use-L-f-t-to-find-laplace-transform-of-d-2-x-dt-2-n-2-x-kcoswt-given-that-x-0-and-dx-dt-0-when-t-0-PLZ-HELP- Tinku Tara June 4, 2023 Differential Equation 0 Comments FacebookTweetPin Question Number 16475 by Sai dadon. last updated on 22/Jun/17 useL{f′(t)}tofindlaplacetransformofd2x/dt2+n2x=kcoswtgiventhatx=0anddx/dt=0whent=0PLZHELP. Answered by sma3l2996 last updated on 23/Jun/17 weknowthatL{f′(t)}=sL{f(t)}−f(0)andL{f″(t)}=sL{f′(t)}−f′(0)soL{f″(t)}=s2L{f(t)}−sf(0)−f′(0)letx(t)=f(t)L{d2x(t)dt2}=s2L{x(t)}−sx(0)−dx(0)dt=s2L{x(t)}andalsowehave:L{d2xdt2+n2x}=L{kcos(wt)}⇔L{d2xdt2}+n2L{x}=kL{cos(wt)}=k.ss2+w2s2L{x}+n2L{x}=k.ss2+w2(s2+n2).L{x}=k.ss2+w2L{x}=k.s(s2+w2)(s2+n2)s(s2+w2)(s2+n2)=as+bs2+w2+cs+ds2+n2a=−c,bw2=−dn2⇔d=−(nw)2b1(1+w2)(1+n2)=a+b1+w2−a+(nw)2b1+n2⇔(a+b)(1+n2)−(a+(nw)2b)(1+w2)=1a(n2−w2)+b(1+n2−(nw)2(1+w2))=1a(n2−w2)=1+b(n2−w2w2)(i):a=bw2+1n2−w22(w2+4)(n2+4)=2a+bw2+4−2a+(nw)2bn2+4(2a+b)(n2+4)−(2a+(nw)2b)(w2+4)=22a(n2−w2)+b(4−4(nw)2)=24b(w2−n2w2)=2+2a(w2−n2)(ii):bw2=12(w2−n2)+a2from(i)and(ii)a=12a+12(w2−n2)+1n2−w2a=1n2−w2=−cbw2=12(w2−n2)+12(n2−w2)=0a=−c=1n2−w2andb=d=0so:L{x}=kn2−w2(ss2+w2−ss2+n2)=kn2−w2(L{cos(wt)}−L{cos(nt)})L{x(t)}=L{kn2−w2(cos(wt)−cos(nt))}so:x(t)=k(cos(wt)−cos(nt))n2−w2 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-16466Next Next post: A-toroid-core-has-N-1200-turns-length-L-80cm-cross-section-area-A-60cm-2-current-I-1-5A-Compute-B-and-H-Assume-an-empty-core- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.