Math 246 Notes

Math 246 Notes May 7, 2021 This note may contain some typos. Feel free to message me if you see any typos. Contents 1 Week 1 3 1.1 First-Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.1.1 Explicit First-Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.1.2 First-Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2 Week 2 6 2.1 Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2.2 General Theory for dy dt = f(t, y) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3 Phase-Line Portraits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3 Week 3 10 3.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.1.1 Tank Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.1.2 Population Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.1.3 Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.1.4 Loans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.2 Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.2.1 Explicit Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.3 Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 4 Week 4 13 4.1 Exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.2 Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1 4.3 Higher-Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.4 Linear Differential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 5 Week 5 16 5.1 Wronskian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 5.2 Natural Fundamental Set of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 5.3 Sample Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 5.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 6 Week 6 18 6.1 Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 6.2 Homogeneous Linear Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . . 19 6.3 Non-homogeneous Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 6.4 Finding YP for Linear Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . 20 6.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 7 Week 7 21 7.1 Undetermined Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 7.2 Green Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 7.3 Applications: Mechanical Viberations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 7.3.1 Unforced, Undamped Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 7.3.2 Unforced, Damped Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 7.3.3 Forced, Undamped Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 7.3.4 Forced, Damped Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 7.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 8 Week 8 25 8.1 Variable Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 8.2 Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 8.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 9 Week 9 27 9.1 Evaluating Green Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 9.2 First-Order Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 9.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 10 Week 10 30 10.1 Explicit Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 10.2 Tank Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2 10.3 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 10.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 11 Week 11 31 11.1 Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 11.2 Linear Homogeneous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 11.3 Non-homogeneous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 11.4 Homogeneous Linear Systems with Constant Coefficients . . . . . . . . . . . . . . . . . . . . . 34 11.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 12 Week 12 36 12.1 Eigen Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 12.2 Method of Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 12.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 13 Week 13 38 13.1 Phase-Plane Portraits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 13.2 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 14 Week 14 40 14.1 Non-linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 14.1.1 Stationary and Semi-Stationary Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 40 14.2 Phase-Plane Portrait for Non-linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 14.3 Orbit Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 14.