Let y be the solution of the initial value problem y′′+y=−sin(2x),y(0)=0,y′(0)=0. y″+y=−sin⁡(2x),y(0)=0,y′(0)=0. The maximum value of y is , careful, ypyp probably does not satisfy the ivp. Find solution yy and then find the max value (e. g. by critical value y′(x)=0y′(x)=0 or maybe using calculator if you do not see how to solve y′=0y′=0)