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

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

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

\frac{t 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

\frac{t 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}

\frac{d}{d x} (\frac{( x + 3 ) ^{2}}{e ^{x} \cdot cos ( x )})

f (x) = \frac{1}{5} (9 x + 8)^{3}

\frac{d}{d x} (- 7 ln (x))

\int (sin^{4} (x)) d x

x \to 1 lim (2 x^{2} - 4 x + 7)

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