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 A Appendix: Complex Numbers 42 1 Week 1 Differential Equation: An equation involving derivatives of one or more functions. Example 1.1. (a)  dx dt 2 + x sin t = cos x. (b) ∂y ∂t ∂y ∂s + y ∂z ∂t = sin(st). (c) y 00 + ty0 + y = cost. Ordinary Differential Equation (ODE): A differential equation that involves no partial derivatives. Those that involve partial derivatives are called Partial Differential Equations or PDE’s. Order of a differential equation is the largest derivative that appears in the differential equation. 3 Example 1.2. Determine if each equation in the previous example is an ODE or a PDE. Determine their orders. Linear Differential Equation: A differential equation that can be written in a way that each of its terms either does not contain any unknown variables or is the product of some derivative of only one of the unknown variables and a function of the independent variables. Example 1.3. Determine which one of the equations in Example 1.1 are linear and why. Example 1.4. Write down the general form of an n-th order linear ODE with one unknown variable y that is a function of one independent variable t. 1.1 First-Order Equations Recall that first-order equations are the ones that involve only the first derivatives of unknown functions. We focus on solving first-order ODE’s of the form dy dt = f(t, y). Depending on the type of function f(t, y) we will employ different strategies. 1.1.1 Explicit First-Order Equations Explicit first-order ODE’s are the ones of the form dy dt = f(t). Example 1.5. Solve dy dt = sin t. Example 1.6. Solve dy dt = 1 t 2 − t , and y(2) = 1. Differential equations such as the ones in Example 1.6 that involve given “initial values” are called Initial Value Problems or IVP’s. Example 1.7. Solve the IVP dy dt = e t 2 , y(0) = 1. The interval of definition of a solution to an IVP is the largest interval over which the solution is defined. Example 1.8. Find the interval of definition of the solution to the IVP dy dt = t + 1 √3 t 2 − 1 , y(0) = 10. Example 1.9. Suppose y is the solution to the IVP given by dy dt = √5 t 3 − 4t, y(1) = −1. Determine where y(t) is increasing and where it is decreasing. 1.1.2 First-Order Linear Equations The general form of a first-order linear equation is p(t) dy dt + q(t)y = f(t), where p(t) is not the zero function. The functions p(t) and q(t) are called coefficients, and f(t) is called the forcing function. If the forcing is 4 zero we say the equation is homogeneous, otherwise we say it is non-homogeneous. Dividing both sides by the leading coefficient we can re-write any first-order linear equation of the following form, called the normal form: dy dt + q(t)y = f(t) Example 1.10. Solve the equation t dy dt + y = 1. Example 1.11. Solve the IVP y 0 + y = t, y(0) = 1. To solve the equations y 0 + p(t)y = f(t), we need to write the left hand side as a derivative of a single function. To do that we multiply both sides by a factor called an integrating factor. The equation then would look like µy0 + µp(t)y = µf(t). In order for the left hand side to follow the form of the product rule we need to have µ 0 = µp(t). This can be achieved by taking µ = e R p(t)dt . Remember that we only need one integrating factor! Example 1.12. Solve (t 2 + 1)y 0 + ty = 0. Theorem 1.1. Suppose functions p(t) and f(t) are continuous over the interval (a, b). Let t0 be a point in (a, b) and y0 be a real number. Then, the IVP dy dt + p(t)y = f(t), y(t0) = y0 has a unique solution defined over (a, b). Example 1.13. Find the interval of definition of the solution to the IVP (1 − t)y 0 + e ty = sin t, y(0) = 1. Example 1.14. Suppose y1 and y2 are solution of a first-order linear homogeneous equation. Show that for every constant c, the function y1 + cy2 is also a solution of the same equation. For more examples, check the textbook. 1.2 Summary • The solution to the IVP dy dt = f(t), y(t0) = y0 is given by y(t) = y0 + Z t t0 f(x) dx. • To find the interval of definition of the solution to the IVP dy dt = f(t), y(t0) = y0 we find the largest open interval containing t0 where f(t) is defined and continuous. • To solve a first-order linear equation: – Write down the equation in the normal form y 0 + p(t)y = f(t). – Find an integrating factor µ = e R p(t) . – The equation can then be written as d dt(µy) = µf(t). – Solve the equation by integrating both sides. 5 – If there is an initial condition use µy = C + Z t t0 µf(s)ds, and find C by substituting t = t0 and y = y0. • To find the interval of definition of a first-order linear IVP: – Write the equation in the normal form y 0 + p(t)y = f(t). – Find the largest open interval containing the initial value t0 for which p(t) and f(t) are defined and continuous. 