What is the solution for y^{''}+y^'+y=3x+8 ?

The solution for y^{''}+y^'+y=3x+8 is y=e^{-x/2}(c_{1}cos((sqrt(3)x)/2)+c_{2}sin((sqrt(3)x)/2))+3x+5

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

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

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + 3 x + 5

Solution steps

y^{''} (x) + y^{'} (x) + y = 3 x + 8

Solve linear ODE:

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + 3 x + 5

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + 3 x + 5

y^{^{''}} + 14 y^{^{'}} + 49 y = 4 t e^{- 7 t}

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{\frac{x}{2}}

y^{^{''}} - y^{^{'}} = e^{x} sin (x)

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

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 20 e^{3 t}

What is the solution for y^{''}+y^'+y=3x+8 ?
The solution for y^{''}+y^'+y=3x+8 is y=e^{-x/2}(c_{1}cos((sqrt(3)x)/2)+c_{2}sin((sqrt(3)x)/2))+3x+5