1. Why don t we have to worry about absolute values in the general form for first order differential equations with constant coefficients?

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1 1. Why don t we have to worry about absolute values in the general form for first order differential equations with constant coefficients? Let y = ay b with y(0) = y 0 We can solve this as follows y = a(y b/a) (1) y y b/a = a (2) ln y b/a = at + C (3) y b/a = e at+c (4) y b/a = Ce at (5) Now, in class we assumed y b/a 0 so that we could forget about the absolute value and get that at y = b/a + Ce

2 1. Why don t we have to worry about absolute values in the general form for first order differential equations with constant coefficients? What if y b/a < 0? In this case y-b/a =b/a-y. So we have: y b/a = Ce at (6) b/a y = Ce at (7) y = b/a + Ce at (8) y = b/a Ce at (9) y = b/a + Ce at (10) Since C is an arbitrary constant, we can absorb the negative into the C (or replace C with -C) to get the same solution as before. Thus in each case we get the solution: y = b/a + Ce at

3 Modeling Problems 1 Define variables. 2 Write a differential equation. Remember Rate of change=rate of change in - rate of change out 3 Keep units in mind. 4 Write down the initial conditions. 5 Solve the differential equation. 6 Again, check to make sure the units make sense.

4 Example(2.3#1): Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200L of a dye solution with a concentration of 1 g/l. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2L/min, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of the dye in the tank reaches 1% of its original value.

5 Example(2.3#10): A home buyer can afford to spend no more than $800/month on mortgage payments. Suppose that the interest rate is 9% and that the term of the mortgage is 20 years. Assume that the interest is compounded continuously and that the payments are also made continuously. Determine the maximum amount that this buyer can afford to borrow.

6 Autonomous Equations 1 Graph dy/dt as a function of y. Focus on zeros and where the function is positive and negative. 2 Graph the phase line (number line with equilibrium points marked and draw arrows to represent when solutions are increasing or decreasing) 3 Classify the stationary points as stable, unstable or semi-stable. 4 Sketch several solutions based on the increasing and decreasing information from the phase line. Example: dy/dt = y 2 (y 2 1) where < y 0 <

7 Figure: Plot of dy/dt as a function of y 2 unstable 1 semistable 0 stable -1-2 Figure: Phase Line and sketch of solutions

8 Figure: Actual Solutions Follow-up Question: What happens to y as t? This depends on the initial condition! Let y(t 0 ) = y 0. For y 0 > 1, y. For 0 < y 0 < 1, y 0. For y 0 < 0, y 1.

9 Exact and Integrating Factor for Exact A differential equation of the form is exact if M(x, y)dx + N(x, y)dy = 0 M y = N x where M y is the partial derivative of the function M(x, y) with respect to y and N x is the partial derivative of N(x, y) with respect to y. If the equation is exact then we can find some function f such that f x = M and f y = N. Then the solution to the differential equation is f = C.

10 Sometimes an equation is not exact but we can use an integrating factor to make it exact. There are two types of problems you should be able to do. 1 Equations that can be made exact by multiplying by a function µ(x) that only depends on x. This is true only if M y N x N only depends on N. In this case e.g. µ(x) = M y N x µ(x) N y = e 2x + y 1

11 2 Equations that can be made exact using a given integrating factor. e.g. x 2 y 3 + x(1 + y 2 )y = 0 µ(x, y) = 1/(xy 3 )

12 Theorems about guaranteed solution intervals First Order Theorem If p and q are continuous on an interval I : α < t < β containing the point t = t 0, then there exists a unique function y that satisfies the IVP y + p(t)y = g(t) y(t 0 ) = y 0 for each t in I. e.g. Find the largest interval in which the IVP ty + 2y = 4t 2, y(1) = 2 has a unique solution.

13 Theorem Let f (t, y) and f / y be continuous for α < t < β, γ < y < δ containing the point (t 0, y 0 ). Then in some interval t 0 h < t < t 0 + h contained in α < t < β, there is a unique solution to the IVP y = f (t, y), where y(t 0 ) = y 0 e.g. Does the theorem guarantee that y = y 1/3 with the initial condition y(0) = 0 has a unique solution? No, since f / y is undefined at y = 0. In fact, y = (2/3t) 3/2 and y = (2/3t) 3/2 are solutions to the IVP.

14 Theorems for 2nd Order Theorem Consider the IVP y + p(t)y + q(t)y = g(t), y(t 0 ) = y 0, y (t 0 ) = y 0 where p, q, g are continuous on an open interval I containing t 0. Then there exists a unique solution y to this problem and the solution exists throughout I.

