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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-

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Question Number 16475 by Sai dadon. last updated on 22/Jun/17
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.
useL{f′(t)}tofindlaplacetransformofd2x/dt2+n2x=kcoswtgiventhatx=0anddx/dt=0whent=0PLZHELP.
Answered by sma3l2996 last updated on 23/Jun/17
we know that L{f′(t)}=sL{f(t)}−f(0) and L{f′′(t)}=sL{f′(t)}−f′(0) so L{f′′(t)}=s^2 L{f(t)}−sf(0)−f′(0) let x(t)=f(t) L{((d^2 x(t))/dt^2 )}=s^2 L{x(t)}−sx(0)−((dx(0))/dt)=s^2 L{x(t)} and also we have: L{(d^2 x/dt^2 )+n^2 x}=L{kcos(wt)} ⇔L{(d^2 x/dt^2 )}+n^2 L{x}=kL{cos(wt)}=k.(s/(s^2 +w^2 )) s^2 L{x}+n^2 L{x}=k.(s/(s^2 +w^2 )) (s^2 +n^2 ).L{x}=k.(s/(s^2 +w^2 )) L{x}=k.(s/((s^2 +w^2 )(s^2 +n^2 ))) (s/((s^2 +w^2 )(s^2 +n^2 )))=((as+b)/(s^2 +w^2 ))+((cs+d)/(s^2 +n^2 )) a=−c , (b/w^2 )=−(d/n^2 )⇔d=−((n/w))^2 b (1/((1+w^2 )(1+n^2 )))=((a+b)/(1+w^2 ))−((a+((n/w))^2 b)/(1+n^2 ))⇔(a+b)(1+n^2 )−(a+((n/w))^2 b)(1+w^2 )=1 a(n^2 −w^2 )+b(1+n^2 −((n/w))^2 (1+w^2 ))=1 a(n^2 −w^2 )=1+b(((n^2 −w^2 )/w^2 )) (i):a=(b/w^2 )+(1/(n^2 −w^2 )) (2/((w^2 +4)(n^2 +4)))=((2a+b)/(w^2 +4))−((2a+((n/w))^2 b)/(n^2 +4)) (2a+b)(n^2 +4)−(2a+((n/w))^2 b)(w^2 +4)=2 2a(n^2 −w^2 )+b(4−4((n/w))^2 )=2 4b(((w^2 −n^2 )/w^2 ))=2+2a(w^2 −n^2 ) (ii):(b/w^2 )=(1/(2(w^2 −n^2 )))+(a/2) from (i) and (ii) a=(1/2)a+(1/(2(w^2 −n^2 )))+(1/(n^2 −w^2 )) a=(1/(n^2 −w^2 ))=−c (b/w^2 )=(1/(2(w^2 −n^2 )))+(1/(2(n^2 −w^2 )))=0 a=−c=(1/(n^2 −w^2 )) and b=d=0 so : L{x}=(k/(n^2 −w^2 ))((s/(s^2 +w^2 ))−(s/(s^2 +n^2 )))=(k/(n^2 −w^2 ))(L{cos(wt)}−L{cos(nt)}) L{x(t)}=L{(k/(n^2 −w^2 ))(cos(wt)−cos(nt))} so : x(t)=((k(cos(wt)−cos(nt)))/(n^2 −w^2 ))
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

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