Laplace Transformation

Transcription

1 Univerity of Technology Electromechanical Department Energy Branch Advance Mathematic Laplace Tranformation nd Cla Lecture 6 Page of 7

2 Laplace Tranformation Definition Suppoe that f(t) i a piecewie continuou function. The Laplace tranform of f(t) i L f t and defined a { } denoted t { f () t } e () L f t dt () There i an alternate notation for Laplace tranform. For the ake of convenience we will often denote Laplace tranform a, L f t F Example If c { }, evaluate the following integral. e ct dt Solution Remember that you need to convert improper integral to limit a follow, ct e dt lim e n Now, do the integral, then evaluate the limit. n n ct ct ct cn e dt lim e dt lim e lim e c c c n ct dt n n n Now, at thi point, we ve got to be careful. The value of c will affect our anwer. We ve already aumed that c wa non-zero, now we need to worry about the ign of c. If c i poitive the exponential will go to infinity. On the other hand, if c i negative the exponential will go to zero. So, the integral i only convergent (i.e. the limit exit and i finite) provided c<. In thi cae we get, e ct dt provided c< c () Now that we remember how to do thee, let compute ome Laplace tranform. We ll tart off with probably the implet Laplace tranform to compute. Dr.Eng Muhammad.A.R.ya Example Compute L{}. Solution There not really a whole lot do here other than plug the function f(t) into () L {} e t dt Now, at thi point notice that thi i nothing more than the integral in the previou example with c. Therefore, all we need to do i reue () with the appropriate ubtitution. Doing thi give, L {} e t dt provided < Page of 7

3 at Example Compute L { e } Solution Plug the function into the definition of the tranform and do a little implification. L { e ( ) } e e e a t at t at dt dt Once again, notice that we can ue () provided c a. So let do thi. L Example 4 Compute L{in(at)}. at ( ) { e } e a t dt a provided a < -a provided > a Solution Note that we re going to leave it to you to check mot of the integration here. Plug the function into the definition. Thi time let alo ue the alternate notation. L in at F { } n t ( at) dt e t e in lim in n Now, if we integrate by part we will arrive at, n n t t F( ) lim e co( at) co( at) dt n a a e at dt Now, evaluate the firt term to implify it a little and integrate by part again on the integral. Doing thi arrive at, n n n t t F( ) lim ( e co( an) ) e in( at) + in ( at) dt n a a a a e Now, evaluate the econd term, take the limit and implify. n n n t F( ) lim ( co( an) ) in( an) in ( at) dt n e a a e + a a e t in ( at) dt a a a e Dr.Eng Muhammad.A.R.ya a a e t in at dt Now, imply olve for F() to get, L { in( at) } F( ) a provided + a > Page of 7

4 Fact Given f(t) and g(t) then, for any contant a and b. L { af ( t) + bg( t) } af( ) + bg( ) Example Find the Laplace tranform of the given function. f t 6e + e + t 9 t t (a) F! (b) g( t) 4co( 4t) 9in( 4t) + co( t) G (c) ht inh( t) + in( t) ( 4) ( 4) ( ) H t t (d) g( t) e + co( 6t) e co( 6t) G Dr.Eng Muhammad.A.R.ya Dr.Eng Example Find the tranform of each of the following function. (a) f ( t) tcoh( t) L { }, where coh( ) F tg t G g t t * Multiply by t So, we then have, Uing # we then have, + 9 G( ) G ( ) 9 9 F + 9 ( 9) ( ) Page 4 of 7

5 (b) ht t in( t) Thi part will alo ue # in the table. In fact we could ue # in one of two way. We could ue it with n. H L tf t F, where f t tin t { } Or we could ue it with n. H L t f t F, where f t in t { } Since it le work to do one derivative, let do it the firt way. So uing #9 we have, 4 6 F( ) F ( ) The tranform i then, (c) g() t t H 6 ( + 4) Thi part can be done uing either #6 (with n ) or # (along with #). We will ue # o we can ee an example of thi. In order to ue # we ll need to notice that t t vdv t t vdv Now, uing #, π f () t t F( ) we get the following. π π G( ) 4 Thi i what we would have gotten had we ued #6. f t t (d) () For thi part we ll will ue #4 along with the anwer from the previou part. To ee thi note that if () then f ( t) g( t) g t t Therefore, the tranform i. F( ) G (e) f ( t) tg ( t) Thi final part will again ue # from the table a well a #. d L L d Dr.Eng Muhammad.A.R.ya d G g d 4 π π 4 { tg () t } { g } { ( ) } ( G( ) + G ( ) ) G( ) G ( ) Remember that g() i jut a contant o when we differentiate it we will get zero! Page of 7

