What is the solution for y^{''}+2y^'+(1+9/4 pi^2)y=0 ?

The solution for y^{''}+2y^'+(1+9/4 pi^2)y=0 is y=e^{-t}(c_{1}cos((3pit)/2)+c_{2}sin((3pit)/2))

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y^{''}+2y^'+(1+9/4 pi^2)y=0

y^{^{''}} + 2 y^{^{'}} + (1 + \frac{9}{4} π^{2}) y = 0

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + (1 + \frac{9}{4} π^{2}) y = 0

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 0, y (0) = cos (4 x)

y^{^{''}} - y^{^{'}} - y = 0, y^{1} (0) = e^{x}, y^{2} (0) = x

12 y^{^{''}} - 5 y^{^{'}} - 3 y = 0

y^{^{''}} + \frac{49}{\frac{4}{32}} y = 0

y^{^{''}} + 9 y = 0, y^{^{'}} (3 π) = 9, y (0) = 9

What is the solution for y^{''}+2y^'+(1+9/4 pi^2)y=0 ?
The solution for y^{''}+2y^'+(1+9/4 pi^2)y=0 is y=e^{-t}(c_{1}cos((3pit)/2)+c_{2}sin((3pit)/2))