Page 5- Chater 5: Lalace Transforms The Lalace Transform is a useful tool that is used to solve many mathematical and alied roblems. In articular, the Lalace transform is a technique that can be used to solve linear constant coefficient differential equations in one or more equations. This tool is used to transform the given system of differential equations into a system of algebraic equations that are then solved simultaineously. The algebraic system obtained is usually much easier to solve then the given differential equations. Grahically, f(t) or f(x) or f(u) etc Transfer function F() Figure (a) Figure As shown in the Figure (a) the original function f(t) (solution to the given differential equation) is transformed into a new function F (s) via a transfer function. This function F (s) is obtained from solving an algebraic equation in s. Once F (s) is found we use transforms to obtain the original variable t and hence, obtain the solution to the given differential equation. This is described in Figure. The Lalace transform method is used in a wide number of engineering roblems, in articular, the design of control systems in electrical engineering or the vibration of mass-sring system to a unit imulse. For examle, Lalace transforms can be used to determine the resonse of a damed mass-sring system which is initially at rest and is suddenly given a shar hammer blow. The governing equations could be reresented by where y + y + y δ(t a) y() and y (). Our usual methods cannot solve this equation due to the imulse function, δ(t a). However, using Lalace transforms we can solve this equation. It s solution being y(t) ( e (t a) e (t a)) h(t a) where h(t) is the ste function. Discussion on solving differential equations will be given in later in this chater. However, first, we need to become familiar with the Lalace transform and its roerties.
Page 5-5. DEFINITION Let f(t) be a function defined on the interval [, ). Then the Lalace Transform of a function f(t) is defined by Note: L{f(t) F () e t f(t)dt Here the indeendent variable of f is t and the transfer arameter is. This means we are transfering from the t domain to the domain when using Lalace transforms. The arameter can be comlex. However, rovided f(t) is a well behaved function, then the Re { must be ositive for the above integral to converge. In some texts s is used as the transfer arameter instead of although the same definition holds. That is, L{f(t) F (s) In MATH both arameters may be used. e st f(t)dt. There are a variety of transform functions which can aid the mathematician or engineer to solve roblems. These will not be discussed here. However, it is worth mentioning two of these transforms. Fourier: F {f(t) π Two-sided Lalace: L{f(t) e ist f(t))dt e t f(t)dt 5. LAPLACE TRANSFORM OF FUNCTIONS Let L { f(t) F () and L { g(t) G() then articular results for the Lalace transform are L{αf(t) αl{f(t) where α a constant. L{αf(t) + βg(t) αl{f(t) + βl{g(t) L{f(at) a F ( a ) where α and β are constants. L { e at f(t) F ( + a) L {f(t a) h(t a) e a F () (Shift Theorem) (Second Shift Theorem) L {f(t) h(t a) e a L { f(t + a) (Third Shift Theorem) where a is a constant. L{tf(t) d d F ().
Page 5- Examles Recall that L{f(t) F () (a) Find the Lalace transform of f(t) t. f(t) e t dt. L { f(t) L{ t. t e t dt e t dt relacing f(t) by t using integration by arts Note: Here f is a function of the indeendent variable t therefore we found the Lalace transform of f with resect to the indeendent variable t. However, not all functions will have the indeendent variable being t. For instance, the indeendent variable for f(x) is x. If this is the case then we simly take Lalace transforms with resect to x. In all cases, the Lalace transform function F () will be the same. Convention We usually use caital letters to reresent the Lalace transform of a function. For examle, L { f(x) F (), L { g(u) G() and L { z(w) Z(). For examle, If f(x) x, say, then L { f(x) L { x or if g(u) u then L { g(u) L { u. A grah of the function f(x) x and its transform (ie L { f(x) x F () ) is shown below. f(x)x F()
