What is the solution for 2yy^'+2=y^2+2x,y(0)=7 ?

The solution for 2yy^'+2=y^2+2x,y(0)=7 is y=sqrt(-2x+49e^x)

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2yy^'+2=y^2+2x,y(0)=7

2 y y^{^{'}} + 2 = y^{2} + 2 x, y (0) = 7

y = - 2 x + 49 e^{x}

Solution steps

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

Solve Bernoulli ODE:

y = - 2 x + 49 e^{x}

y = - 2 x + 49 e^{x}

\int 16 arctan (x) d x

\frac{d}{d x} (x^{2}, x + 2 x cos (x) - 2 sec (x))

\frac{d}{d x} (ln (2 cos (x)))

x^{3} - x^{2} - x, x

y = - x, y = - 8 x + 13, [0, \frac{13}{8}]

What is the solution for 2yy^'+2=y^2+2x,y(0)=7 ?
The solution for 2yy^'+2=y^2+2x,y(0)=7 is y=sqrt(-2x+49e^x)