What is the solution for y^'=(x(1+y))/(y(x+2)) ?

The solution for y^'=(x(1+y))/(y(x+2)) is y=-\W(-((x+2)^2)/(e^{-c_{1)+x+2}})-1

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y^'=(x(1+y))/(y(x+2))

y^{^{'}} = \frac{x ( 1 + y )}{y ( x + 2 )}

y = - W (- \frac{( x + 2 ) ^{2}}{e ^{- c_{1} + x + 2}}) - 1

Solution steps

y^{'} (x) = \frac{x ( 1 + y )}{y ( x + 2 )}

Solve separable ODE:

y = - W (- \frac{( x + 2 ) ^{2}}{e ^{- c_{1} + x + 2}}) - 1

y = - W (- \frac{( x + 2 ) ^{2}}{e ^{- c_{1} + x + 2}}) - 1

y^{^{'}} = 3 x (1 - y)

\frac{d y}{d x} = 3 x^{2} y, y = 2, x = 3

\frac{d z}{d t} = 6 t e^{5 z}, t (0) = 0

\frac{d x}{d y} = \frac{4 x ln ( x )}{y}

\frac{d x}{d t} = \frac{3}{x e ^{t + 6 x}}

What is the solution for y^'=(x(1+y))/(y(x+2)) ?
The solution for y^'=(x(1+y))/(y(x+2)) is y=-\W(-((x+2)^2)/(e^{-c_{1)+x+2}})-1