What is the solution for y^{'''}-y^'=3cos(x) ?

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

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y^{'''}-y^'=3cos(x)

y^{^{'''}} - y^{^{'}} = 3 cos (x)

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

Solution steps

y^{'''} (x) - y^{'} (x) = 3 cos (x)

Solve linear ODE:

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

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

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