2.6 #10: Determine if the equation is exact. If so, solve it. Math 315-01 Homework 3 Solutions (1 y sin x) dx + (cos x) dy = 0 Solution: Let P (x, y) = 1 y sin x and Q(x, y) = cos x. Note P = sin x = Q x. Hence, the equation is exact. Thus, its solution is given by F (x, y) = C, where P (x, y) = x (x, y) and Q(x, y) = (x, y). To find F (x, y), we start by integrating P (x, y) = x (x, y) with respect to x. This gives ( F (x, y) = P (x, y) dx = 1 y sin x) dx = x + y cos x + D(y) where D(y) is some function of y and does not depend on x. Taking the derivative with respect to y gives = ( ) x + y cos x + D(y) = cos x + D (y). Since we also know that we get that = Q = cos x cos(x) + D (y) = cos x. Hence, D (y) = 0. Therefore, we can set D(y) = 0 (otherwise D(y) = constant, which will be able to be incorporated into the other constant in the equation F (x, y) = C later). Thus, F (x, y) = x + y cos x. Since the solution is given by F (x, y) = C, the solution is 2.7 #3: Is the initial value problem x + y cos x = C. y = t tan 1 (y), y(0) = 2 guaranteed a unique solution by the hypotheses of Theorem 7.16? Solution: Note both f(t, y) = t tan 1 (y)
and f = t 1 + y 2 are continuous everywhere. (So in this case we don t need to define a particular rectangle R containing the initial condition, since we can just let R = R 2 = entire real plane.) Thus, a unique solution is guaranteed by Theorem 7.16. 2.7 #8 : Consider the initial value problem ty = 2y t, y(0) = 2. (i) Find the general solution. Sketch several members of the family of solutions. Solution: Note ty = 2y t y 2 t y = 1. Now we calculate the integrating factor u(t). This is given by u(t) = e 2 t dt = e 2 ln(t) = e ln(t 2) = 1 t 2. Multiplying the integrating factor on both sides of the differential equation y 2 t y = 1 gives which implies Integrating, we get 1 t 2 y 2 t 3 y = 1 t 2, ( 1 ) t t 2 y = 1 t 2. Thus, the general solution is given by 1 1 t 2 y = t 2 dt = 1 t + C. y(t) = t + Ct 2. Now we need to sketch several members of the family of solutions. Let us sketch solutions when C = 2, C = 1, C = 0, and C = 1. (This corresponds to plotting y = t+2t 2, y = t+t 2, y = t, and y = t t 2.)
We have used the free online graphing calculator Desmos (https://www.desmos.com/calculator) to plot these solutions, shown below. (ii) Show there is no solution satisfying the given initial condition. Explain why this lack of solution does not contradict the existence theorem. Solution: Note from the general solution y(t) = t + Ct 2 we get y(0) = 0 + C(0) 2 = 0. Thus, every solution must go through (0, 0). Hence, there cannot be a solution satisfying y(0) = 2. To show why this lack of solution does not contradict the existence theorem, note that f(t, y) = y = 2 t y 1 is not continuous at t = 0, and so the hypotheses for the existence theorem are not satisfied. Hence, we can not conclude anything from the existence theorem for the case when t = 0. 2.8 #7: Use the nonlinear equation x = x 2 t with initial condition at t 0 = 0. The target is (5, 0). Solve the equation numerically and then experiment with initial conditions at the given value of t 0 until the solution of x = x 2 t with x(t 0, x 0 ) hits the given target. Solution: We use Maple, which you can access online by logging into http://my.math.wsu.edu and clicking on Go to... and then Software. There are several ways to do this problem, but we have entered the code shown below:
Note what we have done here is, first, we have plotted the differential equation y = y 2 t with the initial condition y(5) = 0. So this is the solution which hits the target (5, 0). Now we just need to find what this solution is at t = 0. To do this, we need to numerically solve the initial value problem, which we have done as well (see code). Note we have entered the command eq := diff(y(t),t) = (y(t))ˆ2-t; which defines the differential equation, and we have called it eq. To solve this equation numerically with initial condition y(5) = 0, we enter sol := dsolve(eq, y(5)=0, y(t), numeric); Note dsolve(eq, y(5)=0, y(t), numeric); solves the initial value problem numerically, and then we have called it sol. Then to find the solution at t = 0, we enter sol(0) as the last command. Another way to do this problem would be to just plot the differential equation for a bunch of different initial conditions for y(0) until you find one where you can see from the graph that the solution curve goes approximately through (5, 0).
The code we entered into Maple returns the figure followed by the output Hence, we can conclude that the solution of x = x 2 t will hit the target (5, 0) given the initial condition x(0) 0.7290108.