M340 HW 2 SOLUTIONS. 1. For the equation y = f(y), where f(y) is given in the following plot:
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1 M340 HW SOLUTIONS 1. For the equation y = f(y), where f(y) is given in the following plot: (a) What are the critical points? (b) Are they stable or unstable? (c) Sketch the solutions in the ty plane. (d) Let y p (t) be the particular solution satisfying y p (0) =, find lim t y p (t). Solutions: (a) Critical points occur when f(y) = 0, thus at y = 1, 3, 5 (b) y = 1, 5 are stable, y = 3 is unstable (c) (d) Since the initial point is (0, ), we are in the interval 1 < y < 3, thus the solution tends to the stable equilibrium point y = 1
2 . True or False. If false, explain why or give a counterexample. (a) