What is the solution for y^{''}+y^'+3.25y=0 ?

The solution for y^{''}+y^'+3.25y=0 is y=e^{-t/2}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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y^{''}+y^'+3.25y=0

y^{^{''}} + y^{^{'}} + 3.25 y = 0

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

Solution steps

y^{''} (t) + y^{'} (t) + 3.25 y = 0

Solve linear ODE:

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

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

y^{^{''}} - (2 a - 1) y^{^{'}} + a (a - 1) y = 0

16 y^{^{''}} + 9 y = 0

33 y^{^{''''}} + 323 y^{^{''}} + 400 y = 0

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} + 4 y = 0

y^{^{'''}} + 7 y^{^{''}} - 10 y = 0

What is the solution for y^{''}+y^'+3.25y=0 ?
The solution for y^{''}+y^'+3.25y=0 is y=e^{-t/2}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))