the solution for y^'=(x+y)^2,y(0)=1 ?
The solution for y^'=(x+y)^2,y(0)=1 is y=tan(x+pi/4+pin)-x