Page 5-4 Find the Lalace transform of f(t) sin t. The indeendent variable is t. find the Lalace transform of f(t) with resect to t. That is, L{ sin t s +. sin t e t dt relacing f(t) by sin t. cos t e t dt using integration by arts Note: Evaluation of this integral could have been done by Integral []. A grah of f(t) sin t and its transform F () is shown below. f(t)sin t F() Similarly, the L{cos t is L{ cos t F () cos t e t dt relacing f(t) by cos t. using integration by arts. + f(t)cos t F()
Page 5-5 (c) Find the Lalace transform of f(t) t + sin t. Recall that L{f(t) + g(t) L { f(t) + L { g(t). Let f(t) t and g(t) sin t then L { f(t) + g(t) L { t + sin t L { t + L { sin t + +. (d) Find the Lalace transform of f(t) {, t, t <. Recall that L{ f(t) e t L{f(t) F () f(t) e t dt + e t dt ] e. f(t) e t dt. f(t) e t dt relacing f(t) by its definition. 5.. Shift Theorem The Shift theorem is an imortant theorem and will be used quite frequently throughout this chater. Recall that L { e at f(t) F ( + a) where a is an arbitrary constant. This theorem means that if we want to find L {e at f(t) we simly find the Lalace transform of f(t) which is F () and then relace by + a. Grahically, this theorem is reresented by F () shifted to the right or left deending on the sign of a. F F() F(-a) a, a >
Page 5-6 Mathematically, we can write that L { e at f(t) F ( + a) F () +a. Also, we shall use the shift theorem in Section 5.4 to find the inverse Lalace transform of a given function of. Examles (a) Find the Lalace transform of f(t) e t sin t. L { e t sin t ( e t sin t ) e t dt relacing f(t) by e t sin t sin t e (+)t dt ( + ) +. Alternatively, we can use the shift theorem. That is, integrating by arts L { e at f(t) F ( + a) where F () L { f(t). In our examle, we see that f(t) sin t and a then F () L { f(t) + from examle (d). Hence, F ( + a) ( + a) + ( + ) +. L { e t sin t ( + ) +. where a Below is a grah of the function f(t) e t sin t and the transform function F ( + ) ( + ) +. f(t) F(+)
Page 5-7 Find the Lalace transform of e t cos 4t. L { e t cos 4t e t cos 4t e t dt cos(4t) e ( )t dt ( ) + 6. relacing f(t) by e t cos 4t integrating by arts Alternatively, using the shift theorem we let f(t) cos 4t where a then F () + 6 from Lalace tables where n 4. Hence, F ( ) ( ) + 6 L { e t cos 4t ( ) + 6. where a. The revious two examles show alternative methods of finding the Lalace transform of functions of the form L { e at f(t). Each of these methods is accetable. However, time and lengthy integration is reduced if the shift theorem and the Lalace transform tables are used. 5.. The Convolution Theorem The Convolution Theorem, states { L f(t u) g(u) du F () G() and is often written as L{f g F.G where L { f(t) F () and L { g(t) G(). Alternatively, the Convolution theorem can also be written in the form where the end result is still the same. Note: It should be noted that the given { L f(u) g(t u) du F () G() Lalace transforms we are doing it with resect to t. f(u) g(t u) du is a function of t and therefore, when we take A secial case of the Convolution Theorem is the result for the Lalace transform of an integral. { L f(u) du F ().
Page 5-8 Examles (a) Find the Lalace transform of u sin(t u) du. It can be easily seen that u sin(t u) du is a function of t. let k(t) u sin(t u) du. L{k(t) k(t) e t dt ( ) u sin(t u) du e t dt. Reversing the order of integration. That is, ( ) L{k(t) u e t sin(t u) dt du u ( ) ue u e z sin(z) dz du as u is a constant in the inner integral Let z t u then dz dt; when t u, z ; t, z. The inner integral is L { sin z. That is, L { sin z +. + ue u du This is the L { u which is Γ(). ( + ). Alternatively, use the Convolution theorem and let f(t u) sin(t u) and g(u) u. Thus, f(t) sin t (relacing t u by t ) and g(t) t. Taking Lalace transforms of both f(t) and g(t) with resect to t we have L { f(t) F () + and L { g(t) G(). { L u sin(t u) du + ( + ).
