What is the solution for t^2y^'+2ty-y^3=0 ?

The solution for t^2y^'+2ty-y^3=0 is y=sqrt((5t)/(2+5c_{1)t^5)},y=-sqrt((5t)/(2+5c_{1)t^5)}

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t^2y^'+2ty-y^3=0

t^{2} y^{^{'}} + 2 t y - y^{3} = 0

y = \frac{5 t}{2 + 5 c _{1} t ^{5}}, y = - \frac{5 t}{2 + 5 c _{1} t ^{5}}

Solution steps

t^{2} y^{'} (t) + 2 t y - y^{3} = 0

Solve Bernoulli ODE:

y = \frac{5 t}{2 + 5 c _{1} t ^{5}}, y = - \frac{5 t}{2 + 5 c _{1} t ^{5}}

y = \frac{5 t}{2 + 5 c _{1} t ^{5}}, y = - \frac{5 t}{2 + 5 c _{1} t ^{5}}

What is the solution for t^2y^'+2ty-y^3=0 ?
The solution for t^2y^'+2ty-y^3=0 is y=sqrt((5t)/(2+5c_{1)t^5)},y=-sqrt((5t)/(2+5c_{1)t^5)}