36CH_PHCalter_TechMath_95099 3//007 :8 PM Page Solving Differential Equation by the Laplace Tranform and by Numerical Method OBJECTIVES When you have completed thi chapter, you hould be able to: Find the Laplace tranform of a function by direct integration or by uing a table. Find the Laplace tranform of an expreion containing derivative, and ubtitute initial condition. Determine the invere of a Laplace tranform by completing the quare or by partial fraction. Solve firt-order and econd-order differential equation uing Laplace tranform. Solve electrical application uing Laplace tranform. Solve firt-order differential equation uing Euler method, the modified Euler method, or the Runge-Kutta method. Solve econd-order differential equation uing the modified Euler method or the Runge-Kutta method. The method for olving differential equation that we learned in Chapter 8 and 9 are often called claical method. In thi chapter we learn other powerful technique: the Laplace tranform and numerical method. The Laplace tranform enable u to tranform a differential equation into an algebraic one. We can then olve the algebraic equation and apply an invere tranform to obtain the olution to the differential equation. The Laplace tranform i good for finding particular olution of differential equation. For example, it enable u to olve initial value problem, that i, when the value of the function i known at t 0. Thu the equation that we deal with here are function of time, uch a electric circuit problem, rather than function of x. We firt ue the Laplace tranform to olve firt- and econd-order differential equation with contant coefficient. We then do ome application and compare our reult with thoe obtained by claical method in Chapter 8 and 9. We go on to olve differential equation uing numerical method. Thee method ue
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page Solving Differential Equation by the Laplace Tranform and by Numerical Method ucceive approximation to give a olution in the form of number, rather than equation. One great advantage of numerical method i that they can be programmed on a computer. The Laplace Tranform of a Function Laplace Tranform of Some Simple Function The Laplace tranform i named after the French analyt, probabilit, atronomer, and phyicit Pierre Laplace (749 87). If y i ome function of time, o that y f(t), the Laplace tranform of that function i defined by the following improper integral: Laplace Tranform [f(t)] 0 f(t)et dt The tranformed expreion i a function of, which we call F(). Thu [ f(t)] F() We can write the tranform of an expreion by direct integration, that i, by writing an integral uing Eq. and then evaluating it. Glance back at Sec. 8 in Method of Integration if you have forgotten how to evaluate an improper integral. y Example : If y f(t), then [ f(t)] [] 0 ()et dt To evaluate the integral, we multiply by and compenate with /. y = e u [] 0 et ( dt) A et 0 Figure how a graph of y e u, which will help u to evaluate the limit. Note that a u approache infinity, e u approache zero, and a u approache zero, e u approache one. So we take e t equal to zero at the upper limit t and e t at the lower limit t 0. Thu 0 u [] (0 ) FIGURE Rule from our table of integral in Text Appendix C ay that af(x) dx a f(x) dx. It follow then that [af(t)] a[ f (t)] The Laplace tranform of a contant time a function i equal to the contant time the tranform of the function. Example : If [] /, a above, then [5] 5[] 5 Example 3: If y f(t) t, then [ f(t)] [t] 0 tet dt
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 3 Section The Laplace Tranform of a Function 3 We evaluate the right ide uing Text Rule 37 with u t and a. [t] e t (t ) A 0 At the upper limit, e t approache zero, while t approache infinity. But e t i hrinking fater than t i growing, o their product approache zero. Thi i confirmed by Fig., which how the product t(e t ) firt increaing but then quickly approaching zero. So we get [t] e t )A (t 0 0 y 0.4 0.3 y = te t 0. 0. 0 4 6 8 t FIGURE Example 4: Find [in at]. Solution: Uing Eq., we find that [in at] 0 et in at dt Uing Text Rule 4 give t e [in at] ( in at a co at)a a 0 At the upper limit of infinity, e t approache zero, cauing the entire expreion to have a value of zero at that limit. At the lower limit of zero, e t approache one, in at approache zero, and co at approache one. Our expreion then become a [in at] 0 (a) a a Note that in each cae the tranformed expreion i a function of only. Tranform of a Sum If we have the um of everal term, f(t) a g(t) b h(t) then [ f(t)] [a g(t) b h(t) 0 ]et dt a 0 g(t)et dt b 0 h(t)et dt Thu [a g(t) b h(t) ] a[g(t)] b[h(t)]
