What is the solution for y^{''}+4y^'+29y=0,y(0)=3,y^'(0)=-2 ?

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

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y^{''}+4y^'+29y=0,y(0)=3,y^'(0)=-2

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

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

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

9 (\frac{d ^{2} y}{d x ^{2}}) + 12 (\frac{d y}{d x}) + 4 y = 0

2 y^{^{''}} + 4 y^{^{'}} + k y = 0

\frac{d ^{2} y}{d x ^{2}} (4) + \frac{d y}{d x} (5) + y = 0

y^{^{''}} + 4 y^{^{'}} - 32 y = 0, y (0) = - 3

x^{^{'''}} + x^{^{''}} - 72 x^{^{'}} = 0

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