What is the solution for y^{''}+36y=cos(8x) ?

The solution for y^{''}+36y=cos(8x) is y=c_{1}cos(6x)+c_{2}sin(6x)-1/28 cos(8x)

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y^{''}+36y=cos(8x)

y^{^{''}} + 36 y = cos (8 x)

y = c_{1} cos (6 x) + c_{2} sin (6 x) - \frac{1}{28} cos (8 x)

Solution steps

y^{''} (x) + 36 y = cos (8 x)

Solve linear ODE:

y = c_{1} cos (6 x) + c_{2} sin (6 x) - \frac{1}{28} cos (8 x)

y = c_{1} cos (6 x) + c_{2} sin (6 x) - \frac{1}{28} cos (8 x)

y^{^{'''}} + y^{^{'}} = ln (x)

y^{^{''}} - 2 y^{^{'}} = x + a

y^{^{''}} + y = sin (t) cos (t)

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{( 2 e ^{- 3 x} )}{x ^{2} + 1}

y^{^{''}} + y = x^{2} + 1, y (0) = 7, y (1) = 0

What is the solution for y^{''}+36y=cos(8x) ?
The solution for y^{''}+36y=cos(8x) is y=c_{1}cos(6x)+c_{2}sin(6x)-1/28 cos(8x)