Math 1B Discussion Exercises GSI Theo Johnson Freyd http math berkeley edu theojf 09Summer1B Find two or three classmates and a few feet of chalkboard Be sure to discuss how to solve the exercises how you get the solution is much more important than whether you get the solution If as 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 Berkeley by James Stewart these are marked with an Others are my own are from the mathematical folklore or are independently marked Here s a hint drawing pictures e g sketching graphs of functions will always make the problem easier Linear Differential Equations of First Order An operator is a kind of function that takes functions as inputs and outputs functions For d example the dx which outputs the derivative of its input is an operator the multiplication by x is another operator For now we will be only interested in differential operators which are composed dy xy for example Today we will focus on of the basic operations and differentiation y 7 dx first order differential operators which do not involve second and higher derivatives An operator F is linear if it has the property that F y1 x y2 x F y1 x F y2 x for any two functions y1 and y2 All the example operators we ve listed so far are linear A first order linear differential operator is necessarily of the form F y x a x dy b x y x dx A linear differential equation is a differential equation of the form F y x c x where F is a linear differential operator By dividing by a x and writing P x b x a x and Q x c x a x it suffices to consider linear differential equations of the form dy P x y Q x dx Let s say that F is a linear operator and that y1 x and y2 x are both solutions to the equation F 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 0 Conversely if y1 x is a solution to the equation F y x c x and y0 x is a solution to the equation F y x 0 then y1 x y0 x is a solution to F y x c x The linear differential equation 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 the form 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 operator d d is a first order linear differential operator Its homogeneous differential equation is dx y 0 dx and the solutions are y C where C is a constant Given a function c x the solutions to R d the inhomogeneous equation dx y x c x are given by c x dx as you know this set can be described by finding a single antiderivative of c x and then adding C 1 In 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 y 0 P y Q The trick is to remember that the product rule turns a product into a sum I x y x 0 I x y 0 x I 0 x y x So the idea is to find a function I x so that I x y 0 x I 0 x y x I x y 0 x P x y x I e we want a function I x with I 0 x P x I x But this is a separable differential equation in I x R P x dx Thus we multiply both sides of y 0 P y Q by its solutions are of the form I x e R R R I e P and then we get Iy 0 e I Q or Iy IQ or y x e R P x dx Z e R P x dx Q x dx 1 When Q x 0 the linear differential equation y 0 x P x y x 0 is also a separable differential equation Solve it as a separable equation and also solve it using the above formula for linear differential equations R P x dx 2 To solve R a linear differential equation we had to choose an integratingR factor I x e But P x dx is defined only up to a constant If you add a C to P x dx what happens to the solution to the differential equation R 3 To solve a linear differential equation we had to integrate I x Q x dx What happens to the constant of integration here How does this compare to your answer to problem 1 Interpret the general solution to y 0 P y Q as a particular solution plus a homogeneous solution 4 Find the general solutions to the following differential equations a xy 0 y d x x dy 4y x4 ex dx b y 0 y sin ex e 1 t dy cos x y sin x2 dx dr f t ln t r tet dt c du u 1 t dt sin x 5 Solve the initial value problem a 2xy 0 y 6x x 0 y 4 20 b xy 0 y x2 sin x y 0 6 A tank contains 100 L of water A solution with salt concentration 0 4 kg L is added at a rate 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 differential equation dy 3 L min y 2 kg min dt 100 L 2 L min t Solve this equation and find the concentration after 20 minutes 7 Use the Bernoulli substitution u y 1 n to turn the differential equation y 0 P y Qy n into a linear differential equation in u 8 Use the method in the previous problem to solve the following differential equations a xy 0 y xy 2 b y 0 2 2 y3 y 2 x x
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