Make a Bernoulli substititution to form a linear differential equation:

t²(dy / dt) + 2ty - y³ = 0

t²(dy / dt) + 2ty = y³

dy / dt + 2y / t = y³ / t²

(dy / dt) / y³ + 2 / (ty²) = 1 / t²

Let u = 1 / y²,

du / dx = -2(du / dx) / y³

-(du / dx) / 2 + 2u / t = 1 / t²

Solve this differential equation by using an integrating factor:

du / dx - 4u / t = -2 / t²

du / dx + P(t)u = f(t)

P(t) = -4 / t

f(t) = -2 / t²

I(t) = ℮^[∫ P(t) dt]

I(t) = ℮^(∫ -4 / t dt)

I(t) = ℮^(-4ln|t|)

I(t) = ℮^[ln(1 / t⁴)]

I(t) = 1 / t⁴

I(t)y = ∫ I(t)f(t) dt

y / t⁴ = ∫ -2 / t⁶ dt

y / t⁴ = 2 / (5t⁵) + C

y / t⁴ = C + 2 / (5t⁵)

y = Ct⁴ + 2 / (5t)

Find the general solution by substituting back for the previous variable:

Since u = 1 / y²,

1 / y² = Ct⁴ + 2 / (5t)

1 / y² = (Ct⁵ + 2) / (5t)

y² = 5t / (Ct⁵ + 2)

y = ±√[5t / (Ct⁵ + 2)]

y = √[5t / (Ct⁵ + 2)]

t²dy/dt + 2ty − y³ = 0 ; t>0 , y>0

Because this can be written d(t²y)/dt = y³, the sub u=t²y is suggested

This gives du/dt = (u/t²)³ = u³/t⁶ → du/u³ = dt/t⁶, a straightforward separated ODE