What is the solution for y^'=(ty^2)/(t^2+e) ?

The solution for y^'=(ty^2)/(t^2+e) is y=-2/(ln(t^2+e)+c_{1)}

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y^'=(ty^2)/(t^2+e)

y^{^{'}} = \frac{t y ^{2}}{t ^{2} + e}

y = - \frac{2}{ln ( t ^{2} + e ) + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{ln ( t ^{2} + e ) + c _{1}}

y = - \frac{2}{ln ( t ^{2} + e ) + c _{1}}

y^{^{'}} = 1 + y + y^{2}

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

\frac{d y}{d x} = - \frac{0.1}{cos ( y )}, y (0) = \frac{π}{2}

\frac{d v}{d t} = \frac{t + 2}{v}

\frac{d y}{d x} = \frac{y ( 2 - 3 x )}{x ( 1 - 3 y )}

What is the solution for y^'=(ty^2)/(t^2+e) ?
The solution for y^'=(ty^2)/(t^2+e) is y=-2/(ln(t^2+e)+c_{1)}