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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 - 3 e^{- x} + 4 cos (3 x)

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

y^{^{''}} - 8 y^{^{'}} + 5 y = x e^{x}

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + 5 y = sin (t)

y^{^{''}} - y = x e^{2 x}, y (0) = 0, y^{^{'}} (0) = 1

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