Chapter 8 Part I

# Chapter 8 Part I - CHAPTER 8 Forced Equations and Systems...

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CHAPTER 8 Forced Equations and Systems 8.1 Linear Nonhomogeneous Problems ! Superposing For Problems 1–6, we use the fact that Ly y y () ≡ ′′ += 0 has the general solution of yt c t c t h () =+ 12 sin cos . The particular solution for t f t == cos2 1 is yt t 1 1 3 2 ()=− cos . The particular solution for t f t =+= 2 2 2 is yt t 2 2 () = . 1. yy t 32 cos Here, t 323 1 cos , so y t t p F H I K =− 33 1 3 22 1 cos cos . Hence, the equation has the general solution of c t t 2 sin cos cos . 599

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600 CHAPTER 8 Forced Equations and Systems 2. ′′ += − − yy t 24 2 Here, Ly t t f t ()=− − =− + =− () 242 2 2 22 2 bg , so yt y t t p =− 2 2 . Hence, the equation has the general solution of c t c t t =+ 12 2 2 sin cos . 3. + + tt 62 3 6 2 cos Here, t t t t + = ++ = () + 623 6 6 23 26 3 cos cos , so t t t t p () = F H I K + 63 6 1 3 2 2 3 cos cos . Hence, the equation has the general solution of c t t t + 2 3 sin cos cos . 4. 3 6 2 cos Here, t t t t + = 3 2 2 3 cos cos , so y y t t p + =− − F H I K + 2 1 3 2 3 cos cos . Hence, the equation has the general solution of c t t t + + 2 2 3 sin cos cos . 5. t 32 cos , y 03 , y 01 As seen in Problem 1, the general solution of the DE is yc tc t t 2 sin cos cos .
SECTION 8.1 Linear Nonhomogeneous Problems 601 Plugging into the initial condition gives yc c 01 3 4 22 1 () =−=⇒= == . The solution of the IVP is yt t t t ()=− + sin cos cos 42 . 6. ′′ += − − yy t 24 2 , y 02 , () = y As seen in Problem 2, the general solution of the DE is yc tc t t =+ 12 2 2 sin cos . Plugging into the initial condition gives 2 1 . The solution of the IVP is t t t =− sin cos 2 . ! Superposing Again For Problems 7–12, we use the fact that Ly y y () ≡ −= 40 has the general solution yt c e ce h tt 1 2 2 2 ; t f t = 41 2 2 1 and t f t 2 sin have particular solutions t 1 2 7 2 ()=− − and t 2 1 4 2 sin , respectively. 7. t 45 2 sin Here, t t = 52 5 2 5 2 2 sin sin ,

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602 CHAPTER 8 Forced Equations and Systems so yt y t t p () == F H I K =− 5 2 5 2 1 4 2 5 8 2 2 sin sin . Hence, the equation has the general solution of ce t tt =+ 1 2 2 2 5 8 2 sin . 8. ′′ −= + yy t 41 1 3 2 Here, Ly t t f t = + = 1 1 3 1 12 12 4 1 12 22 1 bg , so y t t p F H I K 1 12 1 12 7 2 1 12 7 24 1 . Hence, the equation has the general solution of t 1 2 2 1 12 7 24 . 9. t t 43 2 6 2 2 sin Here, t t t t = −+ = () − 3262 3 2 1 2 12 4 3 2 1 2 21 sin sin , so y y t t t t p =−= F H I K −− F H I K + + 3 2 1 2 3 2 1 4 2 1 2 7 2 3 8 2 1 2 7 4 sin sin . Hence, the equation has the general solution of t t + + 1 2 2 3 8 2 1 2 7 4 sin . 10. t t 45 2 51 5 2 sin Here, t t t t = = 525 1 5 5 2 5 4 12 4 5 2 5 4 sin sin , so y y t t t t p F H I K F H I K + + 5 2 5 4 5 2 1 4 2 5 4 7 2 5 8 2 5 4 35 8 sin sin .
SECTION 8.1 Linear Nonhomogeneous Problems 603 Hence, the equation has the general solution of yt ce t t tt () =+ + + 1 2 2 22 5 8 2 5 4 35 8 sin .

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Chapter 8 Part I - CHAPTER 8 Forced Equations and Systems...

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