What is the solution for 4y^{''}+y^'+6y=0 ?

The solution for 4y^{''}+y^'+6y=0 is y=e^{-t/8}(c_{1}cos((sqrt(95)t)/8)+c_{2}sin((sqrt(95)t)/8))

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4y^{''}+y^'+6y=0

4 y^{^{''}} + y^{^{'}} + 6 y = 0

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{95 t}{8}) + c_{2} sin (\frac{95 t}{8}))

Solution steps

4 y^{''} (t) + y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{95 t}{8}) + c_{2} sin (\frac{95 t}{8}))

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{95 t}{8}) + c_{2} sin (\frac{95 t}{8}))

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

y^{^{''}} + 3 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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

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

16 y^{^{''}} + 8 y^{^{'}} + 145 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

What is the solution for 4y^{''}+y^'+6y=0 ?
The solution for 4y^{''}+y^'+6y=0 is y=e^{-t/8}(c_{1}cos((sqrt(95)t)/8)+c_{2}sin((sqrt(95)t)/8))