What is the solution for y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=1 ?

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

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y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{2} e^{- t} sin (2 t)

Solution steps

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

Solve linear ODE:

y = \frac{1}{2} e^{- t} sin (2 t)

y = \frac{1}{2} e^{- t} sin (2 t)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 2 y = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 25 y = 0

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

y^{^{''}} - 17 y^{^{'}} + 72 y = 0

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