What is the solution for (dy)/(dt)+0.3ty=9t ?

The solution for (dy)/(dt)+0.3ty=9t is y=(e^{-0.15t^2+c_{1}})/(0.3)+9/(0.3)

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(dy)/(dt)+0.3ty=9t

\frac{d y}{d t} + 0.3 t y = 9 t

y = \frac{e ^{- 0.15 t^{2} + c_{1}}}{0.3} + \frac{9}{0.3}

Solution steps

\frac{d y}{d t} + 0.3 t y = 9 t

Solve separable ODE:

y = \frac{e ^{- 0.15 t^{2} + c_{1}}}{0.3} + \frac{9}{0.3}

y = \frac{e ^{- 0.15 t^{2} + c_{1}}}{0.3} + \frac{9}{0.3}

6 t - 3

\int_{1}^{1000} \frac{9}{x ^{3}} d x

t^{2} + e^{- 2 t} sin (3 t)

x \to - 1 lim (3 x^{3} + 2 x^{2} + 3)

x \to 0 + lim (e^{(\frac{1}{x})})

What is the solution for (dy)/(dt)+0.3ty=9t ?
The solution for (dy)/(dt)+0.3ty=9t is y=(e^{-0.15t^2+c_{1}})/(0.3)+9/(0.3)