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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 3 y = 2 x

y^{^{''}} + 2 y^{^{'}} + y = \frac{ln ( x )}{e ^{x}}

y^{^{''}} - y^{^{'}} - 2 y = \frac{( e ^{2 t} + e ^{- 2 t} )}{2}

\frac{d ^{2} y}{d x ^{2}} - 3 y = e^{2 x} sin (x)

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

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