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

The solution for y^{''}+y^'+y=kx is y=e^{-x/2}(c_{1}cos((sqrt(3)x)/2)+c_{2}sin((sqrt(3)x)/2))+kx-k

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y^{''}+y^'+y=kx

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

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + k x - k

Solution steps

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

Solve linear ODE:

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + k x - k

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + k x - k

y^{^{''}} - y^{^{'}} - 2 y = 2 x^{3} - 3 x^{2} + 8 x + 1

y^{^{''}} + 16 y = 8 e^{3 x}

y^{^{''}} + 4 y = x - \frac{x ^{2}}{20}

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- 2 x}

y^{^{''}} + 4 y = sin (x), y^{^{'}} (π) = 0, y (0) = 0

What is the solution for y^{''}+y^'+y=kx ?
The solution for y^{''}+y^'+y=kx is y=e^{-x/2}(c_{1}cos((sqrt(3)x)/2)+c_{2}sin((sqrt(3)x)/2))+kx-k