6 * Dividing by t Dr.Eng Muhammad.A.R.ya Page 6 of 7

7 Laplace Tranformation of periodic Function Dr.Eng Muhammad.A.R.ya Page 7 of 7

8 Dr.Eng Muhammad.A.R.ya Page 8 of 7

9 laplace Home work. (a) t (b) t + 4t (a) (b) + 4 ]. (a) t t t + (b) 4 t4 + t (a) (b) ] ]. (a) e t (b) e (a) t (b) + 4. (a) 4 in t (b) co t (a) ] + 9 (b) + 4. (a) 7 coh x (b) inh t ] 7 (a) 4 (b) 9 6. (a) co t (b) in x (a) ( + ) ( + 4) 7. (a) coh t (b) inh θ ] 4 (b) ( + 6) ] (a) ( 4) (b) 6 ( 6) 8. 4 in(at + b), where a and b are contant ] 4 (a co b + in b) + a 9. co(ωt α), where ω and α are contant ] ( co α + ω in α) + ω. Show that L(co t in t) + 6. (a) te t (b) t e (a). (a) 4t e t (b) t4 e t. (a) e t co t (b) e t in t (a) + ( ) (b) ] ( ) ] 4 (a) ( + ) 4 (b) ( + ) ] 6 (b) (a) e t co t (b) 4e t in t ] ( + ) 4 (a) (b) (a) e t in t (b) et co t (a) + (b) (a) e t inh t (b) e t coh 4t ] ( ) (a) (b) ( ) 4 7. (a) e t inh t (b) 4 e t coh t Dr.Eng Muhammad.A.R.ya 6 (a) (a) e t ( co t in t) (b) (b) e t ( inh t coh t) ( ) (a) (b) + ] + 4( ) ] 6( + ) ( + 4) Page 9 of 7 K

10 Invere Laplace Tranform L F( ) f t { } A with Laplace tranform, we ve got the following fact to help u take the invere tranform. Fact Given the two Laplace tranform F() and G() then L af + bg al F + bl G for any contant a and b. { } { } { } Example Find the invere tranform of each of the following. (a) F( ) H F () () 8 t 6 4( t e + e ) f t (b) F (c) H. () ht F 6 e + 4e 8t t ( ) 4! 4! ! 9 + 4! 7 9e e + t 4 t t 4 Dr.Eng Muhammad.A.R.ya ( ) ( ) ( ) ( ) 4+ G (d) give, f t t t () 6co( ) + in( ) Page of 7

11 G ( 4)( 7 ) g() t in( t) + inh( 7t) 7 Example Find the invere tranform of each of the following. F (a) F (b) F F + 8+ ( ) F ( ) ( ) () 6co( 7 ) in( 7 ) f t t t ( ) F () t t e co( ) + e in( ) f t t t ( + ) ( + 4) + ( + ) + ( + 4) + ( 4) ( + 4) + Dr.Eng Muhammad.A.R.ya G (c) 6 G ( ) ( ) 4 Page of 7

12 H G (d) ( ) ( ) ( ) + ( ) ( ) ( ) ( ) t t g() t e coh t + inh t e + 7 H H + 7 ( + )( ) A B + + ( ) + B( + ) + 7 A A + B + A( 7) + B( ) A 7 picking. We can B in the ame way if we cho A( ) + B( 7) B 7 t when it doe it will uually implify the work c tranform become, 7 7 H( ) + + () ht e + e 7 7 t t Dr.Eng Muhammad.A.R.ya Dr.Eng Factor in denominator ax+ b ( ax+ b) k Term in partial fraction decompoition A ax+ b A A Ak ax+ b ax+ b ax+ b Ax+ B ax + bx+ c ax + bx+ c Ax k + B Ax + B Ax k + Bk ax + bx+ c ax + bx+ c ax + bx+ c ax + bx+ c k k Page of 7

