What is the solution for y^{''}+3y^'+8y=0 ?

The solution for y^{''}+3y^'+8y=0 is y=e^{-(3t)/2}(c_{1}cos((sqrt(23)t)/2)+c_{2}sin((sqrt(23)t)/2))

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y^{''}+3y^'+8y=0

y^{^{''}} + 3 y^{^{'}} + 8 y = 0

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

Solution steps

y^{''} (t) + 3 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

\frac{d ^{2} x}{d t ^{2}} = x

y^{^{''}} - 6 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 2

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 0

(D^{2} - 3 D + 2) y = 0

y^{^{''}} + 16 π^{2} y = 0

What is the solution for y^{''}+3y^'+8y=0 ?
The solution for y^{''}+3y^'+8y=0 is y=e^{-(3t)/2}(c_{1}cos((sqrt(23)t)/2)+c_{2}sin((sqrt(23)t)/2))