What is the solution for y^'=(t^3+19t^3y)/t ?

The solution for y^'=(t^3+19t^3y)/t is y=(e^{(19t^3+c_{1})/3})/(19)-1/19

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y^'=(t^3+19t^3y)/t

y^{^{'}} = \frac{t ^{3} + 19 t ^{3} y}{t}

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

Solution steps

y^{'} (t) = \frac{t ^{3} + 19 t ^{3} y}{t}

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}

\frac{\partial}{\partial y} (- yz)

x \to - 7 lim (\frac{1}{x})

x \to 9 lim (cos (\frac{π x}{3}))

y = x^{2} - 2 x

\int x^{3} 1 + x^{4} d x

What is the solution for y^'=(t^3+19t^3y)/t ?
The solution for y^'=(t^3+19t^3y)/t is y=(e^{(19t^3+c_{1})/3})/(19)-1/19