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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d t ^{2}} - y = e^{t}

x^{^{''}} - 5 x^{^{'}} + 4 x = 8 e^{t}

y^{^{''}} + 9 y = 10 e^{- t}

y^{^{''}} - 7 y^{^{'}} + 6 y = (x - 2) e^{x}

y^{^{''}} - 2 y^{^{'}} + 37 y = e^{x} cos (6 x)

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