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

The solution for 2y^{''}+3y^'-2y=cos(2x) is y=c_{1}e^{x/2}+c_{2}e^{-2x}+3/68 sin(2x)-5/68 cos(2x)

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

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

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- 2 x} + \frac{3}{68} sin (2 x) - \frac{5}{68} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- 2 x} + \frac{3}{68} sin (2 x) - \frac{5}{68} cos (2 x)

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- 2 x} + \frac{3}{68} sin (2 x) - \frac{5}{68} cos (2 x)

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What is the solution for 2y^{''}+3y^'-2y=cos(2x) ?
The solution for 2y^{''}+3y^'-2y=cos(2x) is y=c_{1}e^{x/2}+c_{2}e^{-2x}+3/68 sin(2x)-5/68 cos(2x)