What is the solution for y^'=ty^3 ?

The solution for y^'=ty^3 is y=sqrt(-1/(t^2+c_{1))},y=-sqrt(-1/(t^2+c_{1))}

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y^'=ty^3

y^{^{'}} = t y^{3}

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

Solution steps

y^{'} (t) = t y^{3}

Solve separable ODE:

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

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

x \to 8 lim (cos (\frac{1}{x - 8}))

x \to \frac{π}{2} lim (2)

\frac{\partial}{\partial y} (- ln (cos (y)))

\int e^{6 x} sin (5 x) d x

What is the solution for y^'=ty^3 ?
The solution for y^'=ty^3 is y=sqrt(-1/(t^2+c_{1))},y=-sqrt(-1/(t^2+c_{1))}