What is the solution for y^{''}+y^'+y=1+x ?

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

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y^{''}+y^'+y=1+x

y^{^{''}} + y^{^{'}} + y = 1 + x

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

Solution steps

y^{''} (x) + y^{'} (x) + y = 1 + x

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{'}} = 3 sin (t)

y^{^{''}} - y = tan^{2} (x)

y^{^{''}} + y^{^{'}} - 6 y = 36 x

y^{^{''}} + 2 y^{^{'}} + 5 y = 20 e^{- t} cos (2 t)

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

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