What is the solution for 2yy^'+2=y^2+2x,y(0)=4 ?

The solution for 2yy^'+2=y^2+2x,y(0)=4 is y=sqrt(-2x+16e^x)

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2yy^'+2=y^2+2x,y(0)=4

2 y y^{^{'}} + 2 = y^{2} + 2 x, y (0) = 4

y = - 2 x + 16 e^{x}

Solution steps

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

Solve Bernoulli ODE:

y = - 2 x + 16 e^{x}

y = - 2 x + 16 e^{x}

\frac{2 x}{( x ^{2} + 1 ) ^{2}}

\frac{d}{d x} (x^{3} + 2 x^{2} - x + 4)

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

\frac{\partial}{\partial x} (ln (2 x - y))

\int_{0}^{9} 9 + 3 x d x

What is the solution for 2yy^'+2=y^2+2x,y(0)=4 ?
The solution for 2yy^'+2=y^2+2x,y(0)=4 is y=sqrt(-2x+16e^x)