 # Exercise 0.2.1: Check that the y given is really a solution to the ...

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Exercise 0.2.1: Check that the y given is really a solution to the equation. Next, take the second order differential equation = - 2 y, dx2 for some constant k > 0. The general solution for this equation is y(x) = C1cos(kx) + C2 sin(kx). Note that because we have a second order differential equation, we have two constants in our general solution. Exercise 0.2.2: Check that the y given is really a solution to the equation. And finally, take the second order differential equation dx² dRy = K2 y, for some constant k > 0. The general solution for this equation is y(x) = Ciekx + Cze or y(x) = D1 cosh(kx) + D2 sinh(kx). For those that do not know, cosh and sinh are defined by ex +e = , sinh X = 2 These functions are sometimes easier to work with than exponentials. They have some nice familiar properties such as cosh 0 = 1, sinh 0 = 0, and of cosh. X = sinh X (no that is not a typo) and sinh X = cosh X. Exercise 0.2.3: Check that both forms of the y given are really solutions to the equation. An interesting note about cosh: The graph of cosh is the exact shape a hanging chain will make. This shape is called a catenary. Contrary to popular belief this is not a parabola. If you invert the graph of cosh it is also the ideal arch for supporting its own weight. For example, the gateway arch in Saint Louis is an inverted graph of cosh-if it were just a parabola it might fall down. The formula used in the design is inscribed inside the arch: y = - 127.7 ft . cosh(x/127.7 ft) + 757.7 ft.0.2.4 Exercises Exercise 0.2.4: Show that X = eti is a solution to X" - 12x" + 48x' - 64x = 0. Exercise 0.2.5: Show that X = e is not a solution to X" - 12x" + 48x' - - 64x = 0. Exercise 0.2.6: Is y = sint a solution to (%) 2 = 1-y2?Justify. - Exercise 0.2.7: Lety" + 2y' - 8y = 0. Now try a solution of the form y = e"x for some (unknown) constant r. Is this a solution for some r? If so, find all such r. Exercise 0.2.8: Verify that X = Ce-2i is a solution to X' = -2x. Find C to solve for the initial condition x(0) = 100. Exercise 0.2.9: Verify that X = Cie- + C2e21 is a solution to X" - X' - 2x = 0. Find C1 and C2 to solve for the initial conditions x(0) = 10 and x' (0) = 0. Exercise 0.2.10: Find a solution to (x')² 2 + x2 = 4 using your knowledge of derivatives of functions that you know from basic calculus. Exercise 0.2.11: Solve: a) - dA dt = - -10A, A(0) = 5 dH b) = 3H, H(0) = 1 dx change = 4y, y(0) = 0, y' (0) = 1 d) X = -9x, x(0) = 1, X' (0) = 0 d dx² dy² Exercise 0.2.12: Is there a solution to y' = y, such that y(0) = y(1) ?Exercise 1.2.1: Sketch slope field for y' = ex-y. How do the solutions behave as x grows? Can you guess a particular solution by looking at the slope field? Exercise 1.2.2: Sketch slope field for y' = x2 1 Exercise 1.2.3: Sketch slope field for y' = y². Exercise 1.2.4: Is it possible to solve the equation y' = cos for y(0) = 1? Justify. Exercise 1.2.5: Is it possible to solve the equation y' = y v[x| for y(0) = 0? Is the solution unique? Justify. Exercise 1.2.6: Match equations y' = 1 - X, y' = x-2y, - y' = x(1 - y) to slope fields. Justify. / / / / - \ / / / / / / / / / a) b) / c) / / Exercise 1.2.7 (challenging): Take y' = f(x,y), y(0) = 0, where f(x,y) > 1 for all x and y. If the solution exists for all X, can you say what happens to y(x) as x goes to positive infinity? Explain. Exercise 1.2.8 (challenging): Take (y - = 0, y(0) = 0. a) Find two distinct solutions. b) Explain why this does not violate Picard's theorem.Exercise 1.3.1: Solve y' = x/y. Exercise 1.3.2: Solve y' = x2y.Exercise 1.3.3: Solve dx dx (x2 1) for x(0) = - t, = 0. dt Exercise 1.3.4: Solve dx dx = X sin(t), for x(0) = 1. dt Exercise 1.3.5: Solve dy xy y 1. Hint: Factor the right hand = + x + + side. dx Exercise 1.3.6: Solve xy' = y + 2x2y, where y(1) = 1. dy y2 + 1 Exercise 1.3.7: Solve = for y(0) = 1. dx x2 + 1 Exercise 1.3.8: Find an