What is the solution for 25y^{''}+10y^'+5y=0 ?

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

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25y^{''}+10y^'+5y=0

25 y^{^{''}} + 10 y^{^{'}} + 5 y = 0

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

Solution steps

25 y^{''} (t) + 10 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 10 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = - 2

- y^{^{''}} + 9 \cdot π^{2} y = 0

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

(y^{^{''}} + 2 y^{^{'}} + 5 y)^{3} = 0

5 y^{^{''}} + 3 y^{^{'}} + 9 y = 0

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