2 Week 2 2.1 Separable Equations A first-order equation is called separable if it can be written in the form dy dt = f(t)g(y). The name “separable” refers to the fact that we can separate the variables and write the differential equation in the form dy g(y) = f(t) dt. The solution can then be obtained by simply integrating both sides. Example 2.1. Solve the equation dy dt = y 2 − 1. Equations of the form dy dt = g(y) are called autonomous since the derivative of y is “self-governed” by the function y and is independent of t. Notice that because we divided both sides by y 2 − 1, there are some solutions that are missing. In fact y = 1 and y = −1 are two constant functions that satisfy the above equation. This brings us to the following definition. Given a separable equation dy dt = f(t)g(y), any constant y0 that satisfies g(y0) = 0 gives a constant solution y = y0 to this differential equation. This solution is called a stationary solution. Other names for stationary solutions are fixed points or equilibrium points. Example 2.2. Find all solutions of the equation including its stationary solutions. dy dt = 3t + ty y + t 2y . Example 2.3. Given a constant y0, solve the initial value problem and find its interval of definition: dy dt = y 2 , y(0) = y0. Example 2.4. Solve the initial value problem: dy dt = 3y 2/3 , y(0) = 0. 6 Surprisingly these are many solution of this IVP that we can not obtain using the methods discussed before! Theorem 2.1 (Existence-Uniqueness Theorem for Separable Equations). Suppose f(t), g(y) are continuous functions over (tL, tR), and (yL, yR), respectively, and let g be differentiable at its zeros in (yL, yR). Then, for every t0 in (tL, tR) and every y0 in (yL, yR), there is a unique solution to the IVP dy dt = f(t)g(y), y(t0) = y0 for as long as t stays in the interval (tL, tR) and y stays in the interval (yL, yR). Example 2.5. For what values of y0 can we guarantee that the following IVP has a unique solution near (0, y0)? dy dt = t(y − 1)1/3 , y(0) = y0. 2.2 General Theory for dy dt = f(t, y) We are interested in initial value problems of the form dy dt = f(t, y), y(t0) = y0 that have a unique solution. First, recall that partial derivative of a multivariable function is its derivative when all other variables are considered constant. Partial derivative of a function f with respect to a variable x is denoted by fx or ∂xf or ∂f ∂x. Example 2.6. Evaluate fx, fy, and fz, where f(x, y, z) = x + x ln y + z 2x. We say a point (t, y) is an interior point of a region S in the ty-plane if there is a circle around (t, y) that is fully contained in S. Theorem 2.2 (Existence-Uniqueness Theorem for First-Order IVP’s). Suppose f(t, y) is a function defined over a region S in the ty-plane such that • f is continuous over S, and • fy exists and is continuous over S. Then, for any initial time t0 and initial y0 for which (t0, y0) is in the interior of S, the initial value problem dy dt = f(t, y), y(t0) = y0 has a unique solution for as long at (t, y) remains in S. Example 2.7. Determine for which values of t0 and y0 the following initial value problem is guaranteed to have a unique solution: dy dt = ln|y| 1 + t 2 − y 2 , y(t0) = y0. 7 2.3 Phase-Line Portraits Example 2.8. Draw a phase line portrait for the equation dy dt = 4y 2 − y 4 . A stationary solution y = y0 is called stable if all solutions near y0 move towards y0. It is called unstable if all solution near y0 move away from y0. It is said to be semistable if some solutions near y0 move towards it while others move away from y0. Example 2.9. Consider the differential equation dy dt = y − y 3 (y + 2)2 . 1. Sketch the phase-line portrait of this equation. 2. Classify all of the stationary solutions. 3. Evaluate limt→∞ y(t), and lim t→−∞ y(t), where y is the solution satisfying y(0) = 5. Example 2.10. Consider the differential equation y dy dt + t = 0. Draw the direction field of this equation. Example 2.11. Draw the direction field of the equation dy dt = sin2 t + ty. 8 Existence-Uniqueness Theorems tell us that solution curves do not intersect. To sketch solutions that are explicit we use the command fplot in MATLAB. To produce counter plots H(X, Y ) = c in MATLAB we use meshgrid or contour. We produce direction fields using quiver or meshgrid. Check the online textbook for more examples and MATLAB examples. 