What is the solution for (dy)/(dt)+2(t+1)y^2=0,y(0)=-1/15 ?

The solution for (dy)/(dt)+2(t+1)y^2=0,y(0)=-1/15 is y=-1/(-t^2-2t+15)

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(dy)/(dt)+2(t+1)y^2=0,y(0)=-1/15

\frac{d y}{d t} + 2 (t + 1) y^{2} = 0, y (0) = - \frac{1}{15}

y = - \frac{1}{- t ^{2} - 2 t + 15}

Solution steps

\frac{d y}{d t} + 2 (t + 1) y^{2} = 0

Solve separable ODE:

y = - \frac{1}{- t ^{2} - 2 t + 15}

y = - \frac{1}{- t ^{2} - 2 t + 15}

10 y^{^{'}} - 2 y - 63 = 9 y^{^{'}}

y^{5} \frac{d y}{d x} = 4 x^{4} y^{6} - 12 x^{4}

y^{^{'}} + tan (4 \cdot x) y = 0

2 y^{^{'}} = y^{3} cos (x), y (0) = (1 - sin (x))^{- \frac{1}{2}}

2 x y + 8 x + (x^{2} - 4) \frac{d y}{d x} = 0

What is the solution for (dy)/(dt)+2(t+1)y^2=0,y(0)=-1/15 ?
The solution for (dy)/(dt)+2(t+1)y^2=0,y(0)=-1/15 is y=-1/(-t^2-2t+15)