What is the solution for 1q^{''}+8q^'+1/(0.05)q=80t ?

The solution for 1q^{''}+8q^'+1/(0.05)q=80t is q(t)=e^{-4t}(c_{1}cos(2t)+c_{2}sin(2t))+4t-8/5

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1q^{''}+8q^'+1/(0.05)q=80t

1 q^{^{''}} + 8 q^{^{'}} + \frac{1}{0.05} q = 80 t

q (t) = e^{- 4 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 4 t - \frac{8}{5}

Solution steps

1 q^{''} (t) + 8 q^{'} (t) + \frac{1}{0.05} q (t) = 80 t

Solve linear ODE:

q (t) = e^{- 4 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 4 t - \frac{8}{5}

q (t) = e^{- 4 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 4 t - \frac{8}{5}

(D + 1)^{2} y = x^{2} e^{- x}

y^{^{''}} + 4 y^{^{'}} + 4 y = (x + 5) e^{- 2 x}

y^{^{''}} + y^{^{'}} - 20 y = 5 + 2 \cdot x

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 2 y = x e^{x}

What is the solution for 1q^{''}+8q^'+1/(0.05)q=80t ?
The solution for 1q^{''}+8q^'+1/(0.05)q=80t is q(t)=e^{-4t}(c_{1}cos(2t)+c_{2}sin(2t))+4t-8/5