13 Example Find the invere tranform of each of the following. (a) G( ) ( + )( 4)( ) G A B C A 4 + B + + C + 4 A with the lat example, we can eaily get the contant by correctly picking value of. 6 A 7 6 A C C 4 66 B 7 9 B So, the partial fraction decompoition for thi tranform i, G( ) G( ) (b) F( ) ( 6)( + ) t t 4t () e + e + e g t So, for the firt time we ve got a quadratic in the denominator. Here the decompoition for thi part. A B+ C F( ) Setting numerator equal give, A + + B+ C 6 A B A + + B+ C A A B 6B C 6C Dr.Eng Muhammad.A.R.ya A+ B + 6B+ C + A 6C : : 6 : 6B C A, B, C A C F () t e + 8co in f t t t 47 Page of 7

14 (c) G( ) ( ) A B C D+ E G Setting numerator equal and multiplying out give. A B C D+ E 4 A+ D + 4A+ B+ E + A+ 4B+ C + B+ 4C + C : A+ D : 4A+ B+ E 4 :A+ 4B+ C A, B 4, C, D, E : B+ 4C : C G G ( ) ( + ) ( + ) !! then. () t co() t g t t+ t e t e in () t Dr.Eng Muhammad.A.R.ya Page 4 of 7

15 Step Function Before proceeding into olving differential equation we hould take a look at one more function. Without Laplace tranform it would be much more difficult to olve differential equation that involve thi function in g(t). The function i the Heaviide function and i defined a, if t < c uc () t if t c Here i a graph of the Heaviide function. Heaviide function are often called tep function. Here i ome alternate notation for Heaviide function. u t u t c H t c c Example Write the following function (or witch) in term of Heaviide function. 4 if t < 6 if 6 t < 8 f () t 6 if 8 t < if t Solution There are three udden hift in thi function and o (hopefully) it clear that we re going to need three Heaviide function here, one for each hift in the function. Here the function in term of Heaviide function. f t 4+ 9u t 9u t 6u t Dr.Eng Muhammad.A.R.ya It fairly eay to verify thi. 6 8 In the firt interval, t < 6 all three Heaviide function are off and the function ha the value f t 4 Notice that when we know that Heaviide function are on or off we tend to not write them at all a we did in thi cae. In the next interval, 6 t < 8 the firt Heaviide function i now on while the remaining two are till off. So, in thi cae the function ha the value. f ( t ) 4+ 9 Page of 7

16 In the third interval, 8 t < the firt two Heaviide function are one while the lat remain off. Here the function ha the value. f ( t ) In the lat interval, t all three Heaviide function are one and the function ha the value. f ( t ) So, the function ha the correct value in all the interval. Uing Heaviide function thi witch can be wrote a g t u t f t c L c t { c() ( )} e c() ( ) u t f t c u t f t c dt t e f t c dt c. Now ue the ubtitution u t c and the integral become, L u ( c) { c() } + e u c u t f t c f u du e e f u du The econd exponential ha no u in it and o it can be factored out of the integral. Note a well that in the ubtitution proce the lower limit of integration went back to. c u L { uc() t f ( t c) } e e f ( u) du Dr.Eng Muhammad.A.R.ya Now, the integral left i nothing more than the integral that we would need to compute if we were going to find the Laplace tranform of f(t). Therefore, we get the following formula c L u t f t c e F () { c } { } L () e c F u t f t c c c We can ue () to get the Laplace tranform of a Heaviide function by itelf. To do thi we will conider the function in () to by f(t). Doing thi give u c c c L{ uc() t } L{ uc() t i} e L {} e e Putting all of thi together lead to the following two formula. L c e { u () t } L c e Page 6 of 7

17 Example Find the Laplace tranform of each of the following. (a) () () () ( t g t u t + t 6 u t 7 e ) u () t 6 4 are uing the following function, () ( 6) ( 6) f t t f t t and thi ha been hifted by the correct amount. The third term ue the following function, t t 4 f t 7 e f t 4 7 e 7 e which ha alo been hifted by the correct amount. t With thee function identified we can now take the tranform of the function. e 6! 7 4 G( ) + e + e + (b) f ( t) tu ( t) + co( t) u ( t). Here they are. 6 e e e + () ( + ) () + co( + ) () f t t u t t u t 4 co( + ) ( ) co( + ) g t t g t t Thi will make our life a little eaier o we ll do it thi way. Now, breaking up the firt term and leaving the econd term along give u, ( 6 9) () co( ) () () f t t + t + u t + t + u t Okay, o it look like the two function that have been hifted here are g t t + 6t+ 9 () co( t+ ) Taking the tranform then give, 6 9 co( ) in( ) F( ) e + e g t 4 t if t < (c) ht () 4 t t + in if t Dr.Eng Muhammad.A.R.ya Thi one in t a bad a it might look on the urface. The firt thing that we need to do i writ in term of Heaviide functi on. 4 t ht () t u() t in + 4 t + u() t in ( t ) 4! ( ) e 4 H All we need to do now i to take the tranform. e + Page 7 of 7

