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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d y}{d x} = \frac{( 2 x ( y - 1 ) )}{x ^{2} + 3}

\int \frac{1}{x ^{2} 16 + x ^{2}} d x

\int \frac{x}{x - 3} d x

\int \frac{1}{2 x ^{4}} d x

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