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

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

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

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

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 13 y^{^{'}} + 40 y = 0

y^{^{'''}} + 6 y^{^{''}} + 13 y^{^{'}} = 0

0.2 y^{^{''}} + 5 y^{^{'}} + \frac{1}{0.002} y = 0

y^{^{''}} + 3 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - 8

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