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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} ((x - 1) e^{x})

\frac{7}{s ^{2} + 42}

\frac{d}{d x} (1 - ln^{2} (x))

\frac{\partial}{\partial x} (x^{2} - 7 x y + y)

\int csc^{2} (\frac{x}{5}) d x

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