What is the solution for y^'= 5/(y+2) ?

The solution for y^'= 5/(y+2) is y=(-4+sqrt(16-4(-c_{1)-10t)})/2 ,y=(-4-sqrt(16-4(-c_{1)-10t)})/2

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y^'= 5/(y+2)

y^{^{'}} = \frac{5}{y + 2}

y = \frac{- 4 + 16 - 4 ( - c _{1} - 10 t )}{2}, y = \frac{- 4 - 16 - 4 ( - c _{1} - 10 t )}{2}

Solution steps

y^{'} (t) = \frac{5}{y + 2}

Solve separable ODE:

y = \frac{- 4 + 16 - 4 ( - c _{1} - 10 t )}{2}, y = \frac{- 4 - 16 - 4 ( - c _{1} - 10 t )}{2}

y = \frac{- 4 + 16 - 4 ( - c _{1} - 10 t )}{2}, y = \frac{- 4 - 16 - 4 ( - c _{1} - 10 t )}{2}

\frac{d}{d x} ((5 x^{6} + 8 x^{3})^{4})

\int csc (3 x) cot (3 x) d x

\int (3 y^{2}) d y

y = x + x, (49, 56)

\int_{1}^{25} \frac{2}{x} d x

What is the solution for y^'= 5/(y+2) ?
The solution for y^'= 5/(y+2) is y=(-4+sqrt(16-4(-c_{1)-10t)})/2 ,y=(-4-sqrt(16-4(-c_{1)-10t)})/2