What is the solution for y^{''}+4y=cos(4x)+2sin(4x) ?

The solution for y^{''}+4y=cos(4x)+2sin(4x) is y=c_{1}cos(2x)+c_{2}sin(2x)-1/6 sin(4x)-1/12 cos(4x)

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y^{''}+4y=cos(4x)+2sin(4x)

y^{^{''}} + 4 y = cos (4 x) + 2 sin (4 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{6} sin (4 x) - \frac{1}{12} cos (4 x)

Solution steps

y^{''} (x) + 4 y = cos (4 x) + 2 sin (4 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{6} sin (4 x) - \frac{1}{12} cos (4 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{6} sin (4 x) - \frac{1}{12} cos (4 x)

What is the solution for y^{''}+4y=cos(4x)+2sin(4x) ?
The solution for y^{''}+4y=cos(4x)+2sin(4x) is y=c_{1}cos(2x)+c_{2}sin(2x)-1/6 sin(4x)-1/12 cos(4x)