What is the solution for (tdy)/(dt)=t^3+23t^3y ?

The solution for (tdy)/(dt)=t^3+23t^3y is y=(e^{(23t^3+c_{1})/3})/(23)-1/23

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(tdy)/(dt)=t^3+23t^3y

\frac{t d y}{d t} = t^{3} + 23 t^{3} y

y = \frac{e ^{\frac{23 t ^{3} + c _{1}}{3}}}{23} - \frac{1}{23}

Solution steps

\frac{t d y}{d t} = t^{3} + 23 t^{3} y

Solve separable ODE:

y = \frac{e ^{\frac{23 t ^{3} + c _{1}}{3}}}{23} - \frac{1}{23}

y = \frac{e ^{\frac{23 t ^{3} + c _{1}}{3}}}{23} - \frac{1}{23}

\frac{( 6 x ^{2} - 5 ) ( x ^{2} - 6 )}{x ^{2} + 4}

y^{^{''}} - \frac{2 3}{3} y^{^{'}} + \frac{2}{9} = 0

\int_{0}^{2} (2 x - 1)^{2} d x

\int sec (3 t) tan (3 t) d t

y^{^{''}} + 5 y^{^{'}} + 10 y = 510 e^{2 t} cos (6 t)

What is the solution for (tdy)/(dt)=t^3+23t^3y ?
The solution for (tdy)/(dt)=t^3+23t^3y is y=(e^{(23t^3+c_{1})/3})/(23)-1/23