Class Notes (806,783)
MATH 263 (26)
Lecture

Math 263 F11.pdf

10 Pages
373 Views

School
McGill University
Department
Mathematics & Statistics (Sci)
Course
MATH 263
Professor
Jian- Jun Xu
Semester
Fall

Description
McGill University December 2011 Faculty of Engineering Final examination Diﬁerential Equations for Engineers Math 263 Wednesday, December 14, 2011 Time: 2pm-5pm Examiner: Prof. J.J. Xu Associate Examiner: P. Reynolds SOLUTIONS OF FINAL EXAMINATION 1. (a) (8 points) Solve the diﬁerential equation 0 2 2 2xyy + x = 3y : (b) (2 points) Give another method that we have learned of solving this equation. You may only brie°y outline the second method without carrying it out in detail. Solution: Re-arranging, we have x ¡ 3y + 2xyy = 0 which is not exact since y = ¡6y and N x 2y, and these are not equal. However, M y N x ¡8y 4 N = 2xy = ¡ x which is a function of x alone. Thus we ﬂnd an integrating factor „(x) by solving 0 4 „ (x) = ¡ „(x) x ¡4 which gives „(x) = x . Thus multiplication of the original DE by this integrating factor gives the DE 1 y2 y ¡ 3 + 2 y = 0 x 2 x4 x3 which is exact (as is easily checked). Thus we seek F(x;y) such that y 1 y2 Fy= 2 3; F x 2 ¡ 3 4 : x x x 2 Thus F(x;y) = y3¡ 1 and so the solutions are given implicitly by x x 2 y 1 x3 ¡ x = C : MATH 263 Final Exam Page 2 December 14, 2011 (b) (2 points) There is another method, that we have learned, of solving this equation. Determine what this other method is and brie°y outline it (you do not need to carry it out in detail). Solution: The equation can be rearranged as 0 3y ¡ x 2 3 y 1 x y = = ¡ 2xy 2 x 2 y which is a homogeneous DE. Applying the substitution u = y=x results in the DE 1 µ 1¶ u = u ¡ 2x u which is separable. So solve it by integrating: Z Z 2u du = 1dx u ¡ 1 x which gives u ¡ 1 = Cx or y 2 1 ¡ = C : x 3 x 2. (10 points) Solve the diﬁerential equation y ¡ (Acost + B)y + y = 0: Solution: 1¡3 ¡2 This is a Bernoulli DE. We apply the substitution u = y = y , and assume y 6= 0: u = ¡2y ¡3y0 = ¡2y ¡3((Acost + B)y ¡ y ) = ¡2(Acost + B)u + 2 which is a ﬂrst order nonhomogeneous DE for u. Writing it as u + 2(Acost + B)u = 2 we compute an integrating factor R 2A cos t+2Bdt 2A sin t+2Bt „(t) = e = e and thus Z 2A sin t+2Bt 2A sin t+2Bt e u = 2 e dt + C so µ Z ¶ ¡2A sin t¡2Bt 2A sin t+2Bt u = 2e e dt + C : MATH 263 Final Exam Page 3 December 14, 2011 Since y = u ¡1=2, we obtain the general solution as follows: £ ¡R ¢⁄¡1=2 y = § 2e ¡2A sin t¡2Bt e2A sin t+2Bdt + C ; y = 0: 3. (15 points) By using the method of diﬁerential operators, solve the Eq. y(4)¡ 2y + y = e + sint: 1. Determine the annihilator of the inhomogeneous term and the form of particular solution; 2. Find the particular solution and the general solution for the equation; 3. Find the solution of the IVP with the IC’s: 0 00 000 y(0) = 0;y (0) = 1;y (0) = 0;y (0) = 1: Solution: t 1. The inhomogeneous term b(t) = b (t) +1b (t), 2here b (x) = 1 , b (t) = s2nt, the corresponding annihilator is Q(D) = Q (D1 – Q (D) 2 where Q (D) = (D ¡ 1), Q (D) = (D + 1): 2 1 2 2. Let the particular solution as y =py p1 + yp2 where ' P(D) y g p1b (t);1 P(D)fy g p2b (t);2 and P(D) = (D ¡ 2D + 1) = (D ¡ 1) (D + 1) .2 2 Note that the operators P(D) Q (D) h1ve a common root D = 1, and the multiplicity this root in P(D) is m = 2, while the operators P(D) Q (D) have2no common root. Therefore, it can be deduced that the form of a particular solution