What is the solution for y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7 ?

The solution for y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7 is y=e^{-4t}(2cos(5t)+1/5 sin(5t))

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y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

y = e^{- 4 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

Solution steps

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

Solve linear ODE:

y = e^{- 4 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

y = e^{- 4 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

What is the solution for y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7 ?
The solution for y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7 is y=e^{-4t}(2cos(5t)+1/5 sin(5t))