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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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