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

The solution for y^{''}+2y+y=0,y^1(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^1(0)=xe^{-x}

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

3 y^{^{''}} - 12 y^{^{'}} - 36 y = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (\frac{1}{6}) = 2, y (0) = 3

16 x^{^{''}} + 40 x^{^{'}} + 169 x = 0

y^{^{''}} + 5 y^{^{'}} + 16 y = 0

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

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