What is the solution for 2yy^'+2=y^2+2x,y(0)=6 ?

The solution for 2yy^'+2=y^2+2x,y(0)=6 is y=sqrt(-2x+36e^x)

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2yy^'+2=y^2+2x,y(0)=6

2 y y^{^{'}} + 2 = y^{2} + 2 x, y (0) = 6

y = - 2 x + 36 e^{x}

Solution steps

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

Solve Bernoulli ODE:

y = - 2 x + 36 e^{x}

y = - 2 x + 36 e^{x}

\frac{2}{x ^{\frac{1}{3}}} - 2

2 y^{^{'}} + t y = 7

\frac{d}{d x} (\frac{3 x - 1}{2 x + 5})

y = 4 x^{2}, y^{2} = \frac{1}{4} x

y^{^{''}} - x^{2} + y^{^{'}} = 0

What is the solution for 2yy^'+2=y^2+2x,y(0)=6 ?
The solution for 2yy^'+2=y^2+2x,y(0)=6 is y=sqrt(-2x+36e^x)