implicit solution for dy - , for y(0) = 1. x2 + 1 dx y² + 1 Exercise 1.3.9: Find an explicit solution for y' = xe->, y(0) = 1. Exercise 1.3.10: Find an explicit solution for xy' y(1) = 1. Exercise 1.3.11: Find an explicit solution for y' = y(0) = 1. It is alright to leave a definite integral in your answer. Exercise 1.3.12: Suppose a cup of coffee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees at t = 20 minutes. Compute the ambient temperature.In the exercises, feel free to leave answer as a definite integral if a closed form solution cannot be found. If you can find a closed form solution, you should give that. Exercise 1.4.4: Solve y' + xy=x. Exercise 1.4.5: Solve y' + 6y = ex. .Exercise 1.4.1: Try it! Add a constant of integration to the integral in the integrating factor and show that the solution you get in the end is the same as what we got above. An advice: Do not try to remember the formula itself, that is way too hard. It is easier to remember the process and repeat it. Since we cannot always evaluate the integrals in closed form, it is useful to know how to write the solution in definite integral form. A definite integral is something that you can plug into a computer or a calculator. Suppose we are given y'+p(x)y=f(x), y(xo) = yo. Look at the solution and write the integrals as definite integrals. = plop(s)ds f(t) dt + yo ) . (1.4) XO You should be careful to properly use dummy variables here. If you now plug such a formula into a computer or a calculator, it will be happy to give you numerical answers.Exercise 1.4.3: Write the solution of the following problem as a definite integral, but try to simplify as far as you can. You will not be able to find the solution in closed form. V' + y = et?-x, y(0) = 10.Exercise 1.4.6: Solve y' + 3x2y = sin(x)e-x³, with y(0) = 1. Exercise 1.4.7: Solve y' + cos(x)y = cos(x). Exercise 1.4.8: Solve , To 1 + xy = 3, with y(0) = 0. Exercise 1.4.9: Suppose there are two lakes located on a stream. Clean water flows into the first lake, then the water from the first lake flows into the second lake, and then water from the second lake flows further downstream. The in and out flow from each lake is 500 liters per hour. The first lake contains 100 thousand liters of water and the second lake contains 200 thousand liters of water. A truck with 500 kg of toxic substance crashes into the first lake. Assume that the water is being continually mixed perfectly by the stream. a) Find the concentration of toxic substance as a function of time in both lakes. b) When will the concentration in the first lake be below 0.001 kg per liter? c) When will the concentration in the second lake be maximal? Exercise 1.4.10: Newton's law of cooling states that dx = -k(x - A) where X is the temperature, t is time, A is the ambient temperature, and k > 0 is a constant. Suppose that A = A0 cos(wt) for some constants A0 and W. That is, the ambient temperature oscillates (for example night and day temperatures). a) Find the general solution. b) In the long term, will the initial conditions make much of a difference? Why or why not? Exercise 1.4.11: Initially 5 grams of salt are dissolved in 20 liters of water. Brine with concentration of salt 2 grams of salt per liter is added at a rate of 3 liters a minute. The tank is mixed well and is drained at 3 liters a minute. How long does the process have to continue until there are 20 grams of salt in the tank? Exercise 1.4.12: Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 liter per minute. The water is mixed well and drained at 1 liter per minute. In 20 minutes there are 15 grams of salt in the tank. What is the concentration of salt in the incoming brine?Hint: Answers need not always be in closed form. Exercise 1.5.1: Solve y' + y(x2 - 1) + xy6 = 0, with y(1) = 1. Exercise 1.5.2: Solve 2yy' + 1 = V2 + X, with y(0) = 1. Exercise 1.5.3: Solve y' + xy = y4, with y(0) = 1. Exercise 1.5.4: Solve yy' + X = V x2 + y2.

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