What is the solution for y^{''}+2y^'+26y=0,y(0)=2,y^'(0)=-1 ?

The solution for y^{''}+2y^'+26y=0,y(0)=2,y^'(0)=-1 is y=e^{-t}(2cos(5t)+1/5 sin(5t))

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y^{''}+2y^'+26y=0,y(0)=2,y^'(0)=-1

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 26 y = 0

Solve linear ODE:

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

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

(D^{4} - 2 D^{3} + 2 D^{2} - 2 D + 1) y = 0

y^{^{'''}} + 5 y^{^{''}} + 7 y^{^{'}} + 3 y = 0

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

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

y^{^{''}} = 6 y

What is the solution for y^{''}+2y^'+26y=0,y(0)=2,y^'(0)=-1 ?
The solution for y^{''}+2y^'+26y=0,y(0)=2,y^'(0)=-1 is y=e^{-t}(2cos(5t)+1/5 sin(5t))