What is the solution for y^{''}+y^'+12y=0 ?

The solution for y^{''}+y^'+12y=0 is y=e^{-t/2}(c_{1}cos((sqrt(47)t)/2)+c_{2}sin((sqrt(47)t)/2))

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y^{''}+y^'+12y=0

y^{^{''}} + y^{^{'}} + 12 y = 0

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2}))

Solution steps

y^{''} (t) + y^{'} (t) + 12 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2}))

9 y^{^{''}} + 36 y^{^{'}} + 100 y = 0, y (1) = 4, y^{^{'}} (1) = 9

y^{^{''}} = 8 y^{^{'}} - 15 y

y^{^{''}} - y^{^{'}} + a y = 0

y^{^{''''}} - 36 y^{^{'''}} + 324 y^{^{''}} = 0

x^{^{''}} + 9 x = 0, x (0) = 3, x^{^{'}} (0) = 4

What is the solution for y^{''}+y^'+12y=0 ?
The solution for y^{''}+y^'+12y=0 is y=e^{-t/2}(c_{1}cos((sqrt(47)t)/2)+c_{2}sin((sqrt(47)t)/2))