2.4 Summary • To solve a separable equation of the form dy dt = f(t)g(y): – Find all stationary solutions by solving g(y) = 0. – For non-stationary solutions: separate the variables and rewrite the equation as dy g(y) = f(t)dt. Then integrate both sides. • For a separable IVP to have a unique solution we need f(t) and g(y) to be continuous, and g(y) to be differentiable at all of its zeros. • To check if the initial value problem dy dt = f(t, y), y(t0) = y0 has a unique solution we need to make sure: – f(t, y) is continuous. – fy exists and is continuous. – (t0, y0) is an interior point. • To draw the phase-line portrait of an autonomous equation dy dt = g(y): – Find all stationary solutions by solving g(y) = 0. – Find all points of discontinuity of g(y) and the points where g(y) is undefined. – Plot these points on a number line. – Determine the sign of g(y) in each interval. Those values determine if y is increasing or decreasing. Indicate that using arrows. • A stationary solution y = y0 is called stable if all solutions near y0 move towards y0. It is called unstable if all solution near y0 move away from y0. It is said to be semistable if some solutions near y0 move towards it while others move away from y0. 9 3 Week 3 3.1 Applications The general strategy for many of these application problems is that Rate = Rate In - Rate Out. 3.1.1 Tank Problems Example 3.1. A tank contains 200 liters of salt solution at concentration of 7 g/L. A solution at concentration of 0.5 g/L flows into the tank at the rate of 3 L/min. At the same time the well-mixed solution flows out of the tank at the rate of 3 L/min. How long will it be before the solution is at concentration 1 g/L? Example 3.2. Use the same settings as in the previous problem except assume the rate of flow out of the tank is 2 L/min. Write down a differential equation that governs the amount of salt in the tank. Example 3.3. A circular cylindrical tank with an open top has a circular base of radius 1 m, and a height of 2 m. The water pours into the tank at the rate of 7 L/min and drains at the rate of 5√ h L/min, where h is the height of water. Will the tank overflow? 3.1.2 Population Dynamics Populations are often modeled by differential equations. One model is the following dp dt = R(p)p − h(t). R(p) is the growth rate and h(t) is due to predators. Example 3.4. In the absence of predators the population of mosquitoes increases at a rate proportional to its population and doubles every three weeks. There are 250,000 mosquitoes initially when a flock of birds arrive that eats 80,000 mosquitoes per week. How many mosquitoes remain after two weeks? 3.1.3 Motions An object falling in air will be governed by the equation dv dt = g − kv2 , where k is a constant depending on the density of air and the shape of the object. Example 3.5. A skydiver with mass 60 kg jumps from an airplane. Assuming k = 0.002 1/m, find the terminal velocity of the skydiver. 10 3.1.4 Loans If B(t) indicates the loan balance after t years, r is the annual interest rate, and P is the annual payment, then the equation governing the loan balance is given by dB dt = rB − P. Example 3.6. A car buyer has a $4,000 down payment and can afford a constant rate of $ 250 per month. 5-year fixed rate loans at an interest rate of 5% per year compounded continuously are available. What is the most expensive car that the buyer can afford? (Assume the payment is done continuously.) 3.2 Numerical Methods Most differential equations cannot be analytically solved. So, we don’t have any choice but to approximate solutions. In this section we will discuss appropximating solutions to first order IVP’s given below:    dy dt = f(t, y) y(t0) = y0 The goal is to find an approximate value for y(t), where t is a point near t0. 3.2.1 Explicit Euler’s Method In this method we will use the fact that the slope at any point is given by f(t, y). To approximate y(t), we divide the interval [t0, t] into N subintervals of equal width h = t − t0 N . We evaluate approximate values for y(tn) where tn = t0 + nh recursively by yn+1 = yn + f(tn, yn)h. Then, yN is an approximation for y(t). Example 3.7. Estimate y(0.2) and y(−0.1), where dy dt = t 2 + y 2 and y(0) = 1. Use step size h = 0.1. Another way of looking at the solutions of the differential equation dy dt = f(t, y) is y(t + h) − y(t) = Z t+h t y 0 (s) ds = Z t+h t f(s, y(s)) ds. We could use methods that we learned in single variable calculus to approximate this integral. Each method would give us a new formula. Using the left endpoint method we obtain Z t+h t f(s, y(s)) ds ≈ f(t, y(t))h, which yields y(t + h) ≈ y(t) + hf(t, y(t)) which is the same formula as in the Euler’s Explicit method. Using the midpoint method, the trapezoidal method, and the Simpson’s method for approximating this integral we obtain different estimates. The formulas for all of these estimates can be found below: 11 Trapezoidal Method: fn = f(tn, yn), yn+1 = yn + h 2 (fn + f(tn+1, yn + hfn)) ‘ Midpoint Method: fn = f(tn, yn), yn+ 1 2 = yn + h 2 fn fn+ 1 2 = f(tn+ 1 2 , yn+ 1 2 ), yn+1 = yn + hfn+ 1 2 Runge-Kutta Method: If we use the Simpson’s method to estimate the integral Z t+h t f(s, y) ds we obtain a much more complicated formula, with far better results. This method is called the Runge-Kutta method. Example 3.8. Estimate y(0.1) with 1 step in Midpoint method for the solution to the IVP: dy dt = t 2 + y 2 , y(0) = 1. 