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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 14 y^{^{'}} + 49 y = 0, y (0) = - 6, y^{^{'}} (0) = 0

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

\frac{1}{4} y^{^{''}} + y = 0

y^{^{''}} + y^{^{'}} + 3 y = 0, y (0) = 0

81 y^{^{''}} + 36 y^{^{'}} + 4 y = 0

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