18 t if t < 6 (d) f () t 8 + ( t 6) if t 6 Again, the firt thing that we need to do i write the function in term of Heaviide function. Here i the corrected function. 6() () ( 6) f t t t t u t 6() ( 8 ( 6) 6 ( 6) ) 6() ( 4 ( 6) ( 6) ) 6() () + 8 ( 6+ 6) + ( 6) f t t t t u t t+ t + t u t t+ t + t u t So, in the econd term it look like we are hifting g t t t 4 The tranform i then, () 4 e 6 + F Example Find the invere Laplace tranform of each of the following. (a) H( ) H Setting numerator equal give, e 4 ( + )( ) e e 4 4 ( + )( ) F F A B + + ( ) A + B + We ll find the contant here by electing value of. Doing thi give, 8 8B B A A 4 4 Dr.Eng Muhammad.A.R.ya So, the partial fraction decompoition become, F( ) ( + ) Notice that we factored a out of the denominator in order to actually do the invere tranform. The invere tranform of thi i then, t t f () t e + e 4 Now, let go back and do the actual problem. The original tranform wa, 4 H e F ht u t f t where, f(t) i, 4 4 () f t e + e 4 t t Page 8 of 7

19 (b) G( ) 6 e e ( + )( + 9) G G 6 e e ( + )( + 9) ( + )( + 9) 6 e e ( + )( + 9) ( + )( + 9) 6 6 ( e e ) ( e e ) F( ) ( + )( + 9 ) G and now we will jut partial fraction F(). Here i the partial fraction decompoition. A B+ C F Setting numerator equal and combining give u, A B+ C A+ B + B+ C + 9A+ C Setting coefficient equal and olving give, : A+ B : B+ C A, B, C :9A+ C Subtituting back into the tranform give and fixing up the numerator a needed give, + F( ) little impler. Taking the invere tranform then give, t f () t e co( t) + in( t) At thi point we can go back and tart thinking about the original problem. 6 G e e F e F e F( ) 6 We ll alo need to ditribute the F() through a well in order to get the correct Recall that in order to ue () to take the invere tranform you mut have a i time a ingle tranform. Thi mean that we mut multiply the F() through t can now take the invere tranform, g t u t f t 6 u t f t where, Dr.Eng Muhammad.A.R.ya 6 f t t t t () e co( ) + in( ) (c) F( ) 4 + e ( )( + ) Page 9 of 7

20 In thi cae, unlike the previou part, we will need to break up the tranform ince one term ha a contant in it and the other ha an. Note a well that we don t conider the exponential in thi, only it coefficient. Breaking up the tranform give, 4 F( ) + e G( ) + e H( ) + + We will need to partial fraction both of thee term up. We ll tart with G(). A B G( ) + + Setting numerator equal give, 4 A + + B Now, pick value of to find the contant. 8 B 8 B 4 A 4 A So G() and it invere tranform i, Now, repeat the proce for H(). Setting numerator equal give, 4 8 G t 8 t g() t e + e H A B + + B( ) A + + Now, pick value of to find the contant. B B A A So H() and it invere tranform i, H( ) Putting all of thi together give the following, where, Dr.Eng Muhammad.A.R.ya t + ht () e e + e () () + () ( ) F G H f t g t u t ht 4 8 e + e e e () and ht () g t t t t t t Page of 7

