What is the solution for y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=-5 ?

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

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y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=-5

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

y = e^{- 2 t} (2 cos (t) - sin (t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- 2 t} (2 cos (t) - sin (t))

y = e^{- 2 t} (2 cos (t) - sin (t))

y^{^{'''}} + 4 y^{^{''}} - 31 y^{^{'}} - 70 y = 0

y^{^{''}} - 18 y^{^{'}} + 81 y = 0, y (0) = 0, y^{^{'}} (0) = 8

4 y^{^{''}} + 8 y^{^{'}} = 0

144 y^{^{'''}} + 264 y^{^{''}} + 121 y^{^{'}} = 0

y^{^{'''}} - y^{^{''}} = 8 y^{^{'''}}

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