What is the solution for y^{''}+y^'+25y=0 ?

The solution for y^{''}+y^'+25y=0 is y=e^{-t/2}(c_{1}cos((3sqrt(11)t)/2)+c_{2}sin((3sqrt(11)t)/2))

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y^{''}+y^'+25y=0

y^{^{''}} + y^{^{'}} + 25 y = 0

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

Solution steps

y^{''} (t) + y^{'} (t) + 25 y = 0

Solve linear ODE:

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

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

(d^{3} - 6 d^{2} + 11 d - 6) \cdot y = 0

f^{^{''}} (t) + 2 f^{^{'}} (t) + 4 f (t) = 0

y^{^{''}} - 7 y^{^{'}} + 15 y = 0

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

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

What is the solution for y^{''}+y^'+25y=0 ?
The solution for y^{''}+y^'+25y=0 is y=e^{-t/2}(c_{1}cos((3sqrt(11)t)/2)+c_{2}sin((3sqrt(11)t)/2))