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 4 4 Solving Differential Equation by the Laplace Tranform and by Numerical Method The Laplace tranform of a um of everal term i the um of the tranform of each term. Thi property allow u to work term by term in writing the tranform of an expreion made up of um or difference. Example 5: If y 4 5t in 3t, then [ y] 4 5 3 9 When we write the Laplace tranform of a function, we often ay that we are taking the tranform of the function. Tranform of a Derivative To olve differential equation, we mut be able to take the tranform of a derivative. Let f (t) be the derivative of ome function f (t). Then [ f (t)] f 0 (t)et dt Integrating by part, we let u e t and d f(t) dt. Sodu e t dt and f (t) dt f(t). Then [ f(t)] f(t)e A t f 0 0 (t)et dt f(t)e t A 0 [ f(t)] Thu: 0 f(0) [ f(t)] Tranform of a Derivative [ f (t)] [ f(t)] f(0) The tranform of a derivative of a function equal time the tranform of the function minu the function evaluated at t 0. Thu if we have a function y f (t), then [ y] [ y] y(0) Note that the tranform of a derivative contain f(0) or y(0), the value of the function at t 0. Thu to evaluate the Laplace tranform of a derivative, we mut know the initial condition. If two function are equal, their tranform are equal. Thu we can take the tranform of both ide of an equation without changing the meaning of the equation. Example 6: Take the Laplace tranform of both ide of the differential equation y 6t, with the initial condition that y(0) 5. Solution: y 6t [y] [6t] 6 [ y] y(0) or, ubtituting, [ y] 5 6
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 5 Section The Laplace Tranform of a Function 5 ince y(0) 5. Later we will olve uch an equation for [y] and then find the invere of the tranform to obtain y. Similarly, we can find the tranform of the econd derivative ince the econd derivative of f(t) i the derivative of f (t). [ f (t)] [ f (t)] f (0) Subtituting, we get [ f (t)] {[ f(t)] f (0)} f (0) Tranform of the Second Derivative [ f (t)] [ f(t)] f(0) f (0) Another way of expreing thi equation, if y f(t), i [ y] [ y] y(0) y(0) Example 7: Tranform both ide of the econd-order differential equation y 3y 4y 5t if y(0) 6 and y(0) 7. Solution: y 3y 4y 5t [ y] 3[ y] 4[ y] [5t] 5 [y] y(0) y(0) 3{[y] y(0)} 4[y] Subtituting y(0) 6 and y(0) 7, we obtain [ y] 6 7 3[ y] 3(6) 4[ y] 5 or [y] 3[y] 4[y] 6 5 Tranform of an Integral Let u now find the tranform of the integral t f(t) dt. By our definition of the 0 Laplace tranform, Eq., t f(t) dt 0 0 t f(t) dt 0 et dt Integrating by part, we let o that Integrating yield u t f(t) dt and d 0 et dt du f(t) dt and e t / t f(t) dt 0 t (/)et f(t) dt A (/) 0 0 0 f(t)et dt At the upper limit of infinity, the quantity in the bracket vanihe becaue e t goe to zero. At the lower limit of zero, the quantity in the bracket vanihe becaue the
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 6 6 Solving Differential Equation by the Laplace Tranform and by Numerical Method upper and lower limit of the integral are equal. Further, the integral in the econd term i the definition (Eq. ) of the Laplace tranform of f(t). Thu: Tranform of an Integral t f(t) dt 0 (/) f(t) We will need thi equation later for olving electrical problem. Example 8: The voltage acro a capacitor of capacitance C i given by (C) t i dt 0 Find the tranform of thi voltage. Solution: Uing the tranform of an integral, we get (C) t i dt 0 (C)[i] Table of Laplace Tranform The tranform of ome common function are given in Table. Intead of tranforming a function tep by tep, we imply look it up in the table. TABLE Short table of Laplace tranform. Here n i a poitive integer. Tranform Number f(t) [ f (t)] F() f(t) F() e t f(t) dt f (t) [ f(t)] f(0) 3 f (t) [ f(t)] f(0) f(0) 4 a g(t) b h(t) a[g(t)] b[h(t)] 5 6 t 7 t n n! n t 8 (n )! n 9 e at a 0 e at a ( a) te at ( a) e at ( at) ( a) n 3 t n e at n! ( a) n 4 i n e at ( n )! n ( a) 5 e at e bt b a ( a)( b) 0
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 7 Section The Laplace Tranform of a Function 7 TABLE Continued Tranform Number f(t) [ f (t)] F() 6 ae at be bt (a b) ( a) ( b) 7 in at a a 8 co at a 9 t in at a ( a ) a 0 t co at ( a ) a co at ( a ) 3 a at in at ( a ) 3 e at b in bt ( a ) b 4 e at a co bt ( a) b a 3 5 in at at co at ( a ) 6 in at at co at a ( a ) 7 co at at in at 3 ( a ) b 8 a (e at b at ) ( a) 9 t f(t) dt [f (t)] 0 Example 9: Find the Laplace tranform of t 3 e t by uing Table. Solution: Our function matche Tranform 3, with n 3 and a, o [t 3 e t 3! 6 ] ( ) 3 ( ) 4 Exercie The Laplace Tranform of a Function Tranform by Direct Integration Find the Laplace tranform of each function by direct integration.. f(t) 6. f(t) t 3. f(t) t 4. f(t) t 5. f(t) co 5t 6. f(t) e t in t Tranform by Table Ue Table to find the Laplace tranform of each function. 7. f(t) t 4 8. f(t) t 3 t 3t 4
