What is the solution for y^{''}+2y^'+26y=0,y(0)=3,y^'(0)=-1 ?

The solution for y^{''}+2y^'+26y=0,y(0)=3,y^'(0)=-1 is y=e^{-t}(3cos(5t)+2/5 sin(5t))

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y^{''}+2y^'+26y=0,y(0)=3,y^'(0)=-1

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 12 y^{^{'}} + 52 y = 0, y (\frac{π}{8}) = 1, y (0) = 8

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

y^{^{'''}} - 2 y^{^{''}} - 29 y^{^{'}} - 42 y = 0

\frac{1}{2} x^{^{''}} + \frac{3}{2} x^{^{'}} + x = 0

y^{^{''}} + 2 y^{^{'}} + 100 y = 0

What is the solution for y^{''}+2y^'+26y=0,y(0)=3,y^'(0)=-1 ?
The solution for y^{''}+2y^'+26y=0,y(0)=3,y^'(0)=-1 is y=e^{-t}(3cos(5t)+2/5 sin(5t))