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

The solution for y^{''}+2y^'-3y=cos(2x) is y=c_{1}e^x+c_{2}e^{-3x}+4/65 sin(2x)-7/65 cos(2x)

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

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

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{4}{65} sin (2 x) - \frac{7}{65} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{4}{65} sin (2 x) - \frac{7}{65} cos (2 x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{4}{65} sin (2 x) - \frac{7}{65} cos (2 x)

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 20 e^{3 t}

y^{^{''}} - 8 y^{^{'}} + 17 y = 10 e^{4 x} sec (x)

y^{^{''}} - 4 y = 12

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

y^{^{''}} - 4 y^{^{'}} + 3 y = e^{t} + t^{2}

What is the solution for y^{''}+2y^'-3y=cos(2x) ?
The solution for y^{''}+2y^'-3y=cos(2x) is y=c_{1}e^x+c_{2}e^{-3x}+4/65 sin(2x)-7/65 cos(2x)