15 Theorems about the Wronskian Theorem Suppose that y 1 and y 2 are two solutions of the differential equation y + p(t)y + q(t)y = 0. Then the family of solutions y = c 1 y 1 (t) + c 2 y 2 )t) with arbitrary coefficients c 1 and c 2 includes every solutions to the above differential equation if and only if there is a point t 0 where the Wronskian of y 1 and y 2 is not zero. If y 1 and y 2 are both solutions to a homogeneous second order differential equation and their Wronskian is nonzero for some point t 0, we call y 1 and y 2 a fundamental solution set.

16 Just because we can write every solutions as y = c 1 y 1 + c 2 y 2 doesn t mean we can solve every initial value problem. (Why? Because every solution can be expressed via infinitely many initial conditions. Say the solution to the IVP is y(t) = 2t. This satisfies the initial conditions y(0) = 0, y(1) = 2, y(7) = 14,.... Some of these t 0 may have W (y 1, y 2 )(t 0 ) = 0 Theorem Let y + p(t)y + q(t)y = 0, have solutions y 1 and y 2. The the IVP with initial conditions y(t 0 ) = y 0, y (t 0 ) = y 0 has a solutions of the form y = c 1 y 1 + c 2 y 2 if and only if W (y 1, y 2 )(t 0 ) 0

17 Second Order, Linear, Homogeneous Differential Equations with Constant Coefficients ay + by + c = 0 Characteristic Polynomial ar 2 + br + c = 0 Let r 1 and r 2 be the roots of the characteristic polynomial. 1 If r 1 and r 2 are real and distinct, then Let y 1 = e r 1t and y 2 = e r 2t. 2 If r 1, r 2 = λ + µi then y 1 = e λt cosµt and y 1 = e λt sinµt. 3 If r 1 = r 2 then y 1 = e r 1t and y 2 = te r 1t. In each case, y 1 and y 2 form a fundamental solution set so the general solution is y = c 1 y 1 + c 2 y 2.

18 Examples 1 y 2y + y = 0 2 y y 6y = 0 3 y 10y + 26y = 0

19 Reduction of Order Given a second order linear homogeneous differential equation where we known one solution y 1 (t) we assume that the other solution is of the form y = v(t)y 1 (t) and plug this in to the differential equation to solve for a second solution. This is called reduction of order because once we plug in to the differential equation, the y terms should cancel. Then we let w = y and can reduce this to a first order differential equation. Example: Assume t 2 y 4ty + 6y = 0, t > 0. The function y 1 (t) = t 2 is a solution. Use reduction of order to find a second solution.

20 Method of Undetermined Coefficients Given the differential equation y + p(t)y + q(t)y = g(t) we can use the method of undetermined coefficients to find the solutions. First we find the fundamental solution set to the corresponding homogeneous equation. Then we guess the form of the particular solutions. 1 If g(t) = A 0 t n + A 1 t n A n 1 t + A n, guess B 0 t n + B 1 t n B n 1 t + B n 2 If g(t) = e αt, guess Ae αt 3 If g(t) = sin(αt) or g(t) = cos(αt), guess Asin(αt) + Bcos(αt). If we have a product of the above, guess the product (make sure to have different constants or functions in front of the sin or cos parts.) If we have a sum, break into different problems, finding a g(t) for each term in the sum.

21 Warning! If the guess is one of the homogeneous solutions, multiply the guess by t. If it is still a solution to the homogeneous solution (in the case of repeated roots), multiply by t again.

22 What should we guess for the following: 1 3te t (At + B)e t 2 t 2 cos(2t) (At 2 + Bt + C)cos(2t) + (Dt 2 + Et + F )sin(2t) 3 e t (sin(5t)) Ae t (cos(5t)) + Be t (sin(5t)) Once we have our guess, we plug it back in to the differential equation. Then we can solve for the coefficients. Example: y y = t 2 + 3e t

23 Given y + p(t)y + q(t)y = g(t) we can use variation of parameters to solve. We first find y 1 and y 2 the fundamental solution set. Then the particular is of the form Y (t) = y 1 y2 (t)g(t) W (y 1, y 2 ) dt + y 2(t) y1 (t)g(t) W (y 1, y 2 ) dt As long as we remember our +C s this will in fact be the general solution. Example: Consider t 2 y 2y = 3t 2 1 Show that the corresponding homogeneous equation has fundamental solution set y 1 (t) = t 2, y 2 (t) = t 1. Use variation of parameters to find the general solution.