What is the solution for 2yy^'+6=y^2+6x ?

The solution for 2yy^'+6=y^2+6x is y=sqrt(-6x+c_{1)e^x},y=-sqrt(-6x+c_{1)e^x}

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2yy^'+6=y^2+6x

2 y y^{^{'}} + 6 = y^{2} + 6 x

y = - 6 x + c_{1} e^{x}, y = - - 6 x + c_{1} e^{x}

Solution steps

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

Solve Bernoulli ODE:

y = - 6 x + c_{1} e^{x}, y = - - 6 x + c_{1} e^{x}

y = - 6 x + c_{1} e^{x}, y = - - 6 x + c_{1} e^{x}

\int_{3}^{4} (5 - x^{2}) - (17 - 7 x) d x

\frac{\partial}{\partial x} (x^{2} tan (x y))

\frac{d y}{d t} = y + t

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

2 y = 4 x, y = 5, 2 y + 3 x = 7

What is the solution for 2yy^'+6=y^2+6x ?
The solution for 2yy^'+6=y^2+6x is y=sqrt(-6x+c_{1)e^x},y=-sqrt(-6x+c_{1)e^x}