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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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