What is the solution for y^{''}+4y^'+29y=0,y(0)=3,y^'(0)=-5 ?

The solution for y^{''}+4y^'+29y=0,y(0)=3,y^'(0)=-5 is y=e^{-2t}(3cos(5t)+1/5 sin(5t))

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y^{''}+4y^'+29y=0,y(0)=3,y^'(0)=-5

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 2 y^{^{''}} - 63 y^{^{'}} = 0

\frac{3}{4} x^{^{''}} + 72 x = 0, x (0) = - \frac{1}{4}, x^{^{'}} (0) = 0

\frac{1}{16} x^{^{''}} + x^{^{'}} + 4 x = 0

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y (1) = 2

(d^{3} + d^{2} - 9 d - 9) y = 0

What is the solution for y^{''}+4y^'+29y=0,y(0)=3,y^'(0)=-5 ?
The solution for y^{''}+4y^'+29y=0,y(0)=3,y^'(0)=-5 is y=e^{-2t}(3cos(5t)+1/5 sin(5t))