21 (d) G( ) 7 + 8e e + 6e ( + ) Thi one look meier than it actually i. Let firt rearrange the numerator a little. 7 ( e ) + ( 8e + 6e ) G( ) + In thi form it look like we can break thi up into two piece that will require partial fraction. When we break thee up we hould alway try and break thing up into a few piece a poible for the partial fractioning. Doing thi can ave you a great deal of unneceary work. Breaking up the tranform a uggeted above give, 7 G( ) ( e ) + ( 8e + 6e ) ( e ) F( ) ( 8e 6e ) H( ) + + Note that we canceled an in F(). You hould alway implify a much a poible before doing the partial fraction. Let partial fraction up F() firt. Setting numerator equal give, F A B + + A + + B Now, pick value of to find the contant. B B A A So F() and it invere tranform i, Dr.Eng Muhammad.A.R.ya F + t f () t e Now partial fraction H(). Setting numerator equal give, H A B C A + + B + + C Page of 7

22 Pick value of to find the contant. 9C C 9 B B 4A+ 4B+ C 4A+ A 9 9 So H() and it invere tranform i, 9 9 H t ht () + t+ e 9 9 Now, let go back to the original problem, remembering to multiply the tranform through the parenthei. 7 G F e F + 8e H + 6e H Taking the invere tranform give, g t f t u t f t + 8u t ht + 6u t ht 7 7 Dr.Eng Muhammad.A.R.ya Dr.Eng Page of 7

23 Invere Laplace Tranformation. (a) 7. (a). (a) 4. (a) (b) (b) (b) + 9 (a) 7 (b) e t ] (a) e t ] (b) co t (a) in t (b) 4 in t ] (b) 6 (a) ] co t (b) 6t. (a) (b) 8 4 (a) t (b) 4 t ] 6. (a) 8 7. (a) 7 7 (b) 6 4 (b) ( ) 8. (a) ( + ) 4 (b) ( ) (a) 6 coh 4t (b) 74 inh 4t ] (a) inh t (b) et t ] (a) 6 e t t (b) 8 et t 4 ] Dr.Eng Muhammad.A.R.ya 9. (a) (b) (a) e t co t (b) e t in t. (a) ( ) 6 + (b) (a) e t co t (b) 7 ] e4t inh t. (a) (b) (a) e t coh t + e t inh t 4t 4 4t Page of 7

24 Solving IVP with Laplace Tranform Fact Suppoe that f, f, f, f (n-) are all continuou function and f (n) i a piecewie continuou function. Then, ( n) n n n n n L f F f f f f { } Since we are going to be dealing with econd order differential equation it will be convenient to have the Laplace tranform of the firt two derivative. L y Y y L { } { y } Y( ) y( ) y ( ) Example Solve the following IVP. y y + 9y, t y y Solution The firt tep in uing Laplace tranform to olve an IVP i to take the tranform of every term in the differential equation. L y L y + 9L y L t { } { } { } { } Uing the appropriate formula from our table of Laplace tranform give u the following. Y( ) y( ) y ( ) ( Y( ) y( ) ) + 9Y( ) Plug in the initial condition and collect all the term that have a Y() in them. ( + 9) Y( ) + Solve for Y (). Y ( ) Combining the two term give, Y + ( 9)( ) The partial fraction decompoition for thi tranform i, A B C D Y( ) Dr.Eng Muhammad.A.R.ya Setting numerator equal give, + A 9 + B 9 + C + D 9 Picking appropriate value of and olving for the contant give, 9B B 6 8D D C C 4 4A+ A Plugging in the contant give, Y( ) Finally taking the invere tranform give u the olution to the IVP. 9t y() t + t+ e e t Page 4 of 7

25 Example Solve the following IVP. t y + y y te, y y Solution A with the firt example, let firt take the Laplace tranform of all the term in the differential equation. We ll the plug in the initial condition to get, ( Y( ) y( ) y ( ) ) + ( Y( ) y( ) ) Y( ) + Now olve for Y(). Y ( ) Y( ) ( )( + ) ( )( + ) ( + ) Now, a we did in the lat example we ll go ahead and combine the two term together a we will have to partial fraction up the firt denominator anyway, o we may a well make the numerator a little more complex and jut do a ingle partial fraction. Thi will give, Y Setting numerator equal give, Y 4( + ) ( )( + ) ( )( + ) 4 6 A B C D ( A B) ( 6A 7B C) ( A 4B C D) A B C D A 4B C D In thi cae it probably eaier to jut et coefficient equal and olve the reulting ytem of equation rather than pick value of. So, here i the ytem and it olution. : A+ B 9 96 A B : 6A+ 7B+ C 4 :A+ 4B+ C+ D 6 C D : 8A 4B C D Dr.Eng Muhammad.A.R.ya We will get a common denominator of on all thee coefficient and factor that out when we go to plug them back into the tranform. Doing thi give,! 9 96! Y ( ) + ( ) + ( + ) ( + ) Taking the invere tranform then give, t y t 96e + 96e te t e () t t t Page of 7

