What is the solution for y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-10 ?

The solution for y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-10 is y=e^{-4t}(3cos(5t)+2/5 sin(5t))

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y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-10

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 3, y^{^{'}} (0) = - 10

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

Solution steps

y^{''} (t) + 8 y^{'} (t) + 41 y = 0

Solve linear ODE:

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

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

(D^{4} - 1) x = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = - 2

- a y^{^{''}} - cy = 0

y^{^{''}} + 5 y + 6 y = 0

y^{^{''}} - 2 y^{^{'}} + 11 y = 0

What is the solution for y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-10 ?
The solution for y^{''}+8y^'+41y=0,y(0)=3,y^'(0)=-10 is y=e^{-4t}(3cos(5t)+2/5 sin(5t))