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 8 8 Solving Differential Equation by the Laplace Tranform and by Numerical Method 9. f(t) 3e t e t 0. f(t) 5te 3t. f(t) in t co 3t. f(t) 3 e 4t 3. f(t) 5e 3t co 5t 4. f(t) e 4t t 5. f(t) 5e t t in 3t 6. f(t) t 3 4t 3e t Tranform of Derivative Find the Laplace tranform of each expreion, and ubtitute the given initial condition. 7. y y, y(0) 8. 3y y, y(0) 3 9. y 4y, y(0) 0 0. 5y 3y, y(0). y 3y y, y(0) and y(0) 3. y y y, y(0) and y(0) 0 3. y 3y y, y(0) and y(0) 3 4. 3y y y, y(0) and y(0) Invere Tranform Before we can ue the Laplace tranform to olve differential equation, we mut be able to tranform a function of back to a function of t. The invere Laplace tranform i denoted by. Thu if [ f(t)] F(), then we have the following equation: We ll ee that finding the invere i often harder than finding the tranform itelf. Invere Laplace Tranform [F()] f(t) We ue Table to find the invere of ome Laplace tranform. We put our given expreion in a form that matche a right-hand entry of Table and then read the correponding entry at the left. Example 0: If F() 4/( 6), find f(t). Solution: We earch Table in the right-hand column for an expreion of imilar form and find Tranform 7, a [in at] a which matche our expreion if a 4. Thu f(t) in 4t Example : Find y if [y] 5/( 7) 4. Solution: From the table we find Tranform 3, [t n e at n! ] ( a) n
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 9 Section Invere Tranform 9 In order to match our function, we mut have a 7 and n 3. The numerator then mut be 3! 3() 6. So we inert 6 in the numerator and compenate with 6 in the denominator. We then rewrite our function a [ y] 5 6 6 ( 7) 4 which now i the ame a Tranform 3 with a coefficient of 5 6. The invere i then y 5 6 t3 e 7t Completing the Square To get our given function to match a right-hand table entry, we may have to do ome algebra. Sometime we mut complete the quare to make our function match one in the table. Example : Find y if [ y]. 4 0 Solution: Thi doe not now match any function in our table, but it will if we complete the quare on the denominator. 4 0 ( 4 ) 0 ( 4 4) 0 4 ( ) 4 The denominator i now of the ame form a in Tranform 4. However, to ue Tranform 4, the numerator mut be, not. So let u add 3 and ubtract 3 from the numerator, o that ( ) 3. Then ( ) 3 [y] 4 0 ( ) 4 ( ) 4 3 ( ) 4 3 ( ) 4 4 4 ( ) 4 Now the firt expreion matche Tranform 4 and the econd matche Tranform 3. We find the invere of thee tranform to be Thi i no different from the method we ued earlier when we completed the quare. y e t co 4t 3 4 et in 4t e t co 4t 3 4 in 4t Partial Fraction We ued partial fraction in Sec. 5 in Method of Integration to make a given expreion match one lited in the table of integral. Now we ue partial fraction to make an expreion match one lited in our table of Laplace tranform. Example 3: Find y if [y] /( 8). Solution: We eparate [y] into partial fraction. [y] 8 A B 4 o A( ) B( 4) (A B) (A 4B) We could alo complete the quare here, but we would find that the reulting expreion would not match a table entry. Some trial-and-error work in algebra may be neceary to get the function to match a table entry.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 0 0 Solving Differential Equation by the Laplace Tranform and by Numerical Method from which A B 0 and A 4B. Solving imultaneouly give A and B, o [ y] 8 4 Uing Tranform 9 twice, we get y e 4t e t Exercie Invere Tranform Find the invere of each tranform... 3 3. 3 4. 3 4 5. 6. 4 ( 6) 3 4 7. 8. 3 9 4 3 9. 0. ( 9) ( 4) 5. ( ) 9. 6 8 3. 4. ( ) 5 5. 6. ( )( ) ( ) ( ) 7. 8. 3 9. 0. 5 6 ( 4)( ) 3 Solving Differential Equation by the Laplace Tranform To olve a differential equation with the Laplace tranform:. Take the tranform of each ide of the equation.. Solve for [y] F(). 3. Manipulate F() until it matche one or more table entrie. 4. Take the invere tranform to find y f(t). We tart with a very imple example. Example 4: Solve the firt-order differential equation y y if y(0) 0.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page Section 3 Solving Differential Equation by the Laplace Tranform Solution:. We write the Laplace tranform of both ide and get [y] [y] [] [y] y(0) [y]. We ubtitute 0 for y(0) and olve for [y]. ( )[y] [y] ( ) 3. Our function will match Tranform 0 if we write it a [y] ( ) 4. Taking the invere give y ( e t ) In our next example we expre F() a three partial fraction in order to take the invere tranform. Example 5: Solve the firt-order differential equation y 3y 4 9t if y(0). Solution:. We take the Laplace tranform of both ide and get [y] [3y] [4] [9t] [y] y(0) 3[y] 4 9. We ubtitute for y(0) and olve for [y]. 9 ( 3)[y] 4 9 4 [y] ( 3) ( 3) 3 3. We match our expreion with Tranform 8, 0, and 9. [y] ( 4. Taking the invere give 9 4 3) 3 3 ( 3) 3 y e 3t 3t 4 3 ( e3t ) e 3t 5 3 e3t 3t 3 We now ue the Laplace tranform to olve a econd-order differential equation. Example 6: Solve the econd-order differential equation y 4y 3 0 where y i a function of t and y and y are both zero at t 0.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page Solving Differential Equation by the Laplace Tranform and by Numerical Method Solution:. Taking the Laplace tranform of both ide, [y] [4y] [3] 0 [y] y(0) y(0) 4[y] 3 0 Subtituting y(0) 0 and y(0) 0 give. Solving for [y], we have 3. We match it to Tranform. 