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

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

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

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

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + (1 + π^{2}) y = 0

Solve linear ODE:

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

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

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

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

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

y \cdot y^{^{''}} + (y)^{2} = 0, y (0) = 4, y^{^{'}} (0) = 6

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = - 3, y^{^{'}} (0) = 33

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