Find a particular solution of

4y′′+y=−4cos⁡(x2)−8xsin⁡(x2)\displaystyle {4}{y}{''}+{y}=-{4}{\cos{{\left(\frac{{x}}{{2}}\right)}}}-{8}{x}{\sin{{\left(\frac{{x}}{{2}}\right)}}}

yp(x)=\displaystyle {y}_{{p}}{\left({x}\right)}=  

License