What is the solution for t(dy)/(dt)=t^3+19t^3y ?

The solution for t(dy)/(dt)=t^3+19t^3y is y=(e^{(19t^3+c_{1})/3})/(19)-1/19

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t(dy)/(dt)=t^3+19t^3y

t \frac{d y}{d t} = t^{3} + 19 t^{3} y

y = \frac{e ^{\frac{19 t ^{3} + c _{1}}{3}}}{19} - \frac{1}{19}

Solution steps

t \frac{d y}{d t} = t^{3} + 19 t^{3} y

Solve separable ODE:

y = \frac{e ^{\frac{19 t ^{3} + c _{1}}{3}}}{19} - \frac{1}{19}

y = \frac{e ^{\frac{19 t ^{3} + c _{1}}{3}}}{19} - \frac{1}{19}

s^{2}

x \to (- 1) + lim (x + 1)

x \to \frac{1}{3} lim (3 x + \frac{1}{4})

y = 6 x^{3} + 11

n = 2 \sum \infty (- 0.875)^{n}

What is the solution for t(dy)/(dt)=t^3+19t^3y ?
The solution for t(dy)/(dt)=t^3+19t^3y is y=(e^{(19t^3+c_{1})/3})/(19)-1/19