26 Example Solve the following IVP. y 6y + y in t, y y 4 Solution Take the Laplace tranform of everything and plug in the initial condition ( 6+ ) Y( ) ( ) Y y y Y y Y Now olve for Y() and combine into a ingle term a we did in the previou two example Y( ) Now, do the partial fraction on thi. Firt let get the partial fraction decompoition. A+ B C+ D Y( ) Now, etting numerator equal give, A+ B 6+ + C+ D + 9 A+ C + 6A+ B+ D + A 6B+ 9C + B+ 9D Setting coefficient equal and olving for the contant give, : A+ C A B : 6A+ B+ D :A 6B+ 9C 9 C D : B+ 9D 4 Now, plug thee into the decompoition, complete the quare on the denominator of the econd term and then fix up the numerator for the invere tranform proce. + + Y ( ) ( ) ( ) ( ) Dr.Eng Muhammad.A.R.ya Finally, take the invere tranform. 8 t co in co 6 t y t t t t in 6t + e e 6 () Page 6 of 7

27 7.. A firt order differential equation involving current i in a erie R L circuit i given by: di dt + i E and i at time t. Ue Laplace tranform to olve for i when (a) E (b) E 4 e t and (c) E in t. (a) i ( e t ) (b) i ( e t e t ) (c) i (e t co t + in t) In Problem to 9, ue Laplace tranform to olve the given differential equation.. 9 d y dy 4 + 6y, given y() dt dt ] and y (). y ( t)e 4 t Home Work d x + x, given x() and dt x (). x co t] d i di + + i, given dt dt i() and i (). i t e t ] d x dt + 6 dx + 8x, given x() 4 and dt x () 8. x 4(e t e 4t )] d y dx dy dx + y e4x, given y() and y () 4 y (4x ) e x + ] e4x Dr.Eng Muhammad.A.R.ya d y + 6y co 4x, giveny() and dx y () 4. y co 4x + in 4x + 4 ] x in 4x 8. d y dx + dy y co x in x, dx given y() and y () 6 y e x e x + in x] d y 9. dx dy dx + y ex co x, given y() and y () y e x ( co x + in x) e x co x ] Page 7 of 7

28 Table of Laplace Tranform f t f t L F( ) F( ) L L F( ) F( ) L f ( t) f t { } { }.. at e. n t, n,,, n! n + 4. p t, p > -. t 7. in ( at ) a + a 9. tin ( at ) a + a. in( at) atco( at). co( at) atin ( at) π a ( + a ) ( a ) ( + a ) in. in ( at+ b) ( b) + aco( b) 7. inh ( at ) at 9. e in ( bt) at. e inh ( bt). n at t e, n,,,. u ( t c ) u ( t c ) Heaviide Function u t f t c + a a a b a + b b a b n! n ( a ) + c e c F 6. t n 8. co( at ). tco( at ) { }, n,,,. in( at) + atco( at) 4. co( at) + atin ( at) a Γ p + p + { } ( n ) n n + + a a ( + a ) a ( + a ) ( + a ) ( + a ) co 6. co( at+ b) ( b) ain ( b) 8. coh ( at ) at. e co( bt ) at. e coh ( bt) 4. f ( ct ) δ 6. ( t c) Dirac Delta Function + a a a a + b a a b F c c e c { } Dr.Eng Muhammad.A.R.ya c 7. c ( ) e 8. uc ( t) g( t ) e L g( t+ c) 9. ct n ( e f ( t) F( c). t f ( t), n,,, ( ) n n F ) ( ) t F. f () t F( u) du t. f ( v) dv ( ) t. f ( t τ) g( τ) dτ F( ) G( ) 4. f ( t T) f ( t) + T t e f () t dt T e t F f f. f ( t) F( ) f ( ) 6. f ( n ) n n n ( n 7. f ( t ) ) ( n F f f f ( ) f ) ( ) π Page 8 of 7

