What is the solution for 25x^{''}+50x^'+169x=0 ?

The solution for 25x^{''}+50x^'+169x=0 is x=e^{-t}(c_{1}cos((12t)/5)+c_{2}sin((12t)/5))

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25x^{''}+50x^'+169x=0

25 x^{^{''}} + 50 x^{^{'}} + 169 x = 0

x = e^{- t} (c_{1} cos (\frac{12 t}{5}) + c_{2} sin (\frac{12 t}{5}))

Solution steps

25 x^{''} (t) + 50 x^{'} (t) + 169 x = 0

Solve linear ODE:

x = e^{- t} (c_{1} cos (\frac{12 t}{5}) + c_{2} sin (\frac{12 t}{5}))

x = e^{- t} (c_{1} cos (\frac{12 t}{5}) + c_{2} sin (\frac{12 t}{5}))

45 y^{^{''}} + 180 y = 0

\frac{1}{8} x^{^{''}} + 25 x = 0

10 y^{^{''}} + 90 y^{^{'}} + 140 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 8 y^{^{'}} + 3 y = 0

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

What is the solution for 25x^{''}+50x^'+169x=0 ?
The solution for 25x^{''}+50x^'+169x=0 is x=e^{-t}(c_{1}cos((12t)/5)+c_{2}sin((12t)/5))