What is the solution for 3y^{''}+4y^'+5y=0 ?

The solution for 3y^{''}+4y^'+5y=0 is y=e^{-(2t)/3}(c_{1}cos((sqrt(11)t)/3)+c_{2}sin((sqrt(11)t)/3))

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3y^{''}+4y^'+5y=0

3 y^{^{''}} + 4 y^{^{'}} + 5 y = 0

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{11 t}{3}) + c_{2} sin (\frac{11 t}{3}))

Solution steps

3 y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{11 t}{3}) + c_{2} sin (\frac{11 t}{3}))

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{11 t}{3}) + c_{2} sin (\frac{11 t}{3}))

y^{^{''}} + \frac{49}{3} y = 0

y^{^{''}} - 5 y^{^{''}} + 24 y^{^{'}} - 20 y = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.449

y^{^{''''}} + 8 y^{^{'''}} + 21 y^{^{''}} + 22 y^{^{'}} + 8 y = 0

y^{^{''}} + 9 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 4

What is the solution for 3y^{''}+4y^'+5y=0 ?
The solution for 3y^{''}+4y^'+5y=0 is y=e^{-(2t)/3}(c_{1}cos((sqrt(11)t)/3)+c_{2}sin((sqrt(11)t)/3))