Page 5-9 Note: It can be seen in this examle that using the Convolution theorem is quicker than doing the double integration. it wise that students learn this theorem. { Find L e u sinh (t u) du. Matching the functions in the Convolution theorem we can see that f(t u) sinh (t u) g(u) e u Thus, f(t) sinh t (relacing t u by t ) and g(t) e t. We take Lalace transforms of both f(t) and g(t) with resect to t. using the Lalace transform tables, we find that L { f(t) F () 9 and L { g(t) G() +. (c) { L e u sinh(t u) du 9 + ( + )( 9). { x Find L (x u) sin 4u du. Here we are taking the Lalace transform with resect to x. Matching the functions in the Convolution theorem, we can see that f(x u) (x u) and g(u) sin 4u. Thus, f(x) x (relacing x u by x ) and g(x) sin 4x Taking Lalace transforms of both f(x) and g(x) with resect to x by using the Lalace transform tables, we obtain L { f(x) F () Γ() 4 and L { g(x) G() + 6. { x L (x u) sin u du Γ() 4 + 6 4Γ() ( + 6) 8 ( + 6).
Page 5-5.. Lalace Transform of a Derivative The Lalace transform of a derivative is: L{f(t) F () L{f (t) F () f() L{f (t) F () f() f () L{f (t) F () f() f () f () These results can be used to solve linear differential equations, and systems of linear differential equations. Examles (a) Find the Lalace transform of dy y where y(). dx Using the above information, we let L { y(x) Y () then { dy L Y () y() dx Y (), where y(). { d y Find L dx dy dx + 4y { dy L dx y Y () Y () ( )Y () where y() and y (). Let L { y(x) Y () then { { dy d y L Y () y() and L dx dx Y () y() y (). Using the given conditions we find that { { dy d y L Y () and L dx dx Y () +. { d y L dx dy { d { dx + 4y y dy L dx L + L { 4y dx Y () + (Y () ) + 4Y () ( + 4)Y () +.
Page 5-5..4 Lalace Transform of Other Tyes Recall that L { tf(t) d F () where L { f(t) F (). d Examles (a) Find the Lalace transform of t sinh t. Let f(t) sinh t then F (). From Lalace tables where n. 4 L { t sinh t d d F () d ( ) d 4 4 ( 4). Let y y(t). Find the Lalace transform of ty +y ty where y() and y (). Let L { y(t) Y () then L { ty d d Y () that is, L { ty Y (). Also, L { y Y () y() that is, L { y Y (). and L { ty d d [ Y () y() y () ] that is, L { ty Y () Y () +. L { ty + y ty L { ty + L { y L { ty Y () Y () + + (Y () ) ( Y ()) ( )Y () Y (). Note: Students should become familiar with Lalace transforms, its roerties and the Table of Lalace transforms which are at the end of this chater, as well as, the inside of the back cover.
Page 5- Exercise 5A This question refers to the table of Basic Lalace Transforms in section 5.6. (a) Derive each of the results on the first 8 lines of the table by using the definition of a Lalace transform. From the definition, evaluate the following Lalace transforms. (i) L { t sin nt (ii) L { t cos nt (iii) L { e t sin nt (iv) L { e t cos nt (v) L { sin nt nt cos nt (vi) L { e t (sin nt + cos nt) (vii) L { t(sinh t + cos t) Find the Lalace transform of the following functions. (a) f(t) t e t + g(t) t sinh t (c) f(x) cos x sin x (d) g(x) cos x { t t > (e) g(t) < t < { (f) cos x x > π f(x) sin x < x < π (g) k(t) (h) w(t) (i) k(t) e t u sin u du sin (t u) sin u du cos(t u) sin u du (a) Verify the following results. (i) (ii) f(t u) du f(t u) g(u)du f(u) du f(u) g(t u) du Note: (i) is a secial case of (ii) 4 Use the table of Lalace transforms to write down the values of the following integrals. (a) (c) (d) (e) (f) e 5t dt e t t 7 dt e t t cos 5tdt e 4t cosh tdt e t e t u sin udu dt x e x sin (x u) cos udu dx { 5 (a) Show that L x f(x) F ()d. Hence determine the Lalace transforms of the following functions: (i) sin ax x (iii) (ii) e ax e bx x cos ax x 6 (a) Use the definition of the Lalace Transform, and integration by arts, to find the Lalace transforms of y (t), y (t). Let L { y(t) Y (). By taking Lalace transforms, find Y () for each of the following equations subject to the given initial conditions. dy (i) dt y e t where y(). (ii) d y dt + dy + 5y sin t dt where y() and y ().
