What is the solution for 2y^{''}+y=0,y(0)=e^{-x/2} ?

The solution for 2y^{''}+y=0,y(0)=e^{-x/2} is y=c_{1}cos(x/(sqrt(2)))+c_{2}sin(x/(sqrt(2)))

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2y^{''}+y=0,y(0)=e^{-x/2}

2 y^{^{''}} + y = 0, y (0) = e^{- \frac{x}{2}}

y = c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2})

Solution steps

2 y^{''} (x) + y = 0

Solve linear ODE:

y = c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2})

y = c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2})

4 y^{^{''}} + 4 y^{^{'}} + 5 y = 0

(D^{3} - 2 D^{2} - D + 2) y = 0

2 y^{^{'''}} + y^{^{''}} - 8 y^{^{'}} + 5 y = 0

\frac{d ^{2} y}{d t ^{2}} = \frac{d y}{d t}

y^{^{''}} + 5 y^{^{'}} - y = 0

What is the solution for 2y^{''}+y=0,y(0)=e^{-x/2} ?
The solution for 2y^{''}+y=0,y(0)=e^{-x/2} is y=c_{1}cos(x/(sqrt(2)))+c_{2}sin(x/(sqrt(2)))