TMA4125 Matematikk 4N Spring 2016
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1 Norwegian Univerity of Science and Technology Department of Mathematical Science TMA45 Matematikk 4N Spring 6 Solution to problem et 6 In general, unle ele i noted, if f i a function, then F = L(f denote it Laplace tranform. a By definition F ( = = e t f(t dt = t e t dt + e t (t + t + dt te t + dt e t dt. Thee are all eay to compute. A an example, the middle integral i [te t] t= te t dt = + t= = [ e t] t= = t= e t dt by integration by part. The firt integral require integration by part twice. We end up with F ( = b Write out the exponential form of co and inh. Then by definition F ( = = 4 = 4 e t f(t dt = ( e t inh(t co(t dt e t (e it + e it (e t e t dt e t ( e t(i+ e t(i + e t( i+ e t( i dt = e t(i+ e t(i + e t( i+ e t( i dt 4 = ( 4 i + + i i + + i = ( ( ( + 4 ( + 4 ( a By table 6.9, L ( n (t = t n /(n!, o by linearity L (F (t = t +. February 6, 6 Page of 5
2 Solution to problem et 6 b Write G( = / 4 and H( = /( +. Then we know by table 6.9 that g(t = t /6 and h(t = co t. Since F ( = G( + H(, the -hifting theorem give c Rewrite f(t = e t g(t + h(t = et t + co t. Define G( = /, o g(t =. Then F ( = Again -hifting give ( F ( = + +. ((G( G( (. + f(t = + (e t e t. We will here ue the Laplace tranform of derivative from ection 6. without further comment. Only the firt point will be done really thoroughly a the remaining one follow the ame procedure. a Laplace tranform both ide of the equation to obtain We rearrange into Y ( + Y ( = Y ( = ( + ( + 4. Partial fraction decompoition will move u forward, o to thi end, write Y ( = A + + B + C + 4 = A( (B + C( + ( + ( + 4 where A, B and C are contant which we now determine. The numerator mut equal, o the coefficient in front of mut, while that in front of and that in front of mut be. Thu we have (A + B =, (C + B = and (4A + C =. From thi we find A = 9/5, B = /5 and C = 8/5. Thu Y ( = = 5 ( Invere tranformation (uing table 6.8/6.9 and what we know from previou problem give y(t = 5 e t/ 5 co(t 9 5 in(t. February 6, 6 Page of 5
3 Solution to problem et 6 b In order to ue the t-hifting theorem on the right hand ide, we rewrite u(π t = u(t π. Then we have the ODE Thu y (t + 4y(t = 4 co(t ( u(t π. Laplace tranformation uing the t-hifting theorem give (after uing the initial condition that Thu we need to invere tranform Y ( = Y (( + 4 = e π +. 4 ( + ( e π ( + ( = 4G( + 4G(e π + 4 with G( a indicated. Partial fraction decompoition give that y(t = 4g(t + 4g(t πu(t π g(t = (co t co t. = 4 (co t co t + 4 (co(t π co((t π u(t π = 4 (co t co t + 4 ( co t co(t u(t π. (You can probably clean up the olution even further if you want to. c In order to ue the t-hifting theorem, we rewrite the right hand ide in term of the Heaviide tep function: { in t co t for t < π in(t co(t for t π = ( in t co t( u(t π + ( in(t co(tu(t π = in t co t + (co t in tu(t π + ( in(t co(tu(t π = in t co t + (co(t π in(t πu(t π + ( in((t π co((t πu(t π. The lat rewriting wa done uing the periodicity of coine (it wa necceary in order to apply the t-hifting theorem. Laplace tranforming the ODE and uing the initial condition and the t-hifting theorem give ( Y ( = e π + e π + e π + e π. Tidying up, we find that we need to invere tranform Y ( = Make ure you undertand why! ( + ( ( e π. February 6, 6 Page of 5
4 Solution to problem et 6 By partial fraction decompoition, we get ( + ( + = ( + ( + ( = ( + ( + + (. Alo, uch that = ( + ( + ( +, Y ( = ( + ( + + ( + e π ( + ( + = ( + ( + + ( e π ( + + e π ( + by uing partial fraction decompoition on the lat term. Then, by uing the hift theorem, we get y(t = in t e t + et + (in t in(t u(t π. 4 The function f i quite imple, o we can compute it Laplace tranform directly from the definition, namely L(f( = = e t f(t dt = e t ( t dt e ( x x dx = e e x x dx = + e. Or, if you prefer intead to bring in the big gun, write f(t = ( t( u(t = t ( tu(t = t + (t u(t = g(t + g(t u(t, with g(t = t (o G( = L(g( = /. Now we can ue the t-hifting theorem, from which we get L(f( = F ( = G( + e G( = + e. 5 Since Kreyzig operate in a loppy manner where the domain of L(f i not pecified, we mut make ure we undertand exactly what i being aked. For example, g defined by g(t = t clearly ha a Laplace tranform; we know it given by G( = /. But even G i not defined at. Thi i not the kind of non-exitence we re after. Here we are looking for function f for which F ( doe not exit for any. There are of coure many correct anwer to thi quetion, but here are ome example of function f : R R for which F ( doe not exit for any : February 6, 6 Page 4 of 5
5 Solution to problem et 6. If f(t = e t, then for all R it i well known (ee Matematikk that F ( = doe not exit. e t e t dt. Likewie for omething like f(t = e t, or coure.. Defining { if t i rational f(t = otherwie make f non-integrable, whether alone or multiplied with e t for any. February 6, 6 Page 5 of 5
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