Page 5-5. LAPLACE TRANSFORMS OF SPECIAL FUNCTIONS In this section we will look at the Lalace transform of some imortant secial functions. 5.. Ste (Heaviside) Function Recall the ste function definition: Then the Lalace transform of h(t a) is h(t a) L{h(t a) { t, a a e a. t < a. h(t a)e t dx e x dx the inverse Lalace of e a is h(t a). That is, { e L a h(t a). Proerty L { f(t a)h(t a) e a F () where L { f(t) F (). Examles (a) Find L { h(t ). Recall that L { h(t a) e a then L { h(t ) e, where a. Find L { t h(t ). Reshaing the given function we have t h(t ) (t ) h(t ) + h(t ). L { t h(t ) L { (t ) h(t ) + L { h(t ).
Page 5-4 Consider the first term on the right hand side of the equation. Noting the roerty that L { f(t a)h(t a) e a F () where F () L { f(t) then Hence, f(t ) t where a. f(t) t where t is relaced by t. L { f(t) F () L { t Γ(). Also, As a result we find that L { (t ) h(t ) e, where a. L { h(t a) e L { t h(t ) e where a. + e e ( + ). 5.. Dirac Delta (Imulse) Function Recall that f k (t a) { k a < t < a + k otherwise. The limit of f k (t) as k is denoted by δ(t a). Now L{f k (t a) a+k a e t f k (t a) dt [ e t e t k dt ] a+k k a ( ) e e sa k. k Now lim L{f k(t a) lim e a k k l H lim e a k e a. ( e k ) ks ( ) e k ( )
Page 5-5 Hence, L{δ(t a) e a. Examle Find L { δ(x ). L { δ(x ) e From tables, where a. Note: The Dirac Delta function is sometimes called the unit imulse function. It is not a function in the ordinary sense. The definition of the Dirac Delta function can vary deending on the text. 5.. Error Function Recall the defintion of the error function. That is, erf(z) π z e u du then L { erf(t) is obtained by taking the Lalace transform with resect to t. That is, That is, L { erf(t) π e t erf(t) dt e t ( π π ) e u du ( e u e t dt e u u e t ] u du dt ) du Reverse the order of integration. L { erf(t) π π e 4 e 4 π e (u +u) e 4 erfc( ). du e (u+ ) e z dz Comlete the square. du Let z u + when then dz du u, z ; when u, z
Page 5-6 Exercise 5B Define the functions h(t), h(t a), δ(t a) and find their Lalace transforms. Use the Convolution theorem to show that { L h(t u)x(u) du X(). Find the Lalace transform of the following functions. (a) h(t ) (c) (d) h(t )e (t ) h(t )e t sin(t) (h(t π ) ) h(t) (e) δ(t 5). (f) erfc( t) (g) (h) (i) e t. (j) (k) (l) (m) h(t u) sin u du h(t u) cos u du δ(u t) sin u du for t > δ(u t) cos u du for t > h(t u ) sin u du h(t u ) cos u du Let L{y(t) Y (). Find Y () if (a) dy y δ(x ) dx d y dt 5dy + 4y δ(t ) dt 4 Use transform definitions, and the evaluation of a suitable double integral, to calculate the Lalace transform { of the following integrals. t (i) L x(u) du (ii) 5 Evaluate { L f(t u) g(u) du (Convolution integral). x erf( x t) erf( t) dt. 6 Prove the following results. (a) L{δ(x) L{δ(x a) e a, a 7 (a) Show that the Lalace transform of f(x a)h(x a) is e a F (). Hence find the functions whose transforms are given below. (i) (ii) e a, a > e π +. 5.4 INVERSE LAPLACE TRANSFORM OF A FUNCTION The inversion formula for the Lalace transform is where γ is called the Bromwich Line. f(t) γ+i e t F () d πi γ i
Page 5-7 We usually denote the inverse Lalace transform by L. That is, f(t) L {F () where f(t) is the original function. Note: The assumtion here is that the indeendent variable of the original function f necessary the case. is t but this not In this work, we will rely on using the table in section 5.6 to invert many transforms as the inversion formula requires comlex analysis methods. However, we will use the Lalace transform roerties, the table of Lalace transforms and the Convolution theorem to aid in determining the inverse Lalace transform of a function. 5.4. Proerties of the Inverse Lalace Transform Let f(t) L {F () and g(t) L {G() and let α and β be two arbitrary constants then LL L L I where I is the identity oerator. L {αf () αl {F (). L {αf () + βg() αl {F () + βl {G() L {F ( + a) e at f(t) (Shift Theorem) 5.4. Techniques for Finding the Inverse Lalace Transform There are various techniques to finding the inverse Lalace transform of a function. We shall secifically look at four techniques. Technique : Standard Tables Examles (a) Find the inverse Lalace transform of F (). From the tables it can be seen that L { t r Γ(r + ) r+ for r >. Let L {F () f(t) then { f(t) L r+ tr Γ(r + ).
