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

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

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y=2(x+y)y^'

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

y = \frac{1 + c _{1} x + 1}{c _{1}}, y = \frac{1 - c _{1} x + 1}{c _{1}}

Solution steps

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

Solve exact ODE:

y = \frac{1 + c _{1} x + 1}{c _{1}}, y = \frac{1 - c _{1} x + 1}{c _{1}}

y = \frac{1 + c _{1} x + 1}{c _{1}}, y = \frac{1 - c _{1} x + 1}{c _{1}}

\int 3 sin^{6} (x) cos (x) d x

x \to 0 lim (100)

\frac{d}{d x} (a x^{2} e^{2 x})

h (t) = (t^{\frac{2}{5}} + 3) (3 t^{5} + 5)

x \to + 8 - lim (\frac{x}{x - 8})

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