What is the solution for y^{''}+4y^'+y32=0 ?

The solution for y^{''}+4y^'+y32=0 is y=e^{-2t}(c_{1}cos(2sqrt(7)t)+c_{2}sin(2sqrt(7)t))

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y^{''}+4y^'+y32=0

y^{^{''}} + 4 y^{^{'}} + y 32 = 0

y = e^{- 2 t} (c_{1} cos (27 t) + c_{2} sin (27 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + y 32 = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (27 t) + c_{2} sin (27 t))

y = e^{- 2 t} (c_{1} cos (27 t) + c_{2} sin (27 t))

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y^{^{'}} (0) = 4

36 \frac{d ^{2} y}{d x ^{2}} + 12 \frac{d y}{d x} + y = 0

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d t ^{2}} - 10 \frac{d y}{d t} = 0

y^{^{''}} - 4 y + 13 y = 0

What is the solution for y^{''}+4y^'+y32=0 ?
The solution for y^{''}+4y^'+y32=0 is y=e^{-2t}(c_{1}cos(2sqrt(7)t)+c_{2}sin(2sqrt(7)t))