Page 5-8 In comarison to our examle, it can be seen that r +. r and f(t) L { t Γ() t. Find the inverse Lalace transform of + 4. From the tables it can be seen that L { sin t + 4, where n. { L { + 4 L + 4 (c) Find the inverse Lalace transform of sin t.. From the tables it can be seen that L { e t, where b. then Note: { L e t. The assumtion in these examles is that L {F () f(t). That is, the original function f is a function of the indeendent variable t but this is not necessarily the case. We can assume any indeendent variable if our roblem does not secify. Technique : Rational Functions If F () is a rational function of then we can use artial fractions to simlify the rational function and tables to determine L {F (). Examles (a) Find the inverse Lalace transform of ( ). We can see that ( ) A + B where A and B need to be determined by the usual artial fraction technique. It is easily shown that A and B.
{ L ( ) { Find L 6 (. + 9) Using artial fractions we have L { + L { { + L + e t, using tables. 6 ( + 9) A + B + C + D + 9 Page 5-9 where A, B, C and D need to be determined by the usual artial fraction method. It can be shown that A, B, C and D. { { L 6 ( L + 9) + 9 { L { L + 9 t Γ() { 9 L + 9 using tables t sin t, using tables. 9 Technique : Shift Theorem (a) Find the inverse Lalace transform of ( + ). Recall the Shift theorem. That is, L { e at f(t) F ( + a) or e at f(t) L {F ( + a). That is, L {F ( + a) e at L {F (). In our examle we let F ( + ), where a. ( + ) Now L {F () L { F (). t. From tables where r +. Γ()
Page 5- Hence, { { L ( + ) e t L e t t Γ() te t. Find the inverse Lalace transform of ( ) + 4. Here we let then Recall that then F ( ) ( ) + 4, F () + 4. where a. L { sin t + 4, where n. { L {F () L + 4 sin t. { L ( ) + 4 { e t L + 4 e t sin t et sin t. Using shift theorem. (c) Find the inverse Lalace transform of + 4 +. In this examle we will have to comlete the square before being able to use the shift theorem. That is, + 4 + ( + ). let F ( + ) ( + ) Form the Lalace transform tables we see that then F (). { { L L sinh( t). where n
Page 5- { L ( + ) { e t L e t sinh( t). (d) Find the inverse Lalace transform of + 4 +. Once again we comlete the square in the denominator before using the shift theorem. That is, + 4 + ( + ). The right hand side is not in the form of F ( + a) as yet. This is due to the numerator being only a function of. we maniulate the rational function so that we do obtain the this form. That is, Now + 4 + ( + ) + ( + ), adding and subtracting in the numerator + ( + ) ( + ) F ( + ) F ( + ). F ( + ) + ( + ) From the Lalace transform tables we see that then F (). { L {F () L cosh t. Consider F ( + ). That is, { L + ( + ) e t cosh t. F ( + ) ( + ) then F (). From the Lalace transform tables we see that { L {F () L L { sinh t.
Page 5- { L ( + ) { e t L e t sinh t. Hence, { L e t cosh t e t sinh t. + 4 + { (e) Find L (. + 4 + ) Using artial fractions it is found that ( + 4 + ) + 4 + 4 +. Now { { L ( L + 4 + ) + 4 + 4 + L { { L. L { + 4 + 4 +. The second term requires to be ut into a form so that we can use one of our inverse Lalace transform techniques. This can be done by comleting the square and adding and subtracting constants. That is, + 4 + 4 + + 4 ( + ) By comleting the square. + + ( + ), relacing + 4 by ( + ) +. + ( + ) + ( + ). Consider the first term on the right hand side and let F ( + ) { L {F () L cosh( t). Consider the second term on the right hand side and let F ( + ) { L {F () L L { sinh( t). + ( + ) then F (). Using tables where n. ( + ) Using tables where n. then F (). { L ( + 4 + ) ( + cosh( t) + sinh( ) t) e t.
