(a) For what values ofk does the function y=cos(kt) satisfy the differential equation 4y''=-25y?(Enter your answers as a comma-separated list.)k=(b) For those values ofk, verify that every member of the family of functions y=Asin(kt)+Bcos(kt)is also a solution.We begin by calculating the following.y=Asin(kt)+Bcos(kt)=>y'=Akcos(kt)-Bksin(kt)=>y''=|Note that the given differential equation 4y''=-25yis equivalent to4y''+25y=Now, substituting the expressions for y and y "above and simplifying, we haveLHS=4yn+25y=4( Xour answer cannot be understood or graded. More Information )+25(Asin(kt)+Bcos(kt))=-4()+4Bk2cos(kt)+25Asin(kt)+25Bcos(kt)=(25-4k2)=0since for all value ofk found above, k2=