What is the solution for 4y^{''}+20y^'+29y=0,y(0)=8 ?

The solution for 4y^{''}+20y^'+29y=0,y(0)=8 is y=e^{-(5t)/2}(8cos(t)+c_{2}sin(t))

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4y^{''}+20y^'+29y=0,y(0)=8

4 y^{^{''}} + 20 y^{^{'}} + 29 y = 0, y (0) = 8

y = e^{- \frac{5 t}{2}} (8 cos (t) + c_{2} sin (t))

Solution steps

4 y^{''} (t) + 20 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (8 cos (t) + c_{2} sin (t))

y = e^{- \frac{5 t}{2}} (8 cos (t) + c_{2} sin (t))

y^{^{''}} + 16 y = 0, y (π^{8}) = - 3, y^{^{'}} (π^{8}) = 8

y^{^{''}} - \frac{1}{a ^{2}} y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

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

y^{^{''}} + y = 0, y (\frac{π}{3}) = 2

What is the solution for 4y^{''}+20y^'+29y=0,y(0)=8 ?
The solution for 4y^{''}+20y^'+29y=0,y(0)=8 is y=e^{-(5t)/2}(8cos(t)+c_{2}sin(t))