Page 5- Technique 4: Convolution Theorem Taking the Lalace transform of a roduct of two functions may be easily obtained by using the convolution theorem rather than the revious techniques. This will deend on the given function. We shall solve some of the revious examles using this technique. Recall that or that then { L f(t u) g(u) du F () G() { L f(u) g(t u) du F () G() L {F () G() f(t u) g(u) du. Examles { (a) Find L. ( ) We can see that let ( ). F () { f(t) L {F () L and Now and G() { g(t) L {G() L e t. f(t u), Relacing t bt t u. g(u) e u, Relacing t bt u. Using the Convolution theorem we have { L ( ). e u du substituting for f(t u) and g(u) into the Convolution theorem. e u ] t e t. This is the same answer as that obtained in Examle (a) using Technique two. { Find L 6 (. + 9) We can see that 6 ( + 9) + 9.
Page 5-4 let F () f(t) L {F () L { t and Now and G() + 9 { g(t) L {G() L sin t. + 9 f(t u) (t u), Relacing t bt t u. g(u) sin u, Relacing t bt u. Using the Convolution theorem we have that { L 6 ( + 9) (t u) sin u du substituting for f(t u) and g(u) into the Convolution theorem. (t u) sin u du, Integrate by arts. { ] t cos u t cos u (t u) ( ) du { ] t t sin u 9 sin t t. 9 This is the same as that obtained in Examle using Technique two. 5.5 INVERSE LAPLACE TRANSFORM OF SPECIAL FUNCTIONS Examles (a) Find L { e. Using Lalace transform tables, we find that { e L h(t ), where a. Find L { e 4. Recall that L { f(t a)h(t a) e a F () L { e a F () f(t a)h(t a), where L { f(t) F ().
Page 5-5 Now Hence, So that (c) e 4 e 4. let F () and a 4. f(t) L {F () { e Find L. + t t f(t 4) t 4. Γ() { e L 4 (t 4)h(t 4). Hence, So that e + e. let F () + + and a. f(t) L {F () cos t f(t ) cos(t ). { e L cos(t ) h(t ). + (d) Find L { e. Refer to the Lalace transform tables L { e δ(t ) where a. (e) Find L { e ( + ). Now e ( + ) e ( + ). let F () ( + ) and a. { { L ( + ) e t L e t t Γ(), te t. where r + and b using tables. f(t) L {F () te t f(t ) (t )e (t ). So that { e L ( + ) (t )e (t ) h(t ).
Page 5-6 Exercise 5C Use artial fractions or the convolution theorem and the Table of Lalace Transforms, to find functions of t which have the following Lalace transforms. (a) (c) (d) (e) (f) (g) + + + ( + 4) ( + ) ( + )( + 4) 7 ( + 6 + ) ( 4) (h) (i) (j) ( ) + + 5 ( + 4) (a) Show that the Lalace transform of (i) (iii) f(x a) h(x a) is e a F (). Hence find the functions whose transforms are given below. e a, a > (ii) e π +. e 4 + + 5 (iv) e ( + ) 5.6 INTEGRAL AND DIFFERENTIAL EQUATIONS Lalace transforms can be used to solve a both integral and linear differential equations. Consider the following initial value roblem (IVP) ay + by + cy g(t) subject to y() α and y () β. Here the given differential equation is a constant co-efficient non-homogneous differential equation with given initial conditions. That is, a, b and c are constants. The rocedure is to transform the given differential equation to an algebraic equation which in turn may be easily solved. This rocedure is shown diagramatically below. Aly Lalace Transform Differential Equation becomes an algebraic equation in Y(). Find solution y(t) to the given IVP. Aly Inverse Lalace Tansform Solve algebraic equation for Y() Taking transforms of both sides of the given differential equation we have a( Y () y() y ()) + b(y () y()) + cy () G().
