What is the solution for y^{''}+4y=cos(3x) ?

The solution for y^{''}+4y=cos(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)-1/5 cos(3x)

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y^{''}+4y=cos(3x)

y^{^{''}} + 4 y = cos (3 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{5} cos (3 x)

Solution steps

y^{''} (x) + 4 y = cos (3 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{5} cos (3 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{5} cos (3 x)

2 y^{^{''}} - 16 y + 4 = 0

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} + 1

y^{^{''}} + y = 5 t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 6

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 9

y^{^{''}} - 8 y^{^{'}} + 16 y = 24 x + 2

What is the solution for y^{''}+4y=cos(3x) ?
The solution for y^{''}+4y=cos(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)-1/5 cos(3x)