3.3 Error Analysis A function f(h) is said to be of order h 2 (written as O(h 2 )) if the function is roughly a constant multiple of h 2 for small values of h. This means the function goes to zero as fast as h 2 when h approaches zero. The errors in all of the methods above are listed below: Method Global Error Explicit Euler O(h) Midpoint O(h 2 ) Trapezoidal O(h 2 ) Runge-Kutta O(h 4 ) Example 3.9. If we want to make sure that the error in estimating the value of the solution to a first order equation is divided by 16, what changes do we need to make if we are using each of the following methods? (a) Euler’s method. (b) Trapezoidal method. (c) Runge-Kutta method. 3.4 Summary • The main formula for solving application problems is: Rate=Rate In - Rate Out. • Make sure you know the formulas for Explicit Euler’s, Midpoint and Trapezoidal methods. • Error for Euler’s method is O(h), for midpoint and trapezoidal methods the errors are O(h 2 ), and for the Runge-Kutta method the error is O(h 4 ). 12 4 Week 4 4.1 Exact Equations We will continue focusing on solving equations of the form dy dt = f(t, y). Assume the solution is implicitly given by H(t, y) = c. By the chain rule we get Ht + Hy dy dt = 0. This is often written as Htdt + Hydy = 0. (∗) If we can write a first order equation in the form (∗) then we can solve it. Note that ∂Ht dy = ∂Hy ∂t . An equation M(t, y)dt + N(t, y)dy = 0 is called exact if My = Nt. To solve such an equation we will find H for which Ht = M and Hy = N. Example 4.1. Solve the equation dy dx = − xy2 + y + e x x 2y + x . Example 4.2. Solve the initial value problem dy dt + e ty + 2t 2y + e t = 0, y(0) = 0. 4.2 Integrating Factors Sometimes equations that are not exact can be turned into exact equations by multiplying by a factor called an integrating factor. Example 4.3. Solve the equation (2e x + y 3 )dx + 3y 2dy = 0. Example 4.4. Solve the equation 2ty + (2t 2 − e y ) dy dt = 0 Example 4.5. Solve the IVP (4xy + 3y 3 )dx + (x 2 + 3xy2 )dy = 0. 4.3 Higher-Order Linear Equations An n-th order linear differential equation in normal form is an equation of the form: y (n) + an(t)y (n−1) + · · · + a2(t)y 0 + a1(t)y = f(t). f(t) is called the forcing and ai(t)’s are called coefficients. We would also like to explore initial value problems of the form:    y (n) + an(t)y (n−1) + · · · + a2(t)y 0 + a1(t)y = f(t). y(t0) = y0 y 0 (t0) = y1 . . . y (n−1)(t0) = yn−1 13 Theorem 4.1 (Existence-Uniqueness Theorem for Linear Equations.). Assume all coefficients and the forcing of a linear equation in normal form are continuous over an interval (a, b). Then, for every initial time t0 in (a, b) and real numbers y0, y1, . . . , yn−1, the initial value problem    y (n) + an(t)y (n−1) + · · · + a2(t)y 0 + a1(t)y = f(t). y(t0) = y0 y 0 (t0) = y1 . . . y (n−1)(t0) = yn−1 has a unique solution that is defined over (a, b). Example 4.6. Find the interval of definition of the solution to the initial value problem: (t 2 − 1)y 00 − 1 t 2 − 4 y 0 + y = t 2 − 3, y(0) = y 0 (0) = 1.5. Example 4.7. Show that sin(t 3 ) is not a solution to any linear differential equation of the form y 000+a(t)y 00+ b(t)y 0 + c(t)y = 0, where all coefficients are continuous over (−1, 1). 4.4 Linear Differential Operators Recall that linear equations are those of the form: d ny dtn + an(t) d n−1y dtn−1 + · · · + a2(t) dy dt + a1(t)y = f(t) (∗) To simplify this, we denote d dt by D, so instead of d ky dtk we write Dk [y]. The equation (∗) is then written as L[y] = f(t), where L = Dn + an(t)Dn−1 + · · · + a2(t)D + a1(t). Example 4.8. Let y1 and y2 be two solutions to the linear homogeneous equation y 00 + t 2y 0 + sin(t 3 )y = 0. Show that y = c1y1 + c2y2 is also a solution to this differential equation, where c1, c2 are constants. In general for constants c1, . . . , cn and functions y1, . . . , yn we have L[c1y1 + · · · + cnyn] = c1L[y1] + · · · + cnL[yn]