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Berkeley MATH 1B - Math 1B - Discussion Exercises

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Math 1B: Discussion ExercisesGSI: Theo Johnson-Freydhttp://math.berkeley.edu/~theojf/09Summer1B/Find two or three classmates and a few feet of chalkboard. Be sure to discuss how to solve theexercises — how you get the solution is much more important than whether you get the solution. Ifas a group you agree that you all understand a certain type of exercise, move on to later problems.You are not expected to solve all the exercises: some are very hard.Many of the exercises are from Single Variable Calculus: Early Transcendentals for UC Berkeleyby James Stewart; these are marked with an §. Others are my own, are from the mathematicalfolklore, or are independently marked.Here’s a hint: drawing pictures — e.g. sketching graphs of functions — will always make theproblem easier.Linear Differential Equations of First OrderAn operator is a kind of function that takes functions as inputs and outputs functions. Forexample, theddx, which outputs the derivative of its input, is an operator; the multiplication by x isanother operator. For now, we will be only interested in differential operators, which are composedof the basic operations and differentiation: y 7→dydx+ xy for example. Today we will focus onfirst-order differential operators, which do not involve second- and higher derivatives. An operatorF is linear if it has the property that F[y1(x) + y2(x)] = F[y1(x)] + F[y2(x)] for any two functionsy1and y2. All the example operators we’ve listed so far are linear. A first-order linear differentialoperator is necessarily of the formF[y(x)] = a(x)dydx+ b(x) y(x)A linear differential equation is a differential equation of the form F[y(x)] = c(x), where F is a lineardifferential operator. By dividing by a(x), and writing P (x) = b(x)/a(x) and Q(x) = c(x)/a(x), itsuffices to consider linear differential equations of the formdydx+ P (x) y = Q(x)Let’s say that F is a linear operator, and that y1(x) and y2(x) are both solutions to the equationF[y(x)] = c(x). Since F is linear, we have:F[y1(x) − y2(x)] = F[y1(x)] − F[y2(x)] = c(x) − c(x) = 0Conversely, if y1(x) is a solution to the equation F[y(x)] = c(x) and y0(x) is a solution to theequation F[y(x)] = 0, then y1(x) + y0(x) is a solution to F[y(x)] = c(x). The linear differentialequation F[y(x)] = 0 is called homogeneous, and the solutions to it are the homogeneous solutions.We have seen that the difference of any two solutions to F[y(x)] = c(x) is a homogeneous solution,and conversely, given any particular solution yp(x) to F[y(x)] = c(x), all other solutions are of theform yp(x) + y0(x) where y0(x) ranges over the set of homogeneous solutions.To make this more down-to-earth, let’s consider a case you already understand. The operatorddxis a first-order linear differential operator. Its homogeneous differential equation isddx[y] = 0,and the solutions are y = C, where C is a constant. Given a function c(x), the solutions tothe inhomogeneous equationddx[y(x)] = c(x) are given byRc(x) dx; as you know, this set can bedescribed by finding a single antiderivative of c(x) and then adding “+C”.1In any case, we now describe how to find solutions to all first-order linear differential equations.Recall that it suffices to consider equations of the form y0+ P y = Q. The trick is to rememberthat the product rule turns a product into a sum: (I(x) y(x))0= I(x) y0(x) + I0(x) y(x). So theidea is to find a function I(x) so that I(x) y0(x) + I0(x) y(x) = I(x) (y0(x) + P (x) y(x)). I.e. wewant a function I(x) with I0(x) = P (x)I(x). But this is a separable differential equation in I(x):its solutions are of the form I(x) = eRP (x)dx. Thus, we multiply both sides of y0+ P y = Q byI = eRP, and then we get (Iy)0= eRIQ, or Iy =RIQ, ory(x) = e−RP (x)dxZeRP (x)dxQ(x)dx1. When Q(x) = 0, the linear differential equation y0(x) + P (x)y(x) = 0 is also a separabledifferential equation. Solve it as a separable equation, and also solve it using the aboveformula for linear differential equations.2. To solve a linear differential equation, we had to choose an integrating factor I(x) = eRP (x) dx.But “RP (x) dx” is defined only up to a constant. If you add a +C toRP (x) dx, what happensto the solution to the differential equation?3. To solve a linear differential equation, we had to integrateRI(x) Q(x) dx. What happensto the constant of integration here? How does this compare to your answer to problem 1.?Interpret the general solution to y0+ P y = Q as a “particular solution” plus a “homogeneoussolution”.4. § Find the general solutions to the following differential equations:(a) xy0+ y =√x (b) y0+ y = sin(ex) (c) sin xdydx+ (cos x)y = sin(x2)(d) xdydx− 4y = x4ex(e) (1 + t)dudt+ u = 1 + t (f) t ln tdrdt+ r = tet5. § Solve the initial value problem:(a) 2xy0+ y = 6x, x > 0, y(4) = 20 (b) xy0= y + x2sin x, y(π) = 06. § A tank contains 100 L of water. A solution with salt concentration 0.4 kg/L is added at arate of 5 L/min. The solution is kept mixed and is drained from the tank at a rate of 3 L/min.If y(t) is the amount of salt (in kilograms) after t minutes, show that y satisfies the differentialequationdydt= 2 kg/min −(3 L/min) y(100 L) + (2 L/min) tSolve this equation and find the concentration after 20 minutes.7. § Use the Bernoulli substitution u = y1−nto turn the differential equation y0+ Py = Qyninto a linear differential equation in u.8. § Use the method in the previous problem to solve the following differential equations:(a) xy0+ y = −xy2(b) y0+2xy


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