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}))

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))