Page 5-7 Alying the given initial conditions and solving for Y (), it is found that Y () This is can be reresented in the form a (α + β + G()). + b + c Y () (Q() + G()) P () where P () a + b + c, Q() α + β and G() L { g(t). In this manner we have searated the effects of the initial conditions and those due to the inut function g(t). The recirocal of P (), that is, P () is called the transfer function of the given equation (system). P () is the auxilliary/characteristc equation of the differential equation using normal analytic methods. Uon taking inverse Lalace transforms we have that { { Q() G() y(t) L + L. P () P () { Q() If g(t) then the solution L of the roblem is called the zero inut resonse and when P () { G() the initial conditions are all zero, the solution L is called the zero state resonse. P () Examles (a) Solve the following differential equation y y y e t subject to y() and y (). Take Lalace transforms of the differential equation with resect to t. That is, L { y y y e t gives Y () y() y () (Y () y()) Y (). Using the given initial conditions and simlifying, we have Y () ( )( ) ( )( ). At this stage we can use artial fractions or the Convolution theorem. In this, examle we shall use the Convolution theorem.
Page 5-8 Let F () ( ) and G() ( ). Note: The choice of F () and G() is arbitrary. Now then and { { L {F ()G() L L ( ) ( ) { f(t) L e t ( ) { g(t) L ( sinh t. ) f(t u) e (t u) and g(u) sinh u. Using the Convolution theorem we find that the solution to the differential equation is { y(t) L ( )( ) e (t u) sinh u du et + et + e t 6. Our solution can be checked by using D oerator techniques. We find that the general solution to the differential equation is y(t) Ae t + Be t + et. Using the initial conditions we find that y 6 e t + et et. Solve, using Lalace transforms, the following differential equation dy + y cos x, dx subject to y(). Take Lalace transforms of the differential equation with resect to x. That is, gives { dy L + y cos x dx Y () y() + Y () Using the given initial conditions and simlifying, we have + 4. Y () + + ( + )( + 4).
Now and Let F () + { L ( + )( + 4) and G() + 4. { L e x + { L ( + ) (. + 4) Page 5-9 { { Using the Lalace transform tables we find that L e x and L ( + ) ( cos x. + 4) Then f(x) e x g(x) cos x. Hence, f(x u) e (x u) g(u) cos u. the Convolution theorem roduces { L ( + )( + 4) x e (x u) cos u du x e x e u cos u du. The integral on the right hand side of the above equation is a recurring integral after integrating by arts. The working is as follows. Let That is, and x I e x e u cos u du e x { e u cos u ] x e x { e x cos x integrate by arts x + e u sin u du + eu sin u ] x I e x { e x cos x I e x { e x cos x 4 I. + ex sin x. + ex sin x 4 integrate by arts sin x + cos x e x. 4 { L sin x + cos x e x ( + )(. + 4) 4 Hence, the solution to the given differential equation subject to the given initial condition is y(x) e x + sin x + cos x e x. 4 (c) Solve the integral equation for y(t) where y(t) + y(u) du.
Page 5- Let L { y(t) Y () then the Lalace transform of the integral equation becomes Y () + L { y(u) du. Now we have { { L y(u) du L. y(u) du L { L { y(u) Y () + Y (). Y (). Uon solving for Y () we obtain that Y (). Finding the inverse Lalace transform of Y () we have that { y(t) L {Y () y(t) L e t. (d) Solve the integro-differential equation for f(t) where f (t) sinh t 4 f(u) cosh (t u) du subject to f(). Let L { f(t) F () then L { f (t) F () f() F () using the given initial condition and where { L f(u) cosh (t u) du F () 4 g(t u) cosh (t u) g(u) cosh u G() L { g(u) 4. Also, L{sinh t 4. F () { 4 4 F () 4. That is, ( + 4 ) F () 4 4
Page 5- Uon simlifying we have Uon solving for F () we have 4 F () + 4. F (). using the Lalace transform tables we find that f(t) t. Exercise 5D Use Lalace transforms to solve the following differential equations. (a) x + 7x + x Given that L { e bt x(t) X(+b), (Shift Theorem) where x(), x () 5 x + x + x e t where x(), x () 5 (c) x + x + x (d) where x(), x () x + x x cos t + sin t where x(), x () Use the Convolution theorem to find the inverse transforms of the following functions. (a) (c) (d) (e) (f) (Hint: ( + 9) ( + )( + 4) ( 4) ( + n ) ( + n ) + n + n ). ( 4) s ( 4) L {t n x(t) ( ) n dn d n X(), when n is an integer. Find the following Lalace transforms: (a) t n t cos nt (c) t sin nt (d) t n e bt (e) t n e bt (f) t cos t (g) t sin t (h) e 4t cos t (i) e t sin t 4 (a) If F () and G() are the Lalace transforms of f(x) and g(x), rove that F ()g() d and hence in articular that (i) (ii) f(x)g(x) dx, f(x) dx x F () d. Hence evaluate the following integrals. sin ax x dx, a >. e ax e bx dx. x continued next age...
