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

The solution for 3y^{''}+y^'-2y=2cos(x) is y=c_{1}e^{(2x)/3}+c_{2}e^{-x}+1/13 sin(x)-5/13 cos(x)

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

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

y = c_{1} e^{\frac{2 x}{3}} + c_{2} e^{- x} + \frac{1}{13} sin (x) - \frac{5}{13} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{2 x}{3}} + c_{2} e^{- x} + \frac{1}{13} sin (x) - \frac{5}{13} cos (x)

y = c_{1} e^{\frac{2 x}{3}} + c_{2} e^{- x} + \frac{1}{13} sin (x) - \frac{5}{13} cos (x)

What is the solution for 3y^{''}+y^'-2y=2cos(x) ?
The solution for 3y^{''}+y^'-2y=2cos(x) is y=c_{1}e^{(2x)/3}+c_{2}e^{-x}+1/13 sin(x)-5/13 cos(x)