What is the solution for (dy)/(dt)=2ty^2+3t^2y^2 ?

The solution for (dy)/(dt)=2ty^2+3t^2y^2 is y=-1/(t^3+t^2+c_{1)}

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(dy)/(dt)=2ty^2+3t^2y^2

\frac{d y}{d t} = 2 t y^{2} + 3 t^{2} y^{2}

y = - \frac{1}{t ^{3} + t ^{2} + c _{1}}

Solution steps

\frac{d y}{d t} = 2 t y^{2} + 3 t^{2} y^{2}

Solve separable ODE:

y = - \frac{1}{t ^{3} + t ^{2} + c _{1}}

y = - \frac{1}{t ^{3} + t ^{2} + c _{1}}

\frac{d}{d t} (v (t) e^{k t}) = g e^{k t}

d x = x^{7} y d y

\frac{d y}{d x} = (\frac{( ln ( x ^{6} ) )}{x y ^{2}}), y (1) = 5

\frac{d y}{d x} 5 x^{2} - 4 y^{2} = 16

\frac{d y}{d x} y = e^{x + y}

What is the solution for (dy)/(dt)=2ty^2+3t^2y^2 ?
The solution for (dy)/(dt)=2ty^2+3t^2y^2 is y=-1/(t^3+t^2+c_{1)}