What is the solution for y^{''}+5y^'+13y=0 ?

The solution for y^{''}+5y^'+13y=0 is y=e^{-(5t)/2}(c_{1}cos((3^{3/2}t)/2)+c_{2}sin((3^{3/2}t)/2))

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y^{''}+5y^'+13y=0

y^{^{''}} + 5 y^{^{'}} + 13 y = 0

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

Solution steps

y^{''} (t) + 5 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 10 y^{^{'}} + 23 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 15 y^{^{'}} + 50 y = 0, y (0) = 8, y^{^{'}} (0) = - 55

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

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

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

What is the solution for y^{''}+5y^'+13y=0 ?
The solution for y^{''}+5y^'+13y=0 is y=e^{-(5t)/2}(c_{1}cos((3^{3/2}t)/2)+c_{2}sin((3^{3/2}t)/2))