What is the solution for 2q^{''}+10q^'+12q=e^{-3t} ?

The solution for 2q^{''}+10q^'+12q=e^{-3t} is q(t)=c_{1}e^{-2t}+c_{2}e^{-3t}-(e^{-3t}t)/2

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2q^{''}+10q^'+12q=e^{-3t}

2 q^{^{''}} + 10 q^{^{'}} + 12 q = e^{- 3 t}

q (t) = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} - \frac{e ^{- 3 t} t}{2}

Solution steps

2 q^{''} (t) + 10 q^{'} (t) + 12 q (t) = e^{- 3 t}

Solve linear ODE:

q (t) = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} - \frac{e ^{- 3 t} t}{2}

q (t) = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} - \frac{e ^{- 3 t} t}{2}

\frac{d ^{2} y}{d x ^{2}} + 12 \frac{d y}{d x} + 32 y = 160

y^{^{''}} - 4 y^{^{'}} = 16 t + 36, y (- 2) = 2, y^{^{'}} (- 2) = - 1

\frac{d ^{2} y}{d x ^{2}} + a \frac{d y}{d x} = b

y^{^{''''}} - 5 y^{^{''}} + 4 y = 20 e^{3 x}

y^{^{''}} + 36 y = 12 tan (6 t)

What is the solution for 2q^{''}+10q^'+12q=e^{-3t} ?
The solution for 2q^{''}+10q^'+12q=e^{-3t} is q(t)=c_{1}e^{-2t}+c_{2}e^{-3t}-(e^{-3t}t)/2