What is the solution for y^{''}-y^'-6y=18cos(3x) ?

The solution for y^{''}-y^'-6y=18cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-3/13 sin(3x)-15/13 cos(3x)

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y^{''}-y^'-6y=18cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 18 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{3}{13} sin (3 x) - \frac{15}{13} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 18 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{3}{13} sin (3 x) - \frac{15}{13} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{3}{13} sin (3 x) - \frac{15}{13} cos (3 x)

f (x) = \frac{3 x}{( 1 - x ) ^{2}}

\int - 3 tan (\frac{x}{3}) d x

\int_{3}^{7} (x - 6) x - 3 d x

\int_{1}^{e} 1

\int \frac{x ^{3}}{1 + x ^{2}} d x

What is the solution for y^{''}-y^'-6y=18cos(3x) ?
The solution for y^{''}-y^'-6y=18cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-3/13 sin(3x)-15/13 cos(3x)