What is the solution for t(dy)/(dt)=t^3+8t^3y ?

The solution for t(dy)/(dt)=t^3+8t^3y is y=(e^{(8t^3+c_{1})/3})/(8)-1/8

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t(dy)/(dt)=t^3+8t^3y

t \frac{d y}{d t} = t^{3} + 8 t^{3} y

y = \frac{e ^{\frac{8 t ^{3} + c _{1}}{3}}}{8} - \frac{1}{8}

Solution steps

t \frac{d y}{d t} = t^{3} + 8 t^{3} y

Solve separable ODE:

y = \frac{e ^{\frac{8 t ^{3} + c _{1}}{3}}}{8} - \frac{1}{8}

y = \frac{e ^{\frac{8 t ^{3} + c _{1}}{3}}}{8} - \frac{1}{8}

\int 9 cos (θ) d θ

\frac{d}{d x} (- 2 x^{2} + 7 x + 5)

n = 0 \sum \infty ln (1 + n)

\frac{\partial}{\partial y} (x^{y} - y^{x})

y = (x^{2} - 7 x + 7) e^{\frac{9 x}{7}}

What is the solution for t(dy)/(dt)=t^3+8t^3y ?
The solution for t(dy)/(dt)=t^3+8t^3y is y=(e^{(8t^3+c_{1})/3})/(8)-1/8