is yp1 = t kerfQ (1)g; y p2= kerfQ (D2g; so that we have y = B t e + Acost + Bsint: p 2 Then it follows that y = 2B te + B t e ¡ Asint + Bcost; p 2 2 00 t t 2 t yp= 2B te2+ 2B te +2B t e ¡ 2cost ¡ Bsint; y 00= 6B e + 3B te + B t e + Asint ¡ Bcost; p 2 2 2 (4) t t 2 t yp = 12B e2+ 4B te 2 B t e +2Acost + Bsint: Inserting them into the Eq., we have t t 8B 2 + 4Acost + 4Bsint = e + sint: Equalizing two sides, we get 8B = 1;A = 0;4;B = 1: 2 Then B = 1;B = 1 ; 2 8 4 MATH 263 Final Exam Page 4 December 14, 2011 here B 0 B a1e taken arbitrarily. When we take both B and B0as zero1here, the particular solution is 1 2 t 1 yp= t e + sint: 8 4 The general solution for the equation is expressed as t ¡t 1 2 t 1 y = (C 0 C t1e + (D + 0 t)e1 + 8 t e + 4 sint; where C 0C ;1 ;D0ar1 arbitrary constants. 3. With the given initial conditions, we get C0+ D =00; 1 C 0 C +1D ¡ D0+ 1 = 1; 4 1 C 0 2C +1D ¡ 20 + 1 = 0; 4 1 C 0 3C ¡1D + 30 + 1 2 = 0: We solve the associated equations to ﬂnd the particular solution: 7 1 7 C = ;C = ¡ ;D = ¡ ;D = 0: 0 16 1 8 0 16 1 4.(10 points) By changing variables and using the diﬁerential operator method, ﬂnd the general solution of the following Eq: 3 000 2 00 1 0 1 x y + x y + xy ¡ y = 0: 4 4 Solution: Let x = §e , y(x) = y~(s) D = ds , it is derived that 0 2 00 3 000 xy = Dy ~;x y = D(D ¡ 1)y ~;x y = D(D ¡ 1)(D ¡ 2)y ~: Substituting them into the Eq., we have µ ¶ 1 1 P(D)y~ = D(D ¡ 1)(D ¡ 2)+D(D ¡ 1)+ D ¡ ~ = 0: 4 4 The polynomial P(D)has three roots 1 1 D = 1; ; : 2 2 So, we can write the general solution for the equation s 1s ~ = C1e + (C 2 C s)3 : 2 Returning to the variable x , we get p p y = C 1 + C 2 x + C 3 xlnjxj; where C 1C ;2 a3e arbitrary constants. 5. (10 points) Compute the Laplace transform of the function f(t) whose graph is pictured. MATH 263 Final Exam Page 5 December 14, 2011 Solution: The function f(t) has the following formula: 8 >0 0 • t < 1 > <2t ¡ 2 1 • t < 2 f(t) = > 2 2 • t < 4 > ¡ + 3 4 • t < 6 > 2 : 0 t ‚ 6 which is expressed in terms of the unit step function as follows: f(t) = [1 (t) ¡2u (t)](2t ¡ 2) 2 [u (t4 ¡ u (t)]4 + [u 6t) ¡ u (t)](¡t=2 + 3] 1 1 = 2u1(t)(t ¡ 1) ¡ 22 (t)(t ¡ 2)2¡ (t ¡ 24u (t)2+ (t ¡ 6)u (t) = 2u (t)(t ¡ 1) ¡ 2u (t)(t ¡ 2) ¡ u (t)((t ¡ 4) + 2) + u (t)(t ¡ 6) 1 2 2 4 2 6 and so the Laplace transform is 1 1 Lff(t)g = 2Lfu 1t)(t ¡ 1)g ¡ 2Lfu2(t)(t ¡ 2)g ¡2Lf(u4(t)(t ¡ 4) + 2)g 2 Lfu6(t)(t ¡ 6)g ¡s ¡2s 1 ¡4s 1 ¡6s = 2e Lftg ¡ 2e Lftg ¡ 2 Lft + 2g + 2 Lftg µ ¶ ¡s 1 ¡2s1 1 ¡4s 1 2 1 ¡6s 1 = 2e 2 ¡ 2e 2 ¡ 2 2 + + 2 2 s s s s s 1 ¡ ¡s ¡2s ¡4s ¡6s¢ = 2 4e ¡ 4e ¡ e (1 + 2s) + e 2s 6. (15 points) Solve the following initial value problem: 00 0 0 1 y + 3y + 2y = –(t ¡ 5) + u10t); y(0) = 0; y (0) =2 : Solution: Let us take the Laplace transform of both sides of the DE, denoting Lfy(t)g as Y (s): 00 0 Lfy + 3y + 2yg = f–(t ¡ 5) + u10t)g ¡10s 2 0 ¡5s e s Y (s) ¡ sy(0) ¡ y (0) + 3(sY (s) ¡ y(0)) + 2Y (s) =+e s ¡10s 2 ¡5s e 1 (s + 3s + 2)Y (s) = e + + s 2 e¡5s e¡10s 1 Y (s) = + + (s + 1)(s + 2) s(s + 1)(s + 2) 2(s + 1)(s + 2) MATH 263 Final Exam Page 6 December 14, 2011 Now using partial fractions we ﬂnd 1 1 1 (s + 1)(s + 2) s + 1 ¡ s + 2 and 1 1=2 1 1=2 = ¡ + : s(s + 1)(s + 2) s s + 1 s + 2 Thus the last term of Y (s) above is µ ¶ 1 1 1 1 1 = ¡ = Lfe¡t ¡ e¡2tg 2(s + 1)(s + 2) 2 s + 1 s + 2 2 and the ﬂrst term of Y (s) above
More Less

Related notes for MATH 263

OR

Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.