What is the solution for y^{''}+8y^'+41y=0,y^'(0)=2,y(0)=1 ?

The solution for y^{''}+8y^'+41y=0,y^'(0)=2,y(0)=1 is y=e^{-4t}(cos(5t)+6/5 sin(5t))

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y^{''}+8y^'+41y=0,y^'(0)=2,y(0)=1

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y^{^{'}} (0) = 2, y (0) = 1

y = e^{- 4 t} (cos (5 t) + \frac{6}{5} sin (5 t))

Solution steps

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

Solve linear ODE:

y = e^{- 4 t} (cos (5 t) + \frac{6}{5} sin (5 t))

y = e^{- 4 t} (cos (5 t) + \frac{6}{5} sin (5 t))

36 y^{^{''}} + 25 y = 0

2 y^{^{''}} - 17 y^{^{'}} - 9 y = 0, y (0) = 4, y^{^{'}} (0) = - 7

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

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

u^{^{''}} = H (t)

What is the solution for y^{''}+8y^'+41y=0,y^'(0)=2,y(0)=1 ?
The solution for y^{''}+8y^'+41y=0,y^'(0)=2,y(0)=1 is y=e^{-4t}(cos(5t)+6/5 sin(5t))