4. Taking the invere give [y] 0 0 4[y] 3 0 ( 4)[y] 3 [y] ( [y] ( 3 3 4) 4 3 4) 4 ( ) y 3 4 ( co t) We have een that the hardet part in olving a differential equation by Laplace tranform i often in doing the algebra neceary to get the given function to match a table entry. We mut often ue partial fraction or completing the quare, and ometime both, a in the following example. Example 7: Solve y 4y 5y t, where y(0) and y(0). Solution:. Taking the Laplace tranform of both ide and ubtituting initial condition, we find that [y] y(0) y(0) 4[ y] 4y(0) 5[y]. Solving for [y] give [ y]( 4 5) 6 6 [y] () ( 4 5) 4 5 3. Taking for now jut the firt fraction in Equation (), which we ll call F (), we obtain F () A B ( 4 5) C D 4 5 Multiplying by ( 4 5) give A 3 4A 5A B 4B 5B C 3 D (A C) 3 (4A B D) (5A 4B) 5B
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 3 Section 3 Solving Differential Equation by the Laplace Tranform 3 Equating coefficient give A 4 B 5 5 C 4 D 5 5 So F () 4/ 5 /5 4 /5 /5 4 5 5 5 4 4 4 5 We can ue Tranform 5 and 6 for 4/ and 5/, but there i no match for the third term in the parenthee. However, by completing the quare, we can get the denominator to match thoe in Tranform 3 and 4. Thu 4 5 4 4 4 5 ( ) We now manipulate that third term into the form of Tranform 3 and 4. 4 4 (4) 4 5 ( ) / 4 ( ) (4) ( 3/ 4 ) (4) ( ) (3) ( ) Combining the third term with the firt and econd give 5 F () 5 4 (4) ( ) (3) ( ) Returning to the remaining fraction in Equation (), which we call F (), we complete the quare in the denominator and ue partial fraction to get 6 F () ( ) 6 ( ) ( ) 4 ( ) 4. Taking the invere tranform of F () and F (), we obtain y [5t 4 4e t co t 3e t in t] e t co t 4e t in t 5 which implifie to y (5t 4 9e t co t 03e t in t) 5 Exercie 3 Solving Differential Equation by the Laplace Tranform Firt-Order Equation Solve each differential equation by the Laplace tranform.. y 3y 0, y(0). y y, y(0) 3 3. 4y y t, y(0) 0 4. y 5y e t, y(0) 5. 3y y t, y(0) 3 6. y 3y in t, y(0) 0 7. y y co t, y(0) 0 8. 4y y 3t 3, y(0) 0
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 4 4 Solving Differential Equation by the Laplace Tranform and by Numerical Method Second-Order Equation Solve each differential equation by the Laplace tranform. 9. y y 3y 0, y(0) 0 and y(0) 0. y y y, y(0) 0 and y(0) 0. y 3y 3, y(0) and y(0). y y, y(0) 0 and y(0) 3 3. y y 4t, y(0) 3 and y(0) 0 4. y y 5y t, y(0) and y(0) 5. y 4y 3y t, y(0) and y(0) 6. y 4y 8t 3, y(0) 0 and y(0) 0 7. 3y y in t, y(0) and y(0) 3 8. y y 3, y(0) and y(0) 9. y y y e t, y(0) 0 and y(0) 0 0. y 3y co t, y(0) 0 and y(0). y y 3y te t, y(0) 0 and y(0) 0. 3y y y in 3t, y(0) 0 and y(0) 0 Motion in a Reiting Medium 3. A 5.0-kg object i dropped from ret through air whoe reiting force i equal to.85 time the object peed (in m/). Find it peed after.00. Exponential Growth and Decay 4. A certain teel ingot i 900 F and cool at a rate (in F/min) equal to.5 time it preent temperature ( F). Find it temperature after 5.00 min. 5. The rate of growth (bacteria/h) of a colony of bacteria i equal to.5 time the number preent at any intant. How many bacteria are there after 4 h if there are 5000 at firt? Mechanical Vibration 6. A weight that hang from a pring i pulled down.00 in. at t 0 and then i releaed from ret. The differential equation of motion i x 6.5x 5.8x 0. Write the equation for x a a function of time. 7. An alternating force i applied to a weight uch that the equation of motion i x 6.5x 45.3 co.5t. If and x are zero at t 0, write an equation for x a a function of time. 4 Electrical Application R Many of the differential equation for motion and electric circuit covered in Chapter 8 and 9 are nicely handled by the Laplace tranform. However, ince the Laplace tranform i ued mainly for electric circuit, we emphaize that application here. The method i illutrated by example of everal type of circuit with dc and ac ource. E i C Serie RC Circuit with dc Source FIGURE 3 RC circuit. Example 8: Capacitor Dicharging. A fully charged capacitor, Fig. 3, i dicharged by throwing the witch from poition to poition, at t 0. Write expreion for the current and the voltage acro the capacitor.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 5 Section 4 Electrical Application 5 Solution: If the voltage acro the capacitor i, then the voltage acro the reitor mut be, ince the um of the voltage around the loop mut be zero. Since the current through the reitor (/R) mut equal the current through the capacitor (C d/dt) we write or Taking the tranform of each term or Solving for [i] we get R C d dt 0 RC [i] 0 [i] 0 RC [i] [i] E RC E [i] /RC Uing tranform 9 to find the invere give Voltage acro a Dicharging Capacitor t /RC Ee 3 We next find the current by dividing the voltage by R. Current in a Dicharging Capacitor i E R e t/rc Example 9: Capacitor Charging. A fully dicharched capacitor, Fig. 4, i charged by throwing the witch from poition to poition at t 0. Write expreion for the current and voltage acro the capacitor. 4 Note that the reult of thi example and the two example to follow are the ame a we obtained by claical method for olving differential equation. Solution: Summing voltage around the loop give Drop acro reitor Drop acro capacitor Battery voltage () By Text Eq. 08, the voltage acro the capacitor i C t i dt, o or Ri C t i dt E 0 0 E R i E RC i t i dt EC 0 FIGURE 4
