18.03SC Final Exam Solutions

18.03SC Final Exam Solutions 1. (a) The isocline for slope 0 is the pair of straight lines y = ± x. The direction field along these lines is flat. The isocline for slope 2 is the hyperbola on the left and right of the straight lines. The direction field along this hyperbola has slope 2. The isocline for slope -2 is the hyperbola above and below the straight lines. The direction field along this hyperbola has slope -2. (b) The sketch should have the following features: The curve passes through (-2, 0). The slope at (-2, 0) is (−2)2 − (0)2 = 4. Going backward from (-2, 0), the curve goes down (dy/dx > 0), crosses the left branch of the hyperbola x2 − y2 = 2 with slope 2, and gets closer and closer to the line y = x but never touches it. Going forward from (-2, 0), the curve first goes up, crosses the left branch of the hyperbola x2 − y2 = 2 with slope 2, and becomes flat when it intersects with y = − x. Then the curve goes down and stays between y = − x and the upper branch of the hyperbola x2 − y2 = −2, until it becomes flat as it crosses y = x. Finally, the curve goes up again and stays between y = x and the right branch of the hyperbola x2 − y2 = 2 until it leaves the box. (c) f (100) ≈ 100. (d) It follows from the picture in (b) that f ( x ) reaches a local maximum on the line y = − x. Therefore f ( a) = − a. (e) Since we know f (−2) = 0, to estimate f (−1) with two steps, the step size is 0.5. At each step, we calculate xn = xn−1 + 0.5, yn = yn−1 + 0.5( xn2 −1 − y2n−1 ) The calculation is displayed in the following table. n xn yn 0.5( xn2 − y2n ) 0 -2 0 2 1 -1.5 2 -0.875 2 -1 1.125 The estimate of f (−1) is y2 = 1.125. 2. . (a) The equation is x = x ( x − 1)( x − 2). The phase line has three equilibria x = 0, 1, 2. For x < 0, the arrow points down. For 0 < x < 1, the arrow points up. For 1 < x < 2, the arrow points down. For x > 2, the arrow points up. (b) The horizontal axis is t and the vertical axis is x. There are three constant solu­ tions x (t) ≡ 0, 1, 2. Their graphs are horizontal. Below x = 0, all solutions are decreasisng and they tend to −∞. OCW 18.03SC Between x = 0 and x = 1, all solutions are increasing and they approach x = 1. Between x = 1 and x = 2, all solutions are decreasing and they approach x = 1. Above x = 2, all solutions are increasing and they tend to +∞. .. .. (c) A point of inflection ( a, x ( a)) is where x changes sign. In particular, x ( a) must be zero. Differentiating the given equation with respect to t, we have .. . . . . x = 2x − 6xx + 3x2 x = x (2 − 6x + 3x2 ). . If x (t) is not a constant solution, x ( a) �= 0 so that x ( a) must satisfy 2 − 6x ( a) + 3x ( a)2 = 0 ⇔ 1 x ( a) = 1 ± √ . 3 (d) Let x (t) be the number of kilograms of Ct in the reactor at time t. The rate of . loading is 1 kg per year. Hence x (t) satisfies x = −kx + 1, where k is the decay rate of Ct. Since the half life of Ct is 2 years, e−k·2 = 1/2, so that k = ln(2)/2. Therefore we have . ln 2 x + 1. x=− 2 The initial condition is x (0) = 0. (e) The differential equation is linear. Since we have 3 y� + y = x, x an integrating factor is given by � �� 3 exp dx = exp(3 ln x ) = x3 . x Multiply the above equation by x3 and integrate: ( x3 y)� = x3 y� + 3x2 y = x4 ⇒ x3 y = 1 5 x + c. 5 Since y(1) = 1, we have c = 4/5 and y= 3. 