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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

3 x^{^{''}} + 22 x^{^{'}} + 48 x = 5 cos (4 t)

y^{^{''}} - 4 y^{^{'}} = x \cdot e^{4 x}

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} (t^{3} - 2 t)

(d^{2} - 6 d + 3) y = e^{3} x

y^{^{''}} + 4 y^{^{'}} + 3 y = 6

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