What is the solution for 1y^{''}+5y^'+1/(1/6)y=2e^{2t} ?

The solution for 1y^{''}+5y^'+1/(1/6)y=2e^{2t} is y=c_{1}e^{-2t}+c_{2}e^{-3t}+1/10 e^{2t}

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1y^{''}+5y^'+1/(1/6)y=2e^{2t}

1 y^{^{''}} + 5 y^{^{'}} + \frac{1}{\frac{1}{6}} y = 2 e^{2 t}

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + \frac{1}{10} e^{2 t}

Solution steps

1 y^{''} (t) + 5 y^{'} (t) + \frac{1}{\frac{1}{6}} y = 2 e^{2 t}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + \frac{1}{10} e^{2 t}

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + \frac{1}{10} e^{2 t}

y^{^{''}} + 2 y^{^{'}} - 2 y = 5 sin (x) - 10 x e^{x}

y^{^{''}} - 8 y^{^{'}} + 17 y = - 102, y (0) = - 3, y^{^{'}} (0) = 8

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3} x ln (x)

y^{^{'''}} + 2 y^{^{''}} + 4 y^{^{'}} + 8 = 0

y^{^{''''}} - 16 y^{^{''}} = x^{2} + e^{2 x}

What is the solution for 1y^{''}+5y^'+1/(1/6)y=2e^{2t} ?
The solution for 1y^{''}+5y^'+1/(1/6)y=2e^{2t} is y=c_{1}e^{-2t}+c_{2}e^{-3t}+1/10 e^{2t}