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 6 6 Solving Differential Equation by the Laplace Tranform and by Numerical Method Taking the tranform of each term RC[i] [i] E C Solving for [i] and rearranging give [i] E R /RC Uing tranform 9 to find the invere tranform, we get Current in a Charging Capacitor i E R e trc 5 Returing to () to get the voltage acro the capacitor, Drop acro capacitor = Battery voltage Drop acro reitor E R E e t RC R or Voltage acro a Charging Capacitor E e t/rc 6 Serie RL Circuit with dc Source Example 0: Inductor dicharging. A fully charged inductor i dicharged by throwing the witch, Fig. 5, from poition to poition at t = 0. Write expreion for the current and the voltage acro the inductor. Ω Solution: Summing voltage around the loop we get 4 V i 4 H Battery voltage Drop acro reitor Drop acro inductor () E Ri L d i dt FIGURE 5 or E L R i i L Taking the tranform of each term E L R [i] [i] i(0) L Subtituting 0 for i(0) and olving for [i], [i] E R R/L ( R/L)
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 7 Section 4 Electrical Application 7 Then by tranform 0, Current in a Charging Inductor i E R e Rt /L 7 Now uing () to find the voltage acro the inductor, Voltage acro inductor Battery voltage Voltage acro the reitor E Ri E R E R e Rt /L Voltage acro a Charging Inductor Ee Rt L 8 Serie RL Circuit with ac Source Example : A witch (Fig. 6) i cloed at t 0 when the applied voltage i zero and increaing. Find the current. Solution: Summing the voltage around the loop give Ri L d i dt Subtituting value and rearranging, we find that 4i 0.0 d i 80 in 400t dt Taking the tranform of each term yield 80(400) 4[i] 0.0[[i] i(0)] (400 ) Subtituting 0 for i(0) and olving for [i], we have [i] [i](0.0 4) 3,000 [ (400) ](0.0 4) 3, 000 ( 400) We now ue the method of Sec. 5 in Method of Integration for eparating into partial fraction a rational fraction that ha quadratic factor in the denominator.,600,000 [ (400) ] ( 00),600,000 [ (400) ]( 00) A B C (400) 00 4 Ω 80 in 400t V 0.0 H FIGURE 6 RL circuit with ac ource.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 8 8 Solving Differential Equation by the Laplace Tranform and by Numerical Method Multiplying by [ (400) ]( 00) give,600,000 (A B)( 00) C[ (400) ] which implifie to,600,000 (A C) (00A B) 00B 400 C Equating coefficient give A C 0 00A B 0 00B 400C,600,000 Solving imultaneouly for A, B, and C give A 8 B 600 C 8 Subtituting back, we get 600 8 8 [i] ( 400) 00 400 4 (400) 8 (400) 8 00 Taking the invere tranform uing Tranform 9, 7, and 8, we find that i 4 in 400t 8 co 400t 8e 00t We now combine the firt two term in the form I in(400t v), where I 4 8 8.94 and v tan 8 4 63.4.07 rad o i 8.94 in(400t.07) 8e 00t A A graph of thi wave, a well a the applied voltage wave, i hown in Fig. 7. i (A) 0 5 e (V) 00 i e 0 0 0 30 t (m) 5 50 0 00 5 Ω FIGURE 7 75.4 V.5 H 4.75 F Serie RLC Circuit with dc Source FIGURE 8 RLC circuit. Example : A witch (Fig. 8) i cloed at t 0, and there i no initial charge on the capacitor. Write an expreion for the current.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 9 Section 4 Electrical Application 9 Solution:. The differential equation for thi circuit i Ri L d i dt C t i dt E 0. Tranforming each term and ubtituting value give 5[i].5{ [i] i(0)} 6 0 [ i] 75.4 4. 75 3. Setting i(0) equal to zero and olving for [i] yield [i] 50.3 50 40,000 4. We now try to match thi expreion with thoe in the table. Let u complete the quare in the denominator. 50 40,000 ( 50 565) 40,000 565 ( 75) (34,000) ( 75) j (367) ( 75 j367)( 75 j367) after replacing by j and factoring the difference of two quare. So 50.3 [i] 50 40,000 50.3 ( 75 j367)( 75 j367) We can now find partial fraction. 50.3 ( 75 j367)( 75 j367) A 75 j367 Multiplying through by 75 j367 give 50.3 A B( 75 j367) 7 5 j367 75 j367 Setting 75 j367, we get 50.3 50.3 A j0.0685 75 j367 75 j367 j734 We find B by multiplying through by 75 j367 and then etting j36775, and we get B j0.065 (work not hown). Subtituting give u j0. 0685 j0. 0685 [i] 75 j367 75 j367 Thi till doen t match any table entry. Let u now combine the two fraction over a common denominator. j0.0685 j0.0685 [i] 75 j367 75 j367 j0.0685( 75 j367) j0.0685( 75 j367) ( 75 j367)( 75 j367) j 50.4 367 0.37 50 40,000 ( 75) (367) B 75 j367 We ee that mot of the work when uing the Laplace tranform i in algebra, rather than in calculu.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 0 0 Solving Differential Equation by the Laplace Tranform and by Numerical Method 5. We finally have a match, with Tranform 3. Taking the invere tranform give i 37e 75t in 367t ma which agree with the reult found in Chapter 9, Example 9. Exercie 4 Electrical Application. The current in a certain RC circuit atifie the equation 7i 750i 5. If i i zero at t 0, how that i 4.8 e 6t ma. In an RL circuit, R 3750 and L 0.50 H. It i connected to a dc ource of 50 V at t 0. If i i zero at t 0, how that i 66.7 e 5,000t ma 3. The current in an RLC circuit atifie the equation 0.55i 48i 7350i 0 If i and i are zero at t 0, how that i 6.4 e 6t ma 4. In an RLC circuit, R 950, L 0.0 H, and C.55 mf. It i connected to a dc ource of 5 V at t 0. If i i zero at t 0, how that i 39.7e t ma 5. The current in a certain RLC circuit atifie the equation 3.5i 0.03i 680 t i dt If i i zero at t 0, how that i 69e 70.5t in 65t ma 6. In an RL circuit, R 450 and L 0.7 H. It i connected to a dc ource of 8.4 V at t 0. If i i zero at t 0, how that i 6.84 e 3,700t ma 7. In an RL circuit, R 8.37 and L 0.50 H. It i connected to a dc ource of 50.5 V at t 0. If i i zero at t 0, how that i 6.03 e 33.5t A 5 Numerical Solution of Firt-Order Differential Equation In Sec. 8, we howed how to ue Euler method to graphically and numerically olve a differential equation, but we mentioned there that it i not the mot accurate method available. It will, however, erve a a good introduction to other method. We will now how both the modified Euler method that include predictor-corrector tep and the Runge-Kutta method. We will apply thee method to both firt- and econd-order equation. Modified Euler Method with Predictor-Corrector Step The variou predictor-corrector method all have two main tep. Firt, the predictor tep trie to predict the next point on the curve from preceding value, and then the corrector tep trie to improve the predicted value. 0
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page Section 5 Numerical Solution of Firt-Order Differential Equation Here we modify Euler method of Sec. 8 a the predictor, uing the coordinate and lope at the preent point P to etimate the next point Q. The corrector tep then recompute Q uing the average of the lope at P and Q. Our iteration formula i then the following: Modified Euler Method x q x p x y q y p m p m q x 9 Example 3: Find an approximate olution to y x /y, with the boundary condition that y when x 3. Calculate y for x 3 to 0 in tep of. Solution: Uing data from before, the lope m at (3, ) wa 4.5, the econd point wa predicted to be (4, 6.5), and the lope m at (4, 6.5) wa.46. The recomputed ordinate at the econd point i then, by Eq. 9, y 4.5.46 () 5.48 o our corrected econd point i (4, 5.48). The lope at thi point i We ue thi to predict y 3. m y(4, 5.48).99 y 3 (predicted) y m x 5.48.99() 8.400 The lope at (5, 8.400) i m 3 y(5, 8.400).976 The corrected y 3 i then y 3 (corrected) 5.48.99.976 () 8.49 The remaining value are given in Table. TABLE x Approximate y Exact y Error 3.00000.00000 0.00000 4 5.48077 5.3543 0.664 5 8.4850 8.3666 0.084 6.496.4076 0.08950 7 4.7399 4.655 0.0848 8 8.6790 8.0936 0.07555 9.7964.7556 0.07086 0 5.6434 5.54734 0.06700 Note that the error at x 0 i about 0.6%, compared with 0.6% before. We have gained in accuracy at the cot of added computation.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page Solving Differential Equation by the Laplace Tranform and by Numerical Method Thi method i named after two German mathematician, Carl Runge (856 97) and Wilhelm Kutta (867 944). Runge-Kutta Method The Runge-Kutta method, like the modified Euler, i a predictor-corrector method. Prediction: Starting at P (Fig. 9), we ue the lope m p to predict R, the lope m r to predict S, and the lope m to predict Q. y y q Q m y S m r y p 0 P x p x x r R m p x x q x x FIGURE 9 Runge-Kutta method. Correction: We then take a weighted average of the lope at P, Q, R, and S, giving twice the weight to the lope at the midpoint R and S a at the endpoint P and Q. m avg 6 (m p m r m m q ) We ue the average lope and the coordinate of P to find Q. y q y p m avg x x q x p x y q y p m avg x Runge- Kutta Method where m avg 6 (m p m r m m q ) and m p f(x p, y p ) m r f x p x, y p m p x m f x p x, y p m r x m q f(x p x, y p m x) 0 Example 4: Repeat Example 3 uing the Runge-Kutta method and a tep of.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 3 Section 5 Numerical Solution of Firt-Order Differential Equation 3 Solution: Starting at P(3, ), with m p 4.5, y r 4.5(0.5) 4.5 m r f(3.5, 4.5) ( 3. 5.88 4. 5 ) y.88(0.5) 3.44 m f(3.5, 3.44) ( 3.5) 3.560 3.44 y q 3.560() 5.560 4 m q f(4, 5.560).878 5. 560 4.5 (.88) (3.560).878 m avg 3.377 6 y q (corrected) y p m avg x 3.377() 5.377 So our econd point i (3, 5.377). The ret of the calculation i given in Table 3. TABLE 3 x Approximate y Exact y Error 3.00000.000000 0.00000 4 5.37703 5.3546 0.090 5 8.344 8.36664 0.0475 6.455.40760 0.0079 7 4.6599 4.6550 0.0084 8 8.0998 8.09360 0.0068 9.735.75560 0.00569 0 5.559 5.547340 0.00484 The error here at x 0 with the Runge-Kutta method i about 0.0%, compared with 0.6% for the modified Euler method. Comparion of the Three Method Table 4 how the error obtained when olving the ame differential equation by our three different method, each with a tep ize of. TABLE 4 x Euler Modified Euler Runge-Kutta 3 0.00000 0.00000 0.00000 4.4587 0.664 0.090 5 0.63487 0.084 0.0475 6 0.34948 0.08950 0.0079 7 0.634 0.0848 0.0084 8 0.099 0.07555 0.0068 9 0.0773 0.07086 0.00569 0 0.583 0.06700 0.00484 Divergence A with mot numerical method, the computation can diverge, with our anwer getting further and further from the true value. The computation hown here can
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 4 4 Solving Differential Equation by the Laplace Tranform and by Numerical Method diverge if either the tep ize x i too large or the range of x over which we compute i too large. Numerical method can alo diverge if the point ued in the computation are near a dicontinuity, a maximum or minimum point, or a point of inflection. When doing the computation on the computer (the only practical way), it i eay to change the tep ize. A good practice i to reduce the tep ize by a factor of 0 each time (, 0., 0.0,...) until the anwer do not change in a given decimal place. Further reduction of the tep ize will eventually reult in number too mall for the computer to handle, and unreliable reult will occur. Exercie 5 Numerical Solution of Firt-Order Differential Equation Your anwer may differ from thoe given for thi exercie, depending on the method ued and roundoff error. Solve each differential equation and find the value of y aked for. Ue either the modified Euler method or the Runge-Kutta method, with a tep ize of 0... y xy 3; y(4). Find y(5).. y 5y 3x; y(0) 3. Find y(). 3. x 4 y ye x xy 3 ; y(). Find y(3). 4. xy y 3 ln x; y() 3. Find y(). 5..94x y 8.3x in y.99y 0; y() 3. Find y(). 6. y ln x 3.99(x y) 4.8y; y().76. Find y(3). 7. (x y)y.76x y 5.33y; y(0). Find y(). 8. y.97xe x.9x co y 0; y(). Find y(3). 9. xy xy 8y lnhxyh; y() 3. Find y(). 0. x y 8.85x ln y 3.7x 0; y() 3.9. Find y(). Computer. Write a program or ue a preadheet to olve a firt-order differential equation by Euler method. Try it on any of the equation above.. Write a program or ue a preadheet to olve a firt-order differential equation by the modified Euler method. Try it on any of the equation above. 3. Write a program or ue a preadheet to olve a firt-order differential equation by the Runge-Kutta method. Try it on any of the equation above. 4. If you have done one of the calculation above uing a preadheet that ha a graphing utility, ue that utility to make a graph of x veru y over the given range. 5. Some computer algebra ytem have built-in function for numerically olving a differential equation. In Derive, the command Euler will invoke Euler method, and the command RK will call for the Runge-Kutta method. In Maple, the command firteuler, impeuler, and rungekutta are ued for olution by Euler method, the modified Euler method, and the Runge-Kutta method. Mathematica ue NDSolve for the numerical olution of a differential equation. Conult your manual for pecific intruction on how to ue thee command, and ue one or more to olve any of the equation in thi exercie et. 6 Numerical Solution of Second-Order Differential Equation To olve a econd-order differential equation numerically, we firt tranform the given equation into two firt-order equation by ubtituting another variable, m, for y.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 5 Section 6 Numerical Solution of Second-Order Differential Equation 5 Example 5: By making the ubtitution y m, we can tranform the econdorder equation into the two firt-order equation and y 3y 4xy 0 y m m 3m 4xy f(x, y, m) A before, we tart with the given point P(x p, y p ), at which we mut alo know the lope m p. We proceed from P to the next point Q(x q, y q ) a follow:. Compute the average firt derivative, m avg, over the interval x.. Compute the average econd derivative m avg over that interval. 3. Find the coordinate x q and y q at Q by uing the equation and x q x p x y q y p m avg x 4. Find the firt derivative m q at Q by uing m q m p m avg x 5. Return to Step and repeat the equence for the next interval. We can find m avg and m avg in Step and either by the modified Euler method or by the Runge-Kutta method. The latter method, a before, require u to find value at the intermediate point R and S. The equation for econd-order equation are imilar to thoe we had for firt-order equation. At P: x p, y p, and m p are given: m p f(x p, y p, m p ). At R: x r x p x y r y p m p x m r m p m p x m r f(x r, y r, m r ) At S: x x p x y y p m r x m m p m r x m f(x, y, m ) At Q: x q x p x y q y p m x m q m p m x m q f(x q, y q, m q ) Example 6: Find an approximate olution to the equation y y y e x with boundary condition y e at (, 0). Find point from to 3 in tep of 0..