1 2 4 −3 x + x . 5 5 (a) Express all compex numbers in polar form: ie2it eiπ/2 e2it 1 1 = √ = √ ei(2t+π/2−π/4) = √ ei(2t−π/4) iπ/4 1+i 2 2 2e The real part is Re � ie2it 1+i � � 1 π� = √ cos 2t + . 4 2 (b) The trajectory is an outgoing, clockwise spiral that passes through 1. (c) The polar form of 8i is 8ei�/2 . Its three cubic roots are � � � + 2i sin = 3 + i, 6 6 � 5� 5� = 2 cos + 2i sin = − 3 + i, 6 6 3i�/2 = 2e = −2i. 2ei�/6 = 2 cos 2ei(�/6+2�/3) 2ei(�/6+4�/3) 4. (a) Let xp (t) = at2 + bt + c. Plug it into the left hand side of the equation ẍ + 2ẋ + 2x = (2a) + 2(2at + b) + 2(at2 + bt + c) = 2at2 + (4a + 2b)t + (2a + 2b + 2c) and compare coefficients 2a = 1, 4a + 2b = 0, 2a + 2b + 2c = 1. 1 2 t − t + 1. 2 (b) The characteristic polynomial is p(s) = s 2 + 2s + 2. Using the ERF and linearity, The solution is a = 1/2, b = −1, c = 1. Therefore x p (t) = xp (t) = e−2t 1 e−2t 1 + = + p(−2) p(0) 2 2 (c) Consider the complex equation z̈ + 2ż + 2z = eit . For any solution zp , its imaginary part xp = Im zp satisfies the real equation ẍ + 2ẋ + 2x = sin t. The ERF provides a particular solution of the complex equation zp (t) = eit eit eit 1 = =� = � ei(t−�) i� p(i) 1 + 2i 5e 5 where λ is the polar angle of 1 + 2i. Take the imaginary part of z p 1 xp (t) = Im zp (t) = � sin(t − λ) 5 � This is a sinusoidal solution of the real equation. Its amplitude is 1/ 5. (d) If x(t) = t3 is a solution, then q(t) = ẍ + 2ẋ + 2x = 6t + 6t 2 + t3 . (e) The general solution is x(t) = t3 + xh (t), where xh (t) is a solution of the associated homogeneous equation. Since the characteristic polynomial s 2 + 2s + 2 has roots −1 ± i, x(t) = t3 + xh (t) = t3 + c1 e−t cos t + c2 e−t sin t. 5. (a) See the formula sheet for the definition of sq(t). The graph of f (t) is a square wave of period 2�. It has a horizontal line segment of height 1 in the range −�/2 < t < �/2 and a horizontal line segment of height −1 in the range �/2 < t < 3�/2. (b) Replace t by t + �/2 in the definition of sq(t) � � � � � � � � �� 4 �� 1 3� 1 5� f (t) = sq t + = sin t + + sin 3t + + sin 5t + + ... 2 � 2 3 2 5 2 � � 4 1 1 = cos t − cos 3t + cos 5t + . . . � 3 5 (c) First consider the complex equation z̈ + z = eint for a positive integer n. The characteristic polynomial is p(s) = s 2 + 1. One of the ERFs provides a particular solution of the complex equation zp (t) = zp (t) = eint eint = , n ∞= 1 p(in) 1 − n2 teit teint = , n=1 p� (i) 2i The imaginary parts of these functions � int � e sin nt = , up (t) = Im 2 1−n 1 − n2 � it � te 1 up (t) = Im = − t cos t , 2i 2 n ∞= 1 n=1 satisfy the imaginary part of the above complex equation, namely ü + u = sin nt. By linearity, a solution of ẍ + x = sq(t) is given by � � 4 1 1 sin 3t 1 sin 5t xp (t) = − t cos t + · + · + ... . � 2 3 1 − 32 5 1 − 52 6. (a) For For For For For (b) t < 0, the graph is flat on t-axis. 0 < t < 1, the graph is flat at 1 unit above t-axis. 1 < t < 3, the graph is flat at 1 unit below t-axis. 3 < t < 4, the graph is flat at 1 unit above t-axis. t > 4, the graph is flat on t-axis. v(t) = = [u(t) − u(t − 1)] − [u(t − 1) − u(t − 3)] + [u(t − 3) − u(t − 4)] u(t) − 2u(t − 1) + 2u(t − 3) − u(t − 4) (c) The graph coincides with t-axis for all t, except for two upward spikes at t = 0, 3 and two downward spikes at t = 1, 4. (d) v̇(t) = �(t) − 2�(t − 1) + 2�(t − 3) − �(t − 4) (e) By the fundamental solution theorem (a.k.a. Green’s formula), x(t) = (q ≡ w)(t) = � t 0 q(t − π )w(π ) dπ = � b(t) w(π ) dπ. a(t) Now q(t − π ) = 1 only for 0 < t − π < 1, or t − 1 < π < t, and it is zero elsewhere. Therefore the upper limit b(t) equals t. The lower limit a(t) is t − 1 if t − 1 > 0, or 0 if t − 1 < 0. In other words, a(t) = (t − 1)u(t − 1). 