EXERCISES FOR SECTION 6.3

Transcription

1 y 6. Secon-Orer Equation t EXERCISES FOR SECTION 6.. We ue integration by part twice to compute Lin ωt Firt, letting u in ωt an v e t t,weget Lin ωt in ωt e t e t lim b in ωt e t t. in ωt ω e t co ωt t, ince the limit of e b in ωb i a b an >. Uing integration by part on e t co ωt t, b e t ω co ωt t + ω e t co ωt t with u co ωt an v e t t,weget e t co ωt t e t co ωt e t ω in ωt t e t b lim co ωt b ω e t in ωt t e b lim co ωb + b ω e t in ωt t ω e t in ωt t,

2 5 CHAPTER 6 LAPLACE TRANSFORMS ince the limit of e b co ωb i a b an >. Thu, in ωt e t t ω ω ω e t co ωt t in ωt e t t So an + ω ω ω in ωt e t t. in ωt e t t ω, in ωt e t t ω + ω.. We ue integration by part twice to compute Lco ωt co ωt e t t. Firt, letting u co ωt an v e t t,weget Lin ωt e t co ωt e t ω in ωt t e t b lim co ωt b ω e t in ωt t ω e t in ωt t, ince the limit of e b co ωb i a b an >. Uing integration by part on e t in ωt t, with u in ωt an v e t t, we get that e t in ωt t e t in ωt lim b e t e t ω co ωt t b in ωt + ω e t co ωt t ω e t co ωt t,

3 ince the limit of e b in ωb i a b an >. Thu co ωt e t t ω 6. Secon-Orer Equation 5 e t in ωt t So an + ω ω co ωt e t t, co ωt e t t e t co ωt t. + ω.. We nee to compute Le at in ωt e at in ωte t t in ωte at t. We can o thi uing integration by part twice an ening up with Le at in ωt on both ie of the equation. Alternately, if we let r a, then in ωt e at t in ωt e rt t The integral on the right i the Laplace tranform of in ωt with r a the new inepenent variable. From Exercie, we know in ωt e rt ω t r + ω. Subtituting back we have Le at ω in ωt a + ω. 4. We nee to compute Le at co ωt e at co ωt e t t co ωt e at t. We can o thi uing integration by part twice to en up with Le at co ωt on both ie of the equation. Alternately, if we let r a, then co ωt e at t co ωt e rt t. The integral on the right i the Laplace tranform of co ωt with r a the new inepenent variable. From the table, we know co ωt e rt r t r + ω.

4 5 CHAPTER 6 LAPLACE TRANSFORMS Then ubtituting back we have Le at co ωt a a + ω. 5. Uing the formula y L t Ly y y, an the linearity of the Laplace tranform, we get that Ly y y + ω Ly. Subtituting the initial conition an olving for Ly give Ly + ω. 6. Since Lco ωt + ω, we can compute that Lco ωt ω ω + ω ω + ω, but Lco ωt L co ωt L t in ωt. ω ω We can bring the erivative with repect to ω inie the Laplace tranform becaue the Laplace tranform i an integral with repect to t,thati, Lco ωt co ωt e t t ω ω Canceling the minu ign on left an right give Lt in ωt ω + ω. co ωt e t t. ω 7. Since we can compute that but So Lin ωt ω + ω, ω Lin ωt ω + ω, Lin ωt L in ωt Lt co ωt. ω ω Lt co ωt ω + ω.

5 54 CHAPTER 6 LAPLACE TRANSFORMS. In thi cae, b, an + b/ + + +, o In thi cae, b 4, an + b/ 4 + 4, o In thi cae, b, an + b/ + / + + /4, o / + /4 + / + /. 4. In thi cae, b 6, an + b/ , o In Exercie, we complete the quare an obtaine ,o L L L + + e t in t. 6. In Exercie, we complete the quare an obtaine ,o L L L + + L + e t co t + e t int e t co t + int. 7. In Exercie, we complete the quare an obtaine / + /,o / + /. We want to put thi fraction in the right form o that we can ue the formula for Le at co ωt an Le at in ωt. We ee that + + / + / + + / + / + + / + / So + / + / + / + 4/ / + / + /. + L L + / / / L / + / + / e t/ co t + 4 e t/ in t.

6 6. Secon-Orer Equation In Exercie 4, we complete the quare an obtaine ,o We want to put thi fraction in the right form o that we can ue the formula for Le at co ωt an Le at in ωt. We ee that So L + e t co t e t in t We compute L e a+ibt e a+ibt e t t e a+ibt t a + ib lim u e au e ibu. The limit i zero a long a > a. Hence, L e a+ibt a + ib if > a an unefine otherwie. Thi i the ame formula a for real exponential. It can alo be written L e a+ibt a + ib a + b.. Thi follow from linearity: Ly Ly re + iy im y re + iy im e t t y re t e t t + i Ly re +ily im. y im t e t t

7 56 CHAPTER 6 LAPLACE TRANSFORMS. We recall that So e at co ωt Ree a+ibt. Le at co ωt ReLe a+ibt a + iω Re a + ω Similarly, a a + ω. Le at in ωt ImLe a+ibt a + iω Im a + ω ω a + ω.. a The root of + + 5are ± i, o the quaratic factor into + i i. b We write A + + i + B + i. So, fining common enominator that i, uual partial fraction but with complex number give A + B A + B + i A + B. Solving, we get A i/4 anb i/4, o c We know that i/4 + + i + i/4 + i. L i/4 i + + i 4 e it i 4 e t co t + ie t in t e t in t + ie t co t 4

8 6. Secon-Orer Equation 57 an L i/4 i + i 4 e +it i e t co t + ie t in t 4 4 e t in t ie t co t. Aing, we get L e t in t.. Uing the quaratic formula, we ee that the root of + + are ± i. Thu i + i. So we want to fin A an B o that + + A + + i + B + i. So, fining common enominator that i, uual partial fraction only with complex number give A + B A + B + i A + B. Solving, we get A i/6 anb i/6, o Thu L L i/6 + + i + i/6 + i. i/6 + + i + i/6 + i i 6 e it i 6 e +it i e t co t + ie t in t i e t co t + ie t in t 6 6 i ie t in t 6 e t in t.

9 58 CHAPTER 6 LAPLACE TRANSFORMS 4. Uing the quaratic formula, we fin that the root of the enominator are ±i. Hence, we can factor the enominator into i i. Uing partial fraction ecompoition, we get A + i + B i, which give the equation A + B ia + ib. Solving for A an B give A / i an B / + i o i + i + + i i. Taking the invere Laplace tranform of the term on the right give i e +it + + i e it. Uing Euler formula to expan the exponential an implifying give e t co t + int. 5. Uing the quaratic formula, the root of the enominator are ± i /. Hence, we can factor the enominator into. +i We then o the partial fraction ecompoition which give rie to the equation A +i +i A + Solving yiel A i an B + i.so i + i i +i B A + B B. i + + i i,.

10 Taking invere Laplace tranform of the right-han ie give i e +i t/ + + i e i t/. Uing Euler formula to replace the complex exponential an implifying yiel e t/ co t + 4 e t/ in t. 6. Secon-Orer Equation Uing the quaratic formula, we fin the root of the enominator are ± i o the enominator can be factore i i. The partial fraction ecompoition i A i + B + i, which lea to the equation A + B ia + + ib. Solving, we fin A i an B + i, o i i + + i + i. Taking invere Laplace tranform of the right-han ie give i e it + + i e +it an uing Euler formula an implifying give e t co t e t in t. 7. a Taking the Laplace tranform of both ie of the equation, we obtain y L t + 4Ly 8, an uing the fact that L y/t Ly y y,wehave + 4Ly y y 8.

11 5 CHAPTER 6 LAPLACE TRANSFORMS b Subtituting the initial conition yiel an olving for Ly we get + 4Ly 5 8, Ly The partial fraction ecompoition of 8/ + 4 i A + B + C + 4. Putting the right-han ie over a common enominator give u A + B + C + 4A 8, an conequently, A, B, an C. In other wor, We obtain Ly c To take the invere Laplace tranform, we rewrite Ly in the form Ly Therefore, yt + 9cot + 5 in t. 8. a Taking the Laplace tranform of both ie of the equation, we obtain y L t Ly, an uing the fact that L y/t Ly y y,wehave Ly y y. b Subtituting the initial conition yiel Ly +, an olving for Ly we get Ly + +..

12 6. Secon-Orer Equation 5 Uing the partial fraction ecompoition , we obtain Ly c Taking the invere Laplace tranform, we have yt et et e t. 9. a Taking the Laplace tranform of both ie of the equation, we obtain y y L t 4L + 5Ly t, an uing the formula for Ly/t an L y/t in term of Ly,wehave 4 + 5Ly y y + 4y. b Subtituting the initial conition yiel 4 + 5Ly +, an olving for Ly we get Ly Uing the partial fraction ecompoition , we obtain Ly c In orer to compute the invere Laplace tranform, we firt write by completing the quare, an then we write Taking the invere Laplace tranform, we have yt e t + e t co t 4e t in t.

13 5 CHAPTER 6 LAPLACE TRANSFORMS. a Taking the Laplace tranform of both ie of the equation, we obtain y y L t + 6L + Ly e 4, t an uing the formula for Ly/t an L y/t in term of Ly,wehave Ly y y 6y e 4. b Subtituting the initial conition yiel Ly 9 e 4, an olving for Ly we get + 9 Ly Uing the partial fraction ecompoition , e 4. we obtain + 9 Ly e c In orer to compute the invere Laplace tranform, we firt write by completing the quare, an then we write an Taking the invere Laplace tranform, we have yt e t co t +5e t in t +u 4 t e t 4 co t 4 e t 4 in t 4... a Note that thi i reonant forcing of an unampe ocillator. We take the Laplace tranform of both ie y L t + 4Ly Lco t an obtain Ly+ + 4Ly + 4.

14 b Solving for Ly,weget Ly c To take the invere Laplace tranform, we note that L cot + 4 an L L + 4 t in t. 4 So yt cot + t in t, 4 which i of the form we woul expect for a reonant repone.. a We take Laplace tranform of both ie to obtain y L t + Ly Lu 4 t co5t 4, which i equivalent to Ly y y + Ly Uing the given initial conition, we have 6. Secon-Orer Equation 5 e Ly+ e b Solving for Ly give Ly + + e c Now to fin the invere Laplace tranform, we firt note that L L + in t. + For the econ term, we firt ue partial fraction to write Hence L e u 4t co t 4 co5t 4. Combining the two reult, we obtain the olution of the initial-value problem yt in t + u 4t co t 4 co5t 4.

15 56 CHAPTER 6 LAPLACE TRANSFORMS 5. a Conier L f F f t e t t. We can calculate F/ by ifferentiating uner the integral ign. That i, F f t e t t Ltft. f t te t t b If we apply thi reult to we obtain Lin ωt ω + ω ω + ω, Lt in ωt ω + ω ω + ω. Compare thi reult with the reult of Exercie 6. EXERCISES FOR SECTION 6.4. Thi i the cae of L Hôpital Rule. Differentiating numerator an enominator with repect to t, we obtain e t e t, which implifie to e t + e t. Since both e t an e t ten to a t, the eire limit i.. Taking Laplace tranform of both ie an applying the rule yiel Ly y y + Ly 5Lδ. Simplifying, uing the initial conition, an the fact that Lδ e,weget + Ly 5e.