(9/8/08)Math 10B. Lecture Examples.Section 11.4. Separation of variables†Example 1 Figure 1 shows the slope field of the differential equationdydx= yand Figure 2 shows the graphs of eight solutions. (a) Use the differential equation toexplain the pattern of the slope lines. (b) Find an equation for all solutions.x1−1 2−2y−11x1−1 2−2y−11FIGURE 1 FIGURE 2Answer: (a) One description and explanation: The lines in th e slope field ofdydx= y in Figure 1 have the sameslope along each horizontal line because the formula on the right does not involve x. • The lines are horizontalalong the x-axis where y = 0, have pos itive slopes above the x-axis where y > 0, and have negative slopes below thex-axis where y < 0, and they become steeper as y increases through positive values or decreases t hrough negativevalues.(b) The solutions are y = Cexwith arbitrary consta nts C.Example 2 Find the solution of the initial-value problemdydx= 2y cos x, y(0) = 4.Answer: y = 4e2 sin xExample 3 Check the result of Example 2.Answer: Set y = 4 e2 sin x. • y(0) = 4e2 sin(0)= 4 • The initial condition is satisfied. •dydx=ddx(4e2 sin x) = 8(cos x) e2 sin x= 2y cos x • The differential equation is satisfied.Example 4 Find the solutions of the differential equationdydx= −2xy2with the initial conditions (a) y(0) = 1 • y(0) = −14.Answer: (a) y =1x2+ 1(b) y =1x2− 4• (Figure A4a shows the slope field for the differential equation(6), and Figure A4b gives the graphs of the solutions.)†Lecture notes to accompany Section 11.4 of Calculus by Hughes-Hallett et al1Math 10B. Lecture Examples. (9/8/08) Section 11.4, p. 2x1 2 3−1−2−3y1x1 2 3−1−2−3y1Slope field ofdydx= −2xy2Solutions with y(0) = 1 and y(0) = −14Figure A4a Figure A4bExample 5 Solve the initial-value problem K0(x) =pxK(x), K(1) = 1Answer: K = (13x3/2+23)2Example 6 Find all nonzero solutions ofdQdx= −3Q1/4.Answer: Q = (C −94x)4/3Example 6 (a) A two-gram object is moving on an s-axis with distances measured in centimete rs.Its velocity in the positive direction is 1 centimeter p er second at t ime t = 0 (seconds)and the force on it at time t > 0 is 4tv2dynes in the positive s-direction if its velocityis v centimeters per second at that time. Give an initial-value problem satisfied byv = v(t ). (b) Give a formula for v for t ≥ 0. (c) What happens to the velocity ast → ∞? (The slope field and graph of th e solution are in Figure 3.)t0.5 1v12345FIGURE 3Answer: (a) Initial-value problem:dvdt= 2tv2, v(0) = 1 (b) v =11 − t2(c) v → ∞ as t → 1−Interactive ExamplesWork the following Interactive Examples on Shenk’s web page, http//www.math.ucsd.edu/˜ashenk/:‡Section 9.1: Examples 1–3, 5, 6, 8‡The chapter and section numbers on Shen k’s web site refer to his calculus manuscript an d not to the chapter s and sectionsof the textbook for the