What is the solution for 4y^{''}+24y^'+37y=0 ?

The solution for 4y^{''}+24y^'+37y=0 is y=e^{-3t}(c_{1}cos(t/2)+c_{2}sin(t/2))

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4y^{''}+24y^'+37y=0

4 y^{^{''}} + 24 y^{^{'}} + 37 y = 0

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

Solution steps

4 y^{''} (t) + 24 y^{'} (t) + 37 y = 0

Solve linear ODE:

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

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

x^{^{''}} - 6 x^{^{'}} + 9 x = 0

y^{^{''}} + 144 y = 0

\frac{5}{8} x^{^{''}} + 40 x = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.451

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

What is the solution for 4y^{''}+24y^'+37y=0 ?
The solution for 4y^{''}+24y^'+37y=0 is y=e^{-3t}(c_{1}cos(t/2)+c_{2}sin(t/2))