What is the solution for y^{''}+6y^'=-108cos(6x) ?

The solution for y^{''}+6y^'=-108cos(6x) is y=c_{1}+c_{2}e^{-6x}-3/2 sin(6x)+3/2 cos(6x)

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y^{''}+6y^'=-108cos(6x)

y^{^{''}} + 6 y^{^{'}} = - 108 cos (6 x)

y = c_{1} + c_{2} e^{- 6 x} - \frac{3}{2} sin (6 x) + \frac{3}{2} cos (6 x)

Solution steps

y^{''} (x) + 6 y^{'} (x) = - 108 cos (6 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 6 x} - \frac{3}{2} sin (6 x) + \frac{3}{2} cos (6 x)

y = c_{1} + c_{2} e^{- 6 x} - \frac{3}{2} sin (6 x) + \frac{3}{2} cos (6 x)

\int \frac{( 5 sin ^{3} ( x ) )}{cos ( x )} d x

\int 5 e^{9 x} d x

f (x) = x^{4}, (2, 16)

x \to 0 lim (+ (x))

\int \frac{1}{( \frac{x}{2} ) ^{2} + 1} d x

What is the solution for y^{''}+6y^'=-108cos(6x) ?
The solution for y^{''}+6y^'=-108cos(6x) is y=c_{1}+c_{2}e^{-6x}-3/2 sin(6x)+3/2 cos(6x)