What is the solution for 25x^{''}+10x^'+226x=0 ?

The solution for 25x^{''}+10x^'+226x=0 is x=e^{-t/5}(c_{1}cos(3t)+c_{2}sin(3t))

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25x^{''}+10x^'+226x=0

25 x^{^{''}} + 10 x^{^{'}} + 226 x = 0

x = e^{- \frac{t}{5}} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

25 x^{''} (t) + 10 x^{'} (t) + 226 x = 0

Solve linear ODE:

x = e^{- \frac{t}{5}} (c_{1} cos (3 t) + c_{2} sin (3 t))

x = e^{- \frac{t}{5}} (c_{1} cos (3 t) + c_{2} sin (3 t))

- 5 y^{^{'}} + 2 y^{^{''}} - y^{^{'}} - 6 y = 0

121 y^{^{''}} - 154 y^{^{'}} + 74 y = 0

\frac{d ^{2} y}{d x ^{2}} = - a^{2} y

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

49 y^{^{''}} - 126 y^{^{'}} + 117 y = 0, y (4) = 2, y^{^{'}} (4) = 2

What is the solution for 25x^{''}+10x^'+226x=0 ?
The solution for 25x^{''}+10x^'+226x=0 is x=e^{-t/5}(c_{1}cos(3t)+c_{2}sin(3t))