What is the solution for y^{''}+4y^'+29y=0,y(0)=0,y^'(0)=5 ?

The solution for y^{''}+4y^'+29y=0,y(0)=0,y^'(0)=5 is y=e^{-2t}sin(5t)

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y^{''}+4y^'+29y=0,y(0)=0,y^'(0)=5

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y = e^{- 2 t} sin (5 t)

Solution steps

y^{''} (t) + 4 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- 2 t} sin (5 t)

y = e^{- 2 t} sin (5 t)

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

x^{^{''}} + 2 x^{^{'}} + 1000 x = 0

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

2 y^{^{''}} - 3 y^{^{'}} = 0

2 x^{^{''}} + 4 x^{^{'}} - 7 x = 0

What is the solution for y^{''}+4y^'+29y=0,y(0)=0,y^'(0)=5 ?
The solution for y^{''}+4y^'+29y=0,y(0)=0,y^'(0)=5 is y=e^{-2t}sin(5t)