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)

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)