What is the solution for 9y^{''}+6y^'+y=e^{-1/3 t} ?

The solution for 9y^{''}+6y^'+y=e^{-1/3 t} is y=c_{1}e^{-t/3}+c_{2}e^{-t/3}t+(e^{-t/3}t^2)/(18)

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9y^{''}+6y^'+y=e^{-1/3 t}

9 y^{^{''}} + 6 y^{^{'}} + y = e^{- \frac{1}{3} t}

y = c_{1} e^{- \frac{t}{3}} + c_{2} e^{- \frac{t}{3}} t + \frac{e ^{- \frac{t}{3}} t ^{2}}{18}

Solution steps

9 y^{''} (t) + 6 y^{'} (t) + y = e^{- \frac{1}{3} t}

Solve linear ODE:

y = c_{1} e^{- \frac{t}{3}} + c_{2} e^{- \frac{t}{3}} t + \frac{e ^{- \frac{t}{3}} t ^{2}}{18}

y = c_{1} e^{- \frac{t}{3}} + c_{2} e^{- \frac{t}{3}} t + \frac{e ^{- \frac{t}{3}} t ^{2}}{18}

\frac{d}{d x} (\frac{1}{x ^{2} x})

\int \frac{1}{( 1 - x ^{2} ) ^{0.5}} d x

\int \frac{x ^{2} - 3}{x ^{8}} d x

\int 2 x^{2} (- x^{3} + 3)^{5} d x

\frac{\partial}{\partial t} (3 arcsin (t))

What is the solution for 9y^{''}+6y^'+y=e^{-1/3 t} ?
The solution for 9y^{''}+6y^'+y=e^{-1/3 t} is y=c_{1}e^{-t/3}+c_{2}e^{-t/3}t+(e^{-t/3}t^2)/(18)