29 The olution of imultaneoudifferential equation uing Laplace tranform It i ometime neceary to olve imultaneou dif-ferential equation. An example occur when two electrical circuit are coupled magnetically where the equation relating the two current i and i are typically: di L + M di di + R i E L + M di + R i dt dt dt dt where L repreent inductance, R reitance, M mutual inductance and E the p.d. applied to one of the circuit. Procedure (i) Take the Laplace tranform of both ide of each imultaneou equation by applying the formulae for the Laplace tranform of derivative (i.e. (ii) Put in the initial condition, i.e. x(), y(), x (), y (). (iii) Solve the imultaneou equation for L{y} and L{x} by the normal algebraic method. (iv) Determine y and x by uing, where neceary, partial fraction, and taking the invere of each term. Example():- olve the following pair of imultaneou differential equation d y dt + x Uing the above procedure: { } dy (i) L + L{x} L{} () dt { } dx L L{y}+4L{e t } () dt Equation () become: dx dt y + 4et given that at t, x and y. Dr Muhammad A R Ya L{y} y()] + L{x} ( ) Equation () become: L{x} x()] L{y} 4 ( ) (ii) x() and y() hence Page 9 of 7 Equation ( ) become:

30 L{y}+L{x} ( ) and equation ( ) become: L{x} L{y} 4 or L{y}+L{x} 4 (iii) equation ( ) and equation ( ) give: L{y}+L{x} L{y}+ L{x} 4 Adding equation () and (4) give: ( + )L{x} 4 ( ) (4) ( ) from which, L{x} 4 + ( )( + ) Uing partial fraction 4 + ( )( + ) A + Hence B ( ) + C + D ( + ) 4 + ( ) A( )( + ) + B( + ) + (C + D)( ) ( )( + ) 4 + A( )( + ) + B( + ) + (C + D)( ) When, A hence A When, 4 B hence B Equating coefficient: A + B + C hence C (ince A and B ) Equating coefficient: 4 A + D C hence D (ince A and C ) Thu L{x} 4 + ( )( + ) ( ) + ( + ) (iv) Hence Dr Muhammad A R Ya { x L ( ) + } { ( + ) L } ( ) + ( + ) ( + ) ( ) () (4) () i.e. x e t + co t in t, From the econd equation given in the quetion, Page of 7

31 dx dt y + 4et from which, y dx dt + 4et d dt ( et + co t in t) + 4e t e t in t co t + 4e t i.e. y e t in t co t Alternatively, to determine y, return to equation ( ) and ( )] Example():- olve the following pair of imultaneou differential equation dx dt dy dt + x 6 dy dt dx y dt given that at t, x 8 and y. Uing the above procedure: { } { } dx dy () (i) L L + L{x} L{6} dt dt { } { } dy dx L L L{y} L{ } () dt dt Equation () become: L{x} x()] L{y} y()] + L{x} 6 L{x} x() L{y} + y() + L{x} 6 ( ). ( + )L{x} x() L{y}+ y() 6 Equation () become: L{y} y()] L{x} x()] L{y} L{y} y() L{x}+ x() L{y} ( )L{y} y() L{x} + x() ( ) (ii) x() 8 and y(), hence equation ( ) become ( + )L{x} (8) L{y}+ () 6 ( ) and equation ( ) become Dr Muhammad A R Ya ( )L{y} () L{x} + 8 ( + )L{x} L{y} ( ) ( ) i.e. ( + )L{x} L{y} ( ) L{x}+( )L{y} ( ) (A) (iii) equation ( ) and (+) equation ( ) Page of 7

32 ive: 6 ( + )L{x} L{y} + 9 () ( + )L{x}+( + )( )L{y} ( + ) ( ) (4) i.e. ( + )L{x} L{y} ( ) ( + )L{x}+(6 + )L{y} 6 7 (4 ) Adding equation ( ) and (4 ) give: ( + )L{y} + from which, L{y} ( + ) Uing partial fraction ( + ) A + + B ( + ) + C A( + )( ) + B( ) + C( + ) ( ) ( + )( ) A( + )( ) + B( ) + C( + ) When, A, hence A When, C, hence C When, 6B, hence B Thu L{y} ( + ) + ( + ) { (iv) Hence y L + } + e t + Returning to equation (A) to determine L{x} and hence x: ( ) equation ( ) and ( ) give: ( )( + )L{x} ( )L{y} ( ) ( ) and Dr Muhammad A R Ya ()L{x} + ( )L{y} ( ) (6 + )L{x} ( )L{y} ( ) () (6) and L{x}+( )L{y} (6 ) Adding equation ( ) and (6 ) give: Page of 7

