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

The solution for t^2y^'+2ty=y^3 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

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

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}

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}}

y^{^{'}} = \frac{2 sec ( y )}{( x - 2 ) ^{2}}

f (x) = \frac{t ^{2} + 3}{t ^{4} - 2 t + 4}

x \to 0 lim (x + 2)

\int z^{31} + z^{8} d z

\int \frac{10}{x ^{0.5}} d x

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