What is the solution for y^{'''}+3y^{''}+y^'-5y=-5x ?

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

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y^{'''}+3y^{''}+y^'-5y=-5x

y^{^{'''}} + 3 y^{^{''}} + y^{^{'}} - 5 y = - 5 x

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

Solution steps

y^{'''} (x) + 3 y^{''} (x) + y^{'} (x) - 5 y = - 5 x

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 4 sin (2 t) + 8 cos (2 t)

y^{^{''}} + 4 y = 2 cos (x)

y^{^{''}} + 7 y = 5 t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 2

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

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 9 + 2 sin (x)

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