What is the solution for (dy)/(dx)= 1/x (y^2+y),y(1)=0.5 ?

The solution for (dy)/(dx)= 1/x (y^2+y),y(1)=0.5 is y= x/(3-x)

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(dy)/(dx)= 1/x (y^2+y),y(1)=0.5

\frac{d y}{d x} = \frac{1}{x} (y^{2} + y), y (1) = 0.5

y = \frac{x}{3 - x}

Solution steps

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

Solve separable ODE:

y = \frac{x}{3 - x}

y = \frac{x}{3 - x}

\int x^{3} ln (x) d x

\int (x^{3} + 1)^{2} x^{2} d x

(x + 1) y^{^{'}} + (x + 2) y = 2 x e^{- x}

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 10, (27, 1)

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

What is the solution for (dy)/(dx)= 1/x (y^2+y),y(1)=0.5 ?
The solution for (dy)/(dx)= 1/x (y^2+y),y(1)=0.5 is y= x/(3-x)