What is the solution for y^{''}+2y^'+y=2x ?

The solution for y^{''}+2y^'+y=2x is y=c_{1}e^{-x}+c_{2}xe^{-x}+2x-4

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y^{''}+2y^'+y=2x

y^{^{''}} + 2 y^{^{'}} + y = 2 x

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 x - 4

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 x - 4

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 x - 4

x^{^{''}} + 8 x^{^{'}} + 16 x = 4 sin (4 t)

y^{^{''}} - 6 y^{^{'}} + 9 y = t, y (0) = 0, y (0) = 1

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 x - 4

y^{^{''}} + 2 y^{^{'}} = 2 x + 5 + e^{- 2 x}

y^{^{''}} + 16 y = 34 e^{- 2 x} cos (4 x)

What is the solution for y^{''}+2y^'+y=2x ?
The solution for y^{''}+2y^'+y=2x is y=c_{1}e^{-x}+c_{2}xe^{-x}+2x-4