What is the solution for t((dy)/(dt))=t^3+18t^3y ?

The solution for t((dy)/(dt))=t^3+18t^3y is y=(e^{6t^3+c_{1}})/(18)-1/18

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t((dy)/(dt))=t^3+18t^3y

t (\frac{d y}{d t}) = t^{3} + 18 t^{3} y

y = \frac{e ^{6 t^{3} + c_{1}}}{18} - \frac{1}{18}

Solution steps

t (\frac{d y}{d t}) = t^{3} + 18 t^{3} y

Solve separable ODE:

y = \frac{e ^{6 t^{3} + c_{1}}}{18} - \frac{1}{18}

y = \frac{e ^{6 t^{3} + c_{1}}}{18} - \frac{1}{18}

\int e^{- t (s - a)} d t

4 t^{3} - t^{2}, t^{4} + 2 t^{2}

f (x) = \frac{6 x}{6 + x}

\int 8 x^{3} - 6 x^{2} + 10 x - 9 d x

\frac{d y}{d x} = x + y

What is the solution for t((dy)/(dt))=t^3+18t^3y ?
The solution for t((dy)/(dt))=t^3+18t^3y is y=(e^{6t^3+c_{1}})/(18)-1/18