hw3sol - {fly #3 2/2 6. First, solve the differential...

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Unformatted text preview: {fly #3 2/2 6. First, solve the differential equation y” + cy’ — 6. Let y ___ ex: in 6),” + y: __ y = 0 to Obtain ay + by3 = A cos wt for the highest derivative of 2 M M M y present in the equation. 6)L e + M —— e = 0 e“(6i2 + i — 1) = 0. I y': ~cyl+ay ——by3+Acoswt Because e“ 7+— 0, we arrive at the characteristic equa- tion Next, set v = y’. Then 6% + A ~ 1 = 0 (3)» -— 1)(2)t + l) = 0, and roots )t = 1/3 and A = ——1/2. Because the u' = y” 2 —cy’ + ay —— by3 + Acoswt = ——cv +ay —— by3 + Acoscot. Thus, we now have the following system of first or- roots are distinct, the solutions M (t) = 6(1/3):‘and def equations‘ V y2(;) = e‘fl/Z)‘ form a fundamental set of solutions y' _ v and the general solution is 3 - /2 v'=~—cv+ay—by3+Acoswt y(t)=Ciet/ +Cze ' ~ // 1 . . . ' 12. {11303171 1: 2y +17y : 0, then the characteristic equa— 26. If my” ~ y, _ 3y = 0, then the characteristic equafi tionis 2 _ A+2A+l7—0. 10A2~A~3=(2}»+1)(5)»~3)=0: The roots of the characteristic equation are —1 :l: 41' , with roots A 2 —~ 1 / 2 and A = 3/5. This leads to the leading to the complex solutions general solution z(t) = e(‘l+4’7’ and at) = e<-1~4i)r. y(t) = Cie_'/2 + C2e3’/5. Using the initial condition y(0) = 1 provides However, by Euler’s identity, 1=C1+C2. z(t) = We“ = e” cos 4: ‘si . . . . ( + l n 4!) ’ Differentiating the general solution, and the, real and imaginary parts of 2 lead to a fun~ 1 — 2 3 3 5 damental set of real solutions y1(t) = e” cos 4t and ya) : —§Cle t/ + gaze l/ ’ yz (t) = e“’ sin 4:. Hence the general solution is ya) = Cle—t cos 4; + Cze-I’ sin 4t. ‘i‘len using the initial condition y’(0) = 0 leads to 1 3 0 = + fliese equations yield C1 = 6/11 and C2 = 5/ 1], 20' If 4)," + my + 9 y : 0, then the characteristic equa— glvmg the partlcular SOImion tion is 6 W 5 a /5 (t = ——e_ + —— ‘t . 4i2+12i+9=(2i+3)2=0. 3’) 11 116 Hence, the characteristic equation has a repeated root, A = —3/2. Therefore, y1(t) = 670/2)! and y2 (t) = Ira/2” form a fundamental set of real so— lutions. Hence, the general solution is y(t) = Clefii3/2)'+C2te”(3/2)t = (C1+C2t)e_(3/2)’. 36. If y” — 4y’ + 13y = 0, then the characteristic equa- tion is A2 # 4}» +13 = 0 with roots z = 2:1: 3i. The complex solution z(t) = e<2+3")' = e2’(cos 3t + i sin 3t) provides a fundamental set of real solutions, y1(t) = e” cos 3t and yz (t) = ez‘ sin 3t, and the general so- lution y(t) =.e2'(C1 cos 3t + C2 sin 31‘). ' The initial condition y(0) = 4 provides 4 = C1. Differentiating the general solution, y/(t) = 2e2t (C1 cos 3t + 02 sin 31) + e2‘(—3Cl sin 3t + 3C2 cos 3:), and the initial condition y’ (0) = 0 provides 0 = 2C1 + 3C2. Thus, C1 = 4 and C2 = —8/3, giving the solution 8 y(t) = e” (4 cos 3t — 3 sin 3t) . 2/5 8. Let yp = (1 cos 3t + b sin 3t. Then y); = “3a sin 3t + 3b cos 3t y; = —-9a cos 3t — 9b sin 3t, and the equation y”+7 y’ + 10y = —4 sin 3! becomes (a + 2117) cos 3t + (—2la + 1;) sin 3t = —4 sin 3:. Thus, a + 211) = 0 ~21a + b = —-4, leading to a = 42/221 and b = —2/221 and the particular solution yp 2 (42/221) cos 3t - (2/221) sin 3t. 14. Let y : at + I). Then y’ = a, y” = O and y" + 5y’ + 4y = 2 + 3t leads to 5a+4(at+b)=2+3t 4at+(5a+4b)=3t+2. Thus, 4a 2 3 and 5a +41) 2 2, which givesa = 3/4 and b := —7/l6. Therefore, a particular solution is y = (3/4)t ~ 7/16. 30. If y” —- 2y’ —- 3y = 0, then the characteristic equation 15 i2—2A—3=(i—3)(A+1)=0, with roots )c = 3 and )t = ——1. This leads to the general solution 1‘ y(t) = C1e3‘ + Cge‘ . Using the initial condition y(0) = 2 provides 2 = C] + C2} Differentiating the general solution, y'(i) = 3Cie3‘ — Cze”’, then using the initial condition y’(0) = —3 leads to —3 = 3C1 — C2. These equations yield C1 = —l/4 and C2 :: 9/4, giving the particular solution 1 9 y(t) = —Ze3’ + 2e". 18. The homogeneous equation y” + 337’ + 2y = 0 has characteristic equation A2 + 3)» + 2 : (A + 1)()i + 2) = 0 with zeros A1 = —l and M = —2. This leads to the homogeneous solution yh = Cie“ + C2e_2’. The particular solution yp = Ae“4‘ has derivatives y; = -—4Ae”4’ and y; = l6Ae‘4’, which when substituted into the equation y” + 3y’ + 2y = 352"“ provides 16Ae_4‘ + 3(—4Ae‘4‘) + zine—4‘ : 3e'4‘ rru *- i nus, a particular salt: 1 y = Cie” + Cze'zt + 56“”. The initial condition y(0) = 1 provides 1 1=C1+C2+r22 Differentiating, y’ = ~C1e”‘ ~ 2C2e—2r — zer‘“. The initial condition y’(()) = 0 provides O=—Ci ~2C2—2. This system has solution C1 = 3 and C2 = —5/2, leading to the solution 5 1 y = 3e" —- if” + 563"". ...
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hw3sol - {fly #3 2/2 6. First, solve the differential...

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