What is the solution for y^'-ty^3=0 ?

The solution for y^'-ty^3=0 is y=sqrt(-1/(t^2+c_{1))},y=-sqrt(-1/(t^2+c_{1))}

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y^'-ty^3=0

y^{^{'}} - t y^{3} = 0

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

Solution steps

y^{'} (t) - t y^{3} = 0

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}}

n = 7 \sum \infty \frac{1}{n}

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\int \frac{5}{16 + 9 cos ^{2} ( x )} d x

\int \frac{tan ^{3} ( x )}{cos ^{4} ( x )} d x

\frac{\partial}{\partial x} (ln ((\frac{y}{x})^{2}))

What is the solution for y^'-ty^3=0 ?
The solution for y^'-ty^3=0 is y=sqrt(-1/(t^2+c_{1))},y=-sqrt(-1/(t^2+c_{1))}