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

The solution for y^{''}-y^'-2y=3cos(2x) is y=c_{1}e^{2x}+c_{2}e^{-x}-3/20 sin(2x)-9/20 cos(2x)

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3}{20} sin (2 x) - \frac{9}{20} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3}{20} sin (2 x) - \frac{9}{20} cos (2 x)

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3}{20} sin (2 x) - \frac{9}{20} cos (2 x)

6 y^{^{'''}} + 4 y^{^{'}} = e^{x} cos (2 x)

4 y^{^{''}} + 25 y = 41 e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 2 y = 3 e^{x}

y^{^{''}} - 8 y^{^{'}} + 2 y = t e^{t}

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

What is the solution for y^{''}-y^'-2y=3cos(2x) ?
The solution for y^{''}-y^'-2y=3cos(2x) is y=c_{1}e^{2x}+c_{2}e^{-x}-3/20 sin(2x)-9/20 cos(2x)