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 6 6 Solving Differential Equation by the Laplace Tranform and by Numerical Method Solution: Replacing y by m and olving for m give m f(x, y, m) e x y m At P: The boundary value are x p, y p 0, and m p 5.4366. Then Then and So and m p f(x p, y p, m p ) e 0 (5.4366) 6.3097 At R: x r x p x 0.05.05 y r y p m p x 0 5.4366(0.05) 0.78 m r m p m p x 5.4366 6.3097(0.05) 6.5 m r e.05 0.78 (6.5) 7.9477 At S: x.05 y y p m r x 0 6.5(0.05) 0.36 m m p m r x 5.4366 7.9477(0.05) 6.3360 m e.05 0.36 (6.3360) 8.0707 At Q: x q x p x. y q y p m x 0 6.3360(0.) 0.6336 m q m p m x 5.4366 8.0707(0.) 7.437 m q e. 0.6336 (7.437) 9.86 m p m r m m q m avg 6 5.4366 (6.5) (6.3360) 7.437 6.3094 6 m p m r m m q m avg 6 6.3097 (7.9477) (8.0707) 9.86 8.0348 6 y q y p m avg x 0 6.3094(0.) 0.63094 m q m p m avg x 5.4366 8.0348(0.) 7.40 The computation i then repeated. The remaining value are given in Table 5, along with the exact y and exact lope, found by analytical olution of the given equation.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 7 Review Problem 7 TABLE 5 x y Exact y Slope Exact Slope.0 0.000 0.000 5.437 5.437. 0.63 0.63 7.40 7.40..46.46 9.49 9.49.3.53.53.07.07.4 3.893 3.893 5.48 5.48.5 5.60 5.60 9.047 9.047.6 7.77 7.77 3.576 3.576.7 0.346 0.346 8.957 8.957.8 3.55 3.55 35.330 35.330.9 7.450 7.450 4.856 4.857.0.67.67 5.73 5.73. 7.846 7.847 6.44 6.45. 34.656 34.656 74.366 74.366.3 4.789 4.789 88.670 88.670.4 5.470 5.470 05.38 05.38.5 63.958 63.958 4.870 4.870.6 77.55 77.55 47.56 47.56.7 93.593 93.593 73.943 73.944.8.480.48 04.570 04.57.9 34.669 34.670 40.079 40.080 3.0 60.683 60.684 8.96 8.97 Exercie 6 Numerical Solution of Second-Order Differential Equation Solve each equation by the Runge-Kutta method, and find the value of y at x. Take a tep ize of 0. unit.. y y xy 3, y(, ). y xy 3xy y, y(, ) 3. 3y y ye x xy 3, y(, ) 4. y xy y 5 ln x, y(, 0) 5. xy.4x y 5.3x in y 7.4y 0, y(, ) 0. 6. y y ln x.69(x y) 6.6y, y(, 3.7) 8.6 7. x y (x y)y 8.6x y.83y, y(, ) 8. y y 0.533xye x 0.36x co y 0, y(, 73.4) 73 9. y xy 59.xy 74.y lnhxyh, y(, ) 0. xy x y.4x ln y 6.x 0, y(, 8) 5. Computer. Write a program or ue a preadheet to olve a econd-order differential equation by the Runge-Kutta method. Try your program on any of the equation above. If your preadheet ha a graphing utility, ue it to graph x veru y over the given range. REVIEW PROBLEMS Find the Laplace tranform by direct integration.. f(t) t. f(t) 3t 3. f(t) co 3t
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 8 8 Solving Differential Equation by the Laplace Tranform and by Numerical Method Find the Laplace tranform by uing Table. 4. f(t) t 3 3t 5. f(t) 3te t 6. f(t) t 3e t 7. f(t) 3e t t 8. f(t) 3t 3 5t 4e t 9. y 3y, y(0) 0. y 6y, y(0). 3y y, y(0) 0. y y 3y, y(0) and y(0) 3 3. y 3y 4y, y(0) and y(0) 3 Find the invere tranform. 6 4. F() 4 5. F() ( 3) 6 6. F() ( ) 7. F() ( 5) 3 8. F() ( 4 ) 5 9. F() 6 8 4 0. F() ( 3) 3. F(). F() 3 4 4 3. F() 4 4 Solve each differential equation by the Laplace tranform. 4. y y 0, y(0) 5. y 3y t, y(0) 6. 3y y t, y(0) 7. y y y, y(0) 0 and y(0) 0 8. y 3y y t, y(0) and y(0) 9. y y 4t, y(0) 3 and y(0) 0 30. y 3y y e t, y(0) and y(0) 3. A.5-lb ball i dropped from ret through air whoe reiting force i equal to.75 time the ball peed in ft/. Write an expreion for the peed of the ball. 3. A weight that hang from a pring i pulled down 0.500 cm at t 0 and releaed from ret. The differential equation of motion i x 3.x 8.5 0. Write the equation for x a a function of time.
36CH_PHCalter_TechMath_95099 3//007 :8 PM Page 9 Review Problem 9 33. The current in a certain RC circuit atifie the equation 8.4i 49i 00. If i i zero at t 0, write an equation for i. 34. The current in an RLC circuit atifie the equation 5.45i 80i 9730i 0. If i and i are zero at t 0, write an equation for i. 35. In an RLC circuit, R 3750, L 0.50 H, and C.5 µf. The circuit i connected to a dc ource of 50 V at t 0. If i i zero at t 0, write an equation for the intantaneou current. 36. In an RL circuit, R 3750 and L 0.50 H. The circuit i connected to an ac ource of 8.4 in 84t V at t 0. If i i zero at t 0, write an equation for the intantaneou current. Solve each differential equation. Ue any numerical method with a tep ize of 0. unit. 37. y xy 4y, y(). Find y(3). 38. xy 3y 5xy, y() 5. Find y(). 39. 3y y xy ln y, y(, ) 8. Find y(). 40. yy 7.yy.8x in x.y 0, y(, 5) 8.. Find y(). 4. y y ln y 5.7(x y).84x, y(, ) 3. Find y(). 4. x y ( y)y 3.67x 5.8x, y(, ) 3. Find y(). 43. y xy 6e x 6 co y 0, y(, ) 5. Find y(). 44. y xy 8 ln y, y() 7. Find y(). 45. 7.35y.85x in y 7.34x 0, y() 4.3. Find y(3). 46. yy y 77.y 8.4x lnhxyh, y(3, 5) 3. Find y(4). Writing 47. Suppoe that your company get a new project requiring you to olve many differential equation. You ee thi a a chance to get that computer you ve alway wanted. Write a memo to your bo explaining how the computer can be ued to olve thoe differential equation. Team Project 48. Given y y e t 0 where y(0), olve thi DE by all of the method at your dipoal: (a) analytically (b) graphically (c) numerically (d) by the Laplace tranform (e) by CAS Find y when t.