33 from which, L{x} 8 6 ( + ) 8 6 ( + )( ) Uing partial fraction 8 6 ( + )( ) A + B ( + ) + C ( ) 8 6 A( + )( ) + B( ) + C( + ) A( + )( ) + B( ) + C( + ) ( + )( ) When, 6 A, hence A When, C, hence C When, 6B, hence B Thu L{x} 8 6 ( + )( ) + ( + ) { Hence x L + } + e t + Therefore the olution of the given imultaneou differential equation are y + e t and x + e t (Thee olution may be checked by ubtituting the expreion for x and y into the original equation.) Example():- olve the following pair of imultaneou differential equation d x dt x y Uing the procedure: d y dt + y x (i) L{x} x() x ()] L{x} L{y} () L{y} y() y ()] + L{y} L{x} () (ii) x(), y(), x () and y () hence L{x} L{x} L{y} ( ) L{y}+ + L{y} L{x} ( ) (iii) Rearranging give: ( )L{x} L{y} () given that at t, x, y, dx dt and dy dt. Dr Muhammad A R Ya K L{x}+( + )L{y} (4) Equation () ( + ) and equation (4) give: ( + )( )L{x} ( + )L{y} ( + ) () Page of 7 L{x}+( + )L{y} (6)

34 ( + )( ) + ]L{x} ( + ) i.e. 4 L{x} + ( + ) from which, L{x} ( + ) 4 + { (iv) Hence x L + } + + i.e. x + t Returning to equation () and (4) to deter-mine y: equation () and ( ) equation (4) give: ( )L{x} L{y} (7) ( )L{x}+( )( + )L{y} ( ) (8) Equation (7) equation (8) give: ( )( + )]L{y} + ( ) i.e. 4 L{y} + and L{y} + 4 { from which, y L } i.e. y t Dr Muhammad A R Ya Page 4 of 7

35 Further Problem Solve the following pair of imultaneou dif- ferential equation:. dx dt + dy dt et anwer. dy dt y + x + dx dt in t anwer dy dt dx dt given that when t, x and y x e t t and y t + e t ] x co t + in t e t e t ] and y e t + e t in t. d x dt + x y anwer d y dt + y x ] x co t + co ( t) and y co t co ( t) dy dt + x y + dx dt et given that at t, x and y given that at t, x 4, y, dx dt Dr Muhammad A R Ya and dy dt Page of 7

36 Home Work. Find the Laplace tranform of the following function: (a) t 4t + (b) e t 4 in t (c) coh t (d) t 4 e t (e) e t co t (f) e t inh 4t. Find the invere Laplace tranform of the fol- lowing function: (a) + (b) (c) (d) 9. Ue partial fraction to determine the following: { } { (a) L (b) L } + 9 ( )( + ) (e) ( + ) 4 (f) 4. In a galvanometer the deflection θ atifie the differential equation: d θ dt + dθ dt + θ 4 θ given that when t, θ Ue Laplace tranform to olve the equation for and dθ dt 4 8 { (c) L } ( ) Dr Muhammad A R Ya (g) Solve the following pair of imultaneou differ- ential equation: dx dt x + y dy dt + x 6y given that when t, x and y Page 6 of 7

37 Table Note. Thi lit i not incluive and only contain ome of the more commonly ued Laplace tranform and formula.. Recall the definition of hyperbolic function. e + e e e coh() t inh () t t t t t. Be careful when uing normal trig function v. hyperbolic function. The only difference in the formula i the + a for the normal trig function become a - a for the hyperbolic function! 4. Formula #4 ue the Gamma function which i defined a If n i a poitive integer then, x t () Γ t e x dx ( n ) Γ + n! The Gamma function i an extenion of the normal factorial function. Here are a couple of quick fact for the Gamma function ( p ) p ( p) Γ + Γ Γ ( p n) ( p) Γ + p( p+ )( p+ ) ( p+ n ) Γ π Dr.Eng Muhammad.A.R.ya Page 7 of 7