What is the solution for y^'=(t^2)/y (1+t^3) ?

The solution for y^'=(t^2)/y (1+t^3) is y=sqrt((2t^3+t^6+c_{1))/3},y=-sqrt((2t^3+t^6+c_{1))/3}

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y^'=(t^2)/y (1+t^3)

y^{^{'}} = \frac{t ^{2}}{y} (1 + t^{3})

y = \frac{2 t ^{3} + t ^{6} + c _{1}}{3}, y = - \frac{2 t ^{3} + t ^{6} + c _{1}}{3}

Solution steps

y^{'} (t) = \frac{t ^{2}}{y} (1 + t^{3})

Solve separable ODE:

y = \frac{2 t ^{3} + t ^{6} + c _{1}}{3}, y = - \frac{2 t ^{3} + t ^{6} + c _{1}}{3}

y = \frac{2 t ^{3} + t ^{6} + c _{1}}{3}, y = - \frac{2 t ^{3} + t ^{6} + c _{1}}{3}

\int_{0}^{13} 8 x d x

(3 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0

- 10 x^{2} y + y^{^{'}} = 5 x^{2}

x \to 5 + lim (lo g_{3} (x - 5))

x \to 0 lim (\frac{4}{e ^{x} + 1})

What is the solution for y^'=(t^2)/y (1+t^3) ?
The solution for y^'=(t^2)/y (1+t^3) is y=sqrt((2t^3+t^6+c_{1))/3},y=-sqrt((2t^3+t^6+c_{1))/3}