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BC Exam Differential equations are among the most powerful tools we have for analyzing the world mathematically. They are used to formulate the fundamental laws of nature (from Newton’s Laws to Maxwell’s equations and the laws of quantum mechanics) and to model the most diverse physical phenomena. This chapter provides an introduction to some elementary techniques and applications of this important subject. A differential equation is an equation that involves an unknown function y and its first or higher derivatives. A solution is a function y = f (x) satisfying the given equation. As we have seen in previous chapters, solutions usually depend on one or more arbitrary constants (denoted A, B, and C in the following examples): The first step in any study of differential equations is to classify the equations according to various properties. The most important attributes of a differential equation are its order and whether or not it is linear. The order of a differential equation is the order of the highest derivative appearing in the equation. The general solution of an equation of order n usually involves n arbitrary constants. For example, y " y 0 has order 2 and its general solution has two arbitrary constants A and B. Just like the order of a Taylor Polynomial! A differential equation is called linear if it can be written in the form an x y n an 1 x y n 1 a1 x y ' a0 x y b x The coefficients aj(x) and b(x) can be arbitrary functions of x, but a linear equation cannot have terms such as y3, yy ' , or siny. Separation of Variables Separable Equations have the form dy f x g y . dx dy sin x y is separable. First-Order dx dy x y is not separable because x y is not the product of f x & g y . dx dy Show that y x 0 is separable but not linear. Then find the dx Here we go ;-) general solution and plot the family of solutions. y 2 x2 ydy xdx 2 2 C y 2 x2 C y x2 C yy ' NL 2 2 y x 2 1 2 a b A differential equation is called linear if it can be written in the form an x y n an 1 x y n 1 a1 x y ' a0 x y b x This is a conic... A family of Hyperbolas Solve for y dy Solutions y x C to y x 0. dx 2 A Degenerate Hyperbola Let c 0 y x2 C Although it is useful to find general solutions, in applications we are usually interested in the solution that describes a particular physical situation. The general solution to a first-order equation generally depends on one arbitrary constant, so we can pick out a particular solution y(x) by specifying the value y(x0) for some fixed x0. This specification is called an initial condition. A differential equation together with an initial condition is called an initial value problem. Family of Solutions to a Particular Differential Equation Initial Value Problem Solve the initial value problem y ' ty, y 0 3 dy dy ty tdt dt y y 3e t2 2 dy t2 tdt ln y C y 2 y e t2 C 2 y Ce y 3e t2 2 t2 C 2 y e e y Ce t2 2 Family of Solutions , y 0 3 3 Ce0 C 3 t 2 /2 Laws of Exponents y ec f t y ec f t or ec f t Since C is arbitrary, eC represents an arbitrary positive number, and ±eC is an arbitrary nonzero number. We replace ±eC by C and write the general solution as In the context of differential equations, the term “modeling” means finding a differential equation that describes a given physical situation. As an example, consider water leaking through a hole at the bottom of a tank. The problem is to find the water level y (t) at time t. We solve it by showing that y (t) satisfies a differential equation. Differential Equations UP The key observation is that the water lost during the interval from t to t + Δt can be computed in two ways. Let v y velocity of water flowing through the hole when the tank is filled to height y B area of the hole Not constant, but close A y area of horizontal cross-section of the tank at height y First, we observe that the water exiting through the hole during a time interval Δt forms a cylinder of base B and height υ(y)Δt. d rt V BV y t Water leaks out of a tank through a hole of area B at the bottom. In the context of differential equations, the term “modeling” means finding a differential equation that describes a given physical situation. As an example, consider water leaking through a hole at the bottom of a tank. The problem is to find the water level y (t) at time t. We solve it by showing that y (t) satisfies a differential equation The key observation is that the water lost during the interval from t to t + Δt can be computed in two ways. Let v y velocity of water flowing through the hole when the tank is filled to height y B area of the hole A y area of horizontal cross-section of the tank at height y Second, we note that the water level drops by an amount Δy during the interval Δt. water lost between t and t t A y y A y y Bv y t y Bv y dy Bv y t A y dt A y Now we can set up our differential equation! Water leaks out of a tank through a hole of area B at the bottom. dy Bv y dt A y Velocity of the water passing through the hole is... To use our differential equation, we need to know the velocity of the water leaving the hole. This is given by Torricelli’s Law with (g = 9.8 m/s2): Given v y 2 gy 4.43 y m/s Now we simply plug in our known values and solve the differential equation using separation of variables. dy Bv y dt A y v y 2 gy 4.43 y m/s Application of Torricelli’s Law A cylindrical tank of height 4 m and radius 1 m is filled with water. Water drains through a square hole of side 2 cm in the bottom. Determine the water level y(t) at time t (seconds). How long does it take for the tank to go from full to empty? Solution We can use units of centimeters 1 m 100 cm . is constant... A y r 2 10, 000 cm 2 2 2 g 9.8 m/s 980 cm/s B 4 cm 2 v y 2 980 y 44.3 y cm/s 4 44.3 y dy Bv y 0.0056 y dt A y 10, 000 dy dy 0.0056 y 0.0056dt dt y y 0 dy 0.0056dt 2 y1/ 2 0.0056t C y y 0.0028t C y C 0.0028t 2 Step 2. Use the initial condition (the tank was full). y 0 400 cm 400 C 2 C 20 400 y t 20 0.0028t 2 200 400 Which sign is correct? y t 20 0.0028t 2 200 10, 000 Separation of Variables This is actually an IVP. 10, 000 te 7143 s Why can't we just find dy dt ? CONCEPTUAL INSIGHT The previous example highlights the need to analyze solutions to differential equations rather than relying on algebra alone. The algebra seemed to suggest that C = ±20, but further analysis showed that C = −20 does not yield a solution for t ≥ 0. Note also that the function y (t) = (20 − 0.0028t)2 is a solution only for t ≤ te—that is, until the tank is empty. This function cannot satisfy our original differential equation for t > te because its derivative is positive for t > te, and solutions of the given differential must have nonpositive derivatives. Only Separation of Variables is tested on the BC Exam.