Page 5-5 Solve the following Volterra integral equations. (a) g(x) sin x + x sin(x u)g(u) du. x g(x) e x + cos(x t)g(t) dt. 6 Solve the following integro-differential equations. (a) (a) x y(u) du y (x) x x f (x) k f(t) cos k(x t) dt. 7 Use Lalace transforms to show that y k x f(u) sin k(x u) du 8 By taking the Lalace transform of the integral x u m (x u) n du, using the convolution theorem, and then letting x, show that B(m, n) Γ(m)Γ(n) Γ(m + n). 9 By taking transforms with resect to the arameter t, evaluating the resultant integral and then inverting, show that (a) sin tx dx x πt, cos tx x dx is a solution of d y dx + k y f(x), when y() y (). and e tx dx π t. 5.7 SYSTEMS OF EQUATIONS Lalace transforms can be used to transform a set of linear system of differential equations with initial conditions into a set of algebraic equations. Examle Use Lalace transforms to solve the following initial value roblem: subject to x, y, dx dt and dy dt d x dt d y dt x + 4y 4x 4y when t. Note There are two indeendent variables x, y and t is the deendent variable. we take Lalace transforms with resect to the deendent variable, t. Let L{x(t) X() and L{y(t) Y (). Hence, the system of differential equations becomes a system of algebraic equations, namely,
Page 5- X x() x () X + 4Y Y y() y () 4X 4Y where x(), x (), y() and y (). Uon substituting the initial conditions, the algebraic equations reduce to ( + ) X 4Y ( + 4 ) Y 4X Note the function of is droed here urely for convenience. Solving these equations simultaneously, we find that X() ( + )( + ) + 6 Y () ( + )( + ) We need now find x(t) L (X()) and y(t) L (Y ()). To do this we have to use artial fractions. Hence, 6 5 X() + + 5 + 5 Y () + 5 + 6 x(t) L (X()) L { 5 + + 5 + and sin t + 5 sin t y(t) L (Y ()) L { 5 + 5 + 5 sin t + sin t The general solution to the initial value roblem can be now written in the following form: x(t) y(t) sin t + 5 sin t. 5 sin t + sin t
Page 5-4 Exercise 5E Use Lalace transforms to solve the sets of simultaneous differential equations: (a) dx dy + x + y ; x + y ; dt dt x, y when t dy x + y ; dt dx + y 4 cos t; dt y, x when t Use Lalace transforms to solve the following initial value roblem. dx (a) dt x + y dy dt x where x() and y(). dx y + et dt dy dt 8x t where x() and y(). (c) d x dt + x y d y dt + y x where x(), x () and y(), y (). (d) dx 4x y + h(t ) dt dy x y + h(t ) dt where x() and y().
Page 5-5 5.8 TABLE OF LAPLACE TRANSFORMS x(t) X() x(t) e t dt t n, n,,,... t r, r > n! n+ Γ(r + ) r+ e bt + b sin nt cos nt sinh nt cosh nt n + n + n n n n e bt f(t) F ( + b) h(t a) e a f(t a)h(t a) e a F () δ(t a) e a δ(t a)f(t) e a F ( a)
Page 5-6 x(t) X() x(t) e t dt f (t) F () f() f (t) F (s) f() f () f(u)du f(t u) g(u)du F () F () G() t f(t) t sin nt t cos nt sin nt nt cos nt sin(at + b) cos(at + b) J (x) d d F () n ( + n ) n ( + n ) n ( + n ) sin b + a cos ( + a ) cos b a sin b ( + a ) ( +