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}

l a pl a ce t r an s f or m 6 t - 3

l a pl a ce t r an s f or m t^{2} + e^{- 2 t} sin (3 t)

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

d er i v a t i v e \frac{y ^{2} - 2}{2 y ^{2}}

\int \frac{x ^{3} - 2 x ^{2} + 1}{x - 3} d 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)