What is the solution for x^{''}+(2*x^')+27x=0 ?

The solution for x^{''}+(2*x^')+27x=0 is x=e^{-t}(c_{1}cos(sqrt(26)t)+c_{2}sin(sqrt(26)t))

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x^{''}+(2*x^')+27x=0

x^{^{''}} + (2 \cdot x^{^{'}}) + 27 x = 0

x = e^{- t} (c_{1} cos (26 t) + c_{2} sin (26 t))

Solution steps

x^{''} (t) + (2 \cdot x^{'} (t)) + 27 x = 0

Solve linear ODE:

x = e^{- t} (c_{1} cos (26 t) + c_{2} sin (26 t))

x = e^{- t} (c_{1} cos (26 t) + c_{2} sin (26 t))

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

\frac{d ^{2} y}{d x ^{2}} (1) - \frac{d y}{d x} (7) + 10 y = 0

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (π) = 8, y (0) = 1

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = - 4, y^{^{'}} (0) = 14

9 \frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + y = 0

What is the solution for x^{''}+(2*x^')+27x=0 ?
The solution for x^{''}+(2*x^')+27x=0 is x=e^{-t}(c_{1}cos(sqrt(26)t)+c_{2}sin(sqrt(26)t))