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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (e^{x} \cdot sin (x))

8 y^{^{'}} = 3 e^{\frac{x}{8}} + y

\frac{d}{d x} (\frac{8 x}{3 - cot ( x )})

\int \frac{1}{e ^{2 x} - 2 e ^{x}} d x

\int e^{- θ} cos (5 θ) d θ

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