1 1 = 2 . p(s) 2s + 8s + 16 (b) The unit impulse response w(t) is the inverse Laplace transform of W (s). In other words, 7. (a) The transfer function is W (s) = L(w(t)) = ∗ L(e2t w(t)) = 1 1 = + 8s + 16 2[(s + 2)2 + 4] 1 1 = L(sin 2t) 2 2(s + 4) 4 2s2 1 1 sin 2t, and w(t) = e−2t sin 2t. 4 4 (c) Take the Laplace transform of Therefore e2t w(t) = p(D)x = 2ẍ(t) + 8ẋ(t) + 16x(t) = sin t with the initial conditions x(0+) = 1, ẋ(0+) = 2. This yields 1 2[s2 X(s) − s − 2] + 8[sX(s) − 1] + 16X(s) = 2 s +1 � � 1 1 ∗ X(s) = 2 + 2s + 12 2s + 8s + 16 s2 + 1 8. (a) The characteristic polynomial of A is � � 2−� 12 det(A − �I) = det = (2 − �)2 − 36 = (� − 8)(� + 4). 3 2−� Therefore the eigenvalues are � = 8, −4. � � � � −6 12 2 (b) For � = 8, solve (A − 8I)v = 0. Since A − 8I = , a solution is v = . 3 −6 1 � � � � 6 12 2 For � = −4, solve (A + 4I)v = 0. Since A + 4I = , a solution is v = . 3 6 −1 (c) The following is a fundamental matrix for u̇ = Bu � t � e −e2t F (t) = et e2t Then etB can be computed as F (t)F (0)−1 . � � � � 1 1 −1 1 1 −1 F (0) = , F (0) = 1 1 2 −1 1 � t � � � � � 1 1 et + e2t et − e2t e −e2t 1 1 tB −1 · e = F (t)F (0) = = et e2t 2 −1 1 2 e t − e 2t e t + e 2t (d) The general solution of u̇ = Bu is � t � � � � � c1 e −e2t u(t) = c1 + c2 = F (t) et e2t c2 The given initial condition implies � � � � 2 c1 = F (0) 1 c2 � � � � � �� � � � 1 2 3/2 c1 2 1 1 −1 = ∗ = F (0) = 1 −1/2 c2 1 2 −1 1 � � 1 3et + e2t Therefore the solution of the initial value problem is u(t) = . 2 3et − e2t 9. (a) The phase portrait has the following features: • • • • All trajectories start at (0, 0) and run off to infinity. There are straight line trajectories along the lines y = ±x. All other trajectories are tangent to y = x at (0, 0). No two trajectories cross each other. (b) Tr A = a + 1, det A = a + 4, � = (Tr A)2 − 4(det A) = (a − 5)(a + 3) (i) det A < 0 √ a < −4 (ii) not for any a (iii) � > 0, Tr A < 0 and det A > 0 √ −4 < a < −3 (iv) � < 0 and Tr A < 0 √ −3 < a < −1; counterclockwise (v) � < 0 and Tr A > 0 √ −1 < a < 5 (vi) � = 0 and Tr A > 0 √ a = 5 10. (a) The equilibria are the solutions of ẋ = x2 − y 2 = 0, ẏ = x2 + y 2 − 8 = 0. This implies (x2 , y 2 ) = (4, 4), so that (x, y) = (2, 2), (2, −2), (−2, 2), (−2, −2). � � � � −4 4 2x −2y (b) The Jacobian is J(x, y) = . In particular, J(−2, −2) = . 2x 2y −4 −4 (c) The linearization of the nonlinear system at (−2, −2) is the linear system u̇ = J(−2, −2)u. A computation shows that the eigenvalues of J(−2, −2) are −4 ± 4i. The first component of u(t) is of the form c1 e−4t cos 4t + c2 e−4t sin 4t = Ae−4t cos(4t − λ). This means x(t) � −2 + Ae−4t cos(4t − λ) near (−2, −2). (d) Let f (x) = 2x − 3x2 + x3 . The phase line in problem 2(a) shows that ẋ = f (x) has a stable equilibrium at x = 1. The linearization of the nonlinear equation at x = 1 is the linear equation u̇ = f � (1)u = −u. Its solutions are u(t) = Ae−t . This means x(t) � 1 + Ae−t near x = 1.