What is the solution for 2yy^'+4=y^2+4x ?

The solution for 2yy^'+4=y^2+4x is y=sqrt(-4x+c_{1)e^x},y=-sqrt(-4x+c_{1)e^x}

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2yy^'+4=y^2+4x

2 y y^{^{'}} + 4 = y^{2} + 4 x

y = - 4 x + c_{1} e^{x}, y = - - 4 x + c_{1} e^{x}

Solution steps

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

Solve Bernoulli ODE:

y = - 4 x + c_{1} e^{x}, y = - - 4 x + c_{1} e^{x}

y = - 4 x + c_{1} e^{x}, y = - - 4 x + c_{1} e^{x}

x (7 x + 3)^{\frac{1}{3}}

\int_{0}^{1} x^{3} (1 + x^{4})^{3} d x

ln (x + 2)

\int y^{2} e^{y^{3}} d y

\int x e^{\frac{x}{10}} d x

What is the solution for 2yy^'+4=y^2+4x ?
The solution for 2yy^'+4=y^2+4x is y=sqrt(-4x+c_{1)e^x},y=-sqrt(-4x+c_{1)e^x}