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

The solution for y^{''}+2y+y=0,y(0)=xe^{-x} is y=c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x)

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

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

y = c_{1} cos (3 x) + c_{2} sin (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x)

y^{^{''}} - 6 y^{^{'}} + 2 y = 0

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (- 4) = 5

y^{^{''}} + y^{^{'}} + 9.25 y = 0, y (0) = 2, y^{^{'}} (0) = 8

y^{^{'''}} + 6 y^{^{''}} + 10 y^{^{'}} + 8 y = 0

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 1, y^{^{'}} (0) = 4

What is the solution for y^{''}+2y+y=0,y(0)=xe^{-x} ?
The solution for y^{''}+2y+y=0,y(0)=xe^{-x} is y=c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x)