What is the solution for y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=3 ?

The solution for y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=3 is y=e^{-2t}(2cos(t)+7sin(t))

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y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=3

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y = e^{- 2 t} (2 cos (t) + 7 sin (t))

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (2 cos (t) + 7 sin (t))

y = e^{- 2 t} (2 cos (t) + 7 sin (t))

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 2, y^{^{'}} (0) = 28

1124 y^{^{''}} + 5802 y^{^{'}} + 58292 y = 0

324 y^{^{''}} + 252 y^{^{'}} + 49 y = 0, y^{^{'}} (0) = 2

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

y^{^{''}} + 7 y = 0, y (0) = 1, y^{^{'}} (0) = 7

What is the solution for y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=3 ?
The solution for y^{''}+4y^'+5y=0,y(0)=2,y^'(0)=3 is y=e^{-2t}(2cos(t)+7sin(t))