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

The solution for y^{''}+4y^'+29y=0,y(0)=4,y^'(0)=-5 is y=e^{-2t}(4cos(5t)+3/5 sin(5t))

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

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

y = e^{- 2 t} (4 cos (5 t) + \frac{3}{5} sin (5 t))

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (4 cos (5 t) + \frac{3}{5} sin (5 t))

y = e^{- 2 t} (4 cos (5 t) + \frac{3}{5} sin (5 t))

y + 2 y^{^{'''}} + 10 y^{^{''}} + 18 y^{^{'}} + 9 y = 0

y^{^{''}} + 16 y = 0, y (0) = 1, y^{^{'}} (1) = 0

y^{^{''}} + 16 y = 0, y (0) = 1, y (1) = 0

2 y^{^{''}} + 9 y^{^{'}} + 7 y = 0

9 y^{^{''}} - 64 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'+29y=0,y(0)=4,y^'(0)=-5 ?
The solution for y^{''}+4y^'+29y=0,y(0)=4,y^'(0)=-5 is y=e^{-2t}(4cos(5t)+3/5 sin(5t))