What is the solution for 4q^{''}+8q^'+80q=88 ?

The solution for 4q^{''}+8q^'+80q=88 is q(t)=e^{-t}(c_{1}cos(sqrt(19)t)+c_{2}sin(sqrt(19)t))+11/10

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4q^{''}+8q^'+80q=88

4 q^{^{''}} + 8 q^{^{'}} + 80 q = 88

q (t) = e^{- t} (c_{1} cos (19 t) + c_{2} sin (19 t)) + \frac{11}{10}

Solution steps

4 q^{''} (t) + 8 q^{'} (t) + 80 q (t) = 88

Solve linear ODE:

q (t) = e^{- t} (c_{1} cos (19 t) + c_{2} sin (19 t)) + \frac{11}{10}

q (t) = e^{- t} (c_{1} cos (19 t) + c_{2} sin (19 t)) + \frac{11}{10}

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

\frac{d ^{2} y}{d x ^{2}} - 25 y = 3 x^{2} - 2 + e^{x}

\frac{d ^{2} y}{d x ^{2}} - 4 (\frac{d y}{d x}) + 8 y = x e^{x}

y^{^{'''}} + 4 y^{^{''}} - 2 y = 2 x^{2} - 3 x + 6

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 t^{2}, y (1) = 0

What is the solution for 4q^{''}+8q^'+80q=88 ?
The solution for 4q^{''}+8q^'+80q=88 is q(t)=e^{-t}(c_{1}cos(sqrt(19)t)+c_{2}sin(sqrt(19)t))+11/10