What is the solution for (dy)/(dt)=t^2+20t^2y ?

The solution for (dy)/(dt)=t^2+20t^2y is y=(e^{(20t^3+c_{1})/3})/(20)-1/20

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(dy)/(dt)=t^2+20t^2y

\frac{d y}{d t} = t^{2} + 20 t^{2} y

y = \frac{e ^{\frac{20 t ^{3} + c _{1}}{3}}}{20} - \frac{1}{20}

Solution steps

\frac{d y}{d t} = t^{2} + 20 t^{2} y

Solve separable ODE:

y = \frac{e ^{\frac{20 t ^{3} + c _{1}}{3}}}{20} - \frac{1}{20}

y = \frac{e ^{\frac{20 t ^{3} + c _{1}}{3}}}{20} - \frac{1}{20}

\int \frac{1}{9 z ^{2} - 25} d z

y = 2 t (3 t^{2} - 5)^{4}

\int e^{5 x} cos (9 x) d x

f (x) = \frac{2}{x}, at x = 9

\frac{\partial}{\partial z} (- \frac{1}{( y + 2 z ) ^{2}})

What is the solution for (dy)/(dt)=t^2+20t^2y ?
The solution for (dy)/(dt)=t^2+20t^2y is y=(e^{(20t^3+c_{1})/3})/(20)-1/20