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

The solution for 3y^{''}+y^'+y=0 is y=e^{-t/6}(c_{1}cos((sqrt(11)t)/6)+c_{2}sin((sqrt(11)t)/6))

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

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

y = e^{- \frac{t}{6}} (c_{1} cos (\frac{11 t}{6}) + c_{2} sin (\frac{11 t}{6}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{6}} (c_{1} cos (\frac{11 t}{6}) + c_{2} sin (\frac{11 t}{6}))

y = e^{- \frac{t}{6}} (c_{1} cos (\frac{11 t}{6}) + c_{2} sin (\frac{11 t}{6}))

y^{^{''}} + 16 y = 0, y (0) = 2, y (0) = - 2

a y^{^{''}} + 2 y^{^{'}} + b y = 0, ab = 1

\frac{d ^{2} x}{d t ^{2}} + 25 x = 0

\frac{d ^{2} y}{d x ^{2}} - 1 \frac{d y}{d x} - 6 y = 0

y^{^{''}} + 4 y^{^{'}} - 12 y = 0, y^{0} = 5, y^{^{'}} 0 = 18

What is the solution for 3y^{''}+y^'+y=0 ?
The solution for 3y^{''}+y^'+y=0 is y=e^{-t/6}(c_{1}cos((sqrt(11)t)/6)+c_{2}sin((sqrt(11)t)/6))