What is the solution for y^{'''}-y^'=sin^2(x) ?

The solution for y^{'''}-y^'=sin^2(x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x+(-10x+sin(2x))/(20)

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y^{'''}-y^'=sin^2(x)

y^{^{'''}} - y^{^{'}} = sin^{2} (x)

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} + \frac{- 10 x + sin ( 2 x )}{20}

Solution steps

y^{'''} (x) - y^{'} (x) = sin^{2} (x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} + \frac{- 10 x + sin ( 2 x )}{20}

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} + \frac{- 10 x + sin ( 2 x )}{20}

What is the solution for y^{'''}-y^'=sin^2(x) ?
The solution for y^{'''}-y^'=sin^2(x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x+(-10x+sin(2x))/(20)