What is the solution for y^{'''}-y^'=2sin(3x) ?

The solution for y^{'''}-y^'=2sin(3x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x+1/15 cos(3x)

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y^{'''}-y^'=2sin(3x)

y^{^{'''}} - y^{^{'}} = 2 sin (3 x)

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{15} cos (3 x)

Solution steps

y^{'''} (x) - y^{'} (x) = 2 sin (3 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{15} cos (3 x)

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{15} cos (3 x)

What is the solution for y^{'''}-y^'=2sin(3x) ?
The solution for y^{'''}-y^'=2sin(3x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x+1/15 cos(3x)