[1] (20 points) Find the general solutions of y y 2y = sin(t) + e t. Solution: y(t) = y c (t) + y p (t). Complementary Solutions: y

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2 [] (2 points) Find th gnral solutions of y y 2y = sin(t) + t. y(t) = y c (t) + y p (t). Complmntary Solutions: y c y c 2y c =. = λ 2 λ 2 = (λ + )(λ 2), λ =, λ 2 = 2 y c = C t + C 2 2t. A Particular Solution (by undtrmind cofficints): St y p = A cost + B sin t + Ct t and substitut in th nonhomog diff q: y p y p 2y p = sin t + t = ( A cost B sin t + Ct t 2C t ) ( A sin t + b cos t Ct t + C t ) 2(A cost + B sin t + Ct t ) = ( 3A B) cost + (A 3B) sin t 3C t Compar th cofficints: 3A B = A 3B = 3C = = A = / B = 3/ C = /3 = y p = cost 3 sin t 3 t t Gnral Solutions: y = C t + C 2 2t + cos t 3 sin t 3 t t

3 [2] (2 = + points) Considr x = (a) Find th gnral solutions. (b) Sktch th phas portrait. 2 x. 2 2 λ (a) Eignvalus: dt(a λi) = dt = λ 2 λ 2 4λ + 3 = (λ 3)(λ ) = λ = 3, λ 2 = Eignvctors [ for λ = 3: ][ (A] 3I)v = v v = = = v v 2 v 2 2 Eignvctors for [ λ 2 = ][ : ](A[ I)v ] = v v = = = v v 2 v 2 2 Gnral Solutions: x(t) = C 3t + C 2 t (b) Not for th gradr: Th phas portrait should show: straight lins for solution trajctoris on th ignspacs, curvd trajctoris for solutions that ar not on th ignspacs, at th origin, all th curvd trajctoris ar tangnt to th ignspac of λ 2 =, and ndlss to say, th corrct dirctions of arrows on all solution curvs.

4 3 2 [3] (2 points) Solv x t = x t, x() =. 3 2 You may us th fact that th matrix has 4 3 ignvalu λ = with ignvctor, and 2 ignvalu λ 2 = with ignvctor. t A Fundamntal M(t) = t St x(t) = M(t)u(t). 2 t Th Equation for u(t): u (t) = M(t) f(t) u t (t) = t t 2 2t 2 t t t = 3 u() = M() x() = Intgrat: u(t) = Th 2t + 3t t [ t x(t) = M(t)u(t) = t 2 t t = t + t + 3t t 2 t + 2 t + 3t t ] 2t + 3t

5 [4] (2 points) Solv y 6ty = 3t, y() = 2. Solution : (Intgrating Factor) Prpar 6tdt = 3t 2 + C. Multiply th diff q by 3t2 : ) (y 3t2 6ty = 3t 3t2, ( 3t2 y) = 3t 3t 2, 3t2 y = 3t 3t2 dt (substitut s = 3t 2, ds = 6tdt) = 2 s ds = 2 s + C = + C. 2 3t2 Thus, th gnral solutions ar: y = 2 +. C3t2 Th initial condition y() = 2 implis: + C = 2, C = Thrfor, th solution to th initial valu problm is: y = t2 Solution 2: (Sparation of Variabls) Rwrit th diff q as: dy dt = 6ty + 3t = 6t ( y + ). 2 Sparat th variabls and intgrat: dy dy y + = 6tdt, y + = 6tdt, 2 2 ln y + 2 = 3t2 + C, y + 2 = +C = C ±3t2 3t2 (whr C = ± C ) Th gnral solutions ar y = 2 +. C3t2 Th initial condition y() = 2 implis: + C = 2, C = Thrfor, th solution to th initial valu problm is: y = t2

6 [5] (2 = + points) Considr: dx dt = 2x x2 xy, dy dt = x y. (a) Giv th linar approximating systm nar th quilibrium (, ). (b) Dtrmin whthr th quilibrium (, ) is stabl, asymptotically stabl, or unstabl with rspct to th nonlinar systm ( ). 2 2x y x (a) Th Jacobian matrix: J = At point (, ): J = Th linar approximating systm nar (, ) is: x x = y (b) Th ignvalus of th Jacobian matrix y : λ dt = ( λ) 2 + =, λ λ = ± i. Sinc R λ = <, th quilibrium (, ) is asymptotically stabl with rspct to th nonlinar systm ( ). ( )

7 [6] (2 = points) Considr th diffrntial quation: dy dt = (y + 2)2 (y )(y 3). (a) Find all quilibria. (b) Sktch th phas portrait. (c) Dtrmin whthr ach quilibrium is stabl, asymptotically stabl, or unstabl. (a) (y + 2) 2 (y )(y 3) = y = 2,, 3. (b) (c) y = 2 is unstabl (or smistabl). y = is asymptotically stabl. y = 3 is unstabl.

