What is the solution for y^{''}+2y^'+0.75x=0 ?

The solution for y^{''}+2y^'+0.75x=0 is y=c_{1}+c_{2}e^{-2x}-0.1875x^2+0.1875x

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y^{''}+2y^'+0.75x=0

y^{^{''}} + 2 y^{^{'}} + 0.75 x = 0

y = c_{1} + c_{2} e^{- 2 x} - 0.1875 x^{2} + 0.1875 x

Solution steps

y^{''} (x) + 2 y^{'} (x) + 0.75 x = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} - 0.1875 x^{2} + 0.1875 x

y = c_{1} + c_{2} e^{- 2 x} - 0.1875 x^{2} + 0.1875 x

y^{^{''}} + 6 y^{^{'}} + 10 y = 3 x - 3

x^{^{''}} + 9 x = 5 cos (t)

y^{^{''}} - 3 y^{^{'}} + 7 y = 3 e^{3 t}

y^{^{''}} + 6 y^{^{'}} = - 9

y^{^{''}} + 4 y^{^{'}} + 8 y = 24, y^{^{'}} (0) = 0, y (0) = 0

What is the solution for y^{''}+2y^'+0.75x=0 ?
The solution for y^{''}+2y^'+0.75x=0 is y=c_{1}+c_{2}e^{-2x}-0.1875x^2+0.1875x