8 [7] (2 = + points) (a) Us th Laplac transform to solv y 3y + 2y =, y() = 2, y () = 5. Convrt th diff q to th s-domain by th Laplac tranform: Solv for Y (s): Find th partial fraction: [s 2 Y (s) 2s 5] 3[sY (s) 2] + 2Y (s) =. Y (s) = 2s s 2 3s + 2 = 2s (s 2)(s ). 2s (s 2)(s ) = a s 2 + b s 2s = a(s ) + b(s 2). St s = 2 3 = a; St s = = b b =. Hnc, Y (s) = 3 s 2 + s. Th invrs Laplac transform convrts this back to th t-domain: y(t) = 3 2t t. (b) Find th Laplac transform of f(t) = { t 2 t < 4, t 4. f(t) = t 2 [ u(t 4)] = t 2 t 2 u(t 4) = t 2 [(t 4) + 4]u(t 4). L{f(t)} = 2 4s L{[t + 4] 2 } s 3 = 2 4s L{t 2 + 8t + 6} s 3 = 2 4s s 3 s 3 s 2 s.

9 [8] (2 = + points) (a) Considr an opn systm of two brin tanks. Tank initially contains 2 gallons of brin with lb of salt dissolvd in it and Tank 2 contains 3 gallons of brin with 5 lb of salt dissolvd in it. Pur watr flows from an outsid sourc into Tank at 5 gal/min; th mixtur flows from Tank into Tank 2 at th sam rat; and it is dischargd out of th systm from Tank 2 with th sam rat. Dnot by Q (t) and Q 2 (t) th amounts of salt in th two tanks at tim t rspctivly. Writ down diffrntial quations for Q (t) and Q 2 (t). Pur watr in 5 gal/min = TANK Brin: 2 gal Salt: Q (t) lb Q () = lb 5 gal/min = TANK 2 Brin: 3 gal Salt: Q 2 (t) lb Q 2 () = 5 lb = Brin out 5 gal/min Q (t) = 5 Q (t) 2 Q 2(t) = 5 Q (t) 2 5 Q 2(t) 3 Q (t) = Q (t), 4 Q 2(t) = Q (t) 4 Q 2(t). 6 (b) On a smooth horizontal surfac, a mass m = 2 kg is attachd to a fixd wall by a spring with spring constant k = 4 N/m. Anothr mass m 2 = 5 kg is attachd to th first objct by a spring with spring constant k 2 = 3 N/m. Ignor th damping. Writ down diffrntial quations for th displacmnts x (t) and x 2 (t). Us Nwton s law: Mass Acclration = Nt forc. { m x = k x + k 2 (x 2 x ) m 2 x 2 = k 2(x 2 x ) { 2x = 7x + 3x 2 5x 2 = 3x 3x 2

10 4 3 3 [9] (2 points) Find th gnral solutions of x = Ax, whr A = 5 2. You may us th fact that dt(a λi) = (λ 7)(λ )(λ + 2), A = 7. Eignvalus of A ar in th givn info: λ = 7, λ 2 =, λ 3 = 2. An Eignvctor for λ = 7 is also in th givn info:. Eignvctors for λ 2 = : (A I)v = v v 2 = = v 3 Eignvctors for λ 3 = 2: (A + 2I)v = v v 2 = = 2 v 3 Gnral Solutions: x(t) = C 7t + C 2 t v v 2 v 3 = v 3 v v 2 = v 3 v 3 + C 3 2t... Turn ovr for on mor problm. Kp going!

11 [ [] (2 points) Lt A = 3 ]. Evaluat ta. You may us th fact that matrix A has [ ignvalu λ = 2 with ignvctor 3 [ ignvalu λ 2 = 2 with ignvctor ], and From th ignvalus and ignvctors, w obtain a fundamntal solution: 2t M(t) = 2t. It follows that ]. 3 2t 2t ta = M(t)M() 2t = 2t 3 2t 2t 3 2t = 2t 3 2t 2t 4 3 = [ 2t + 3 2t 2t 2t 4 3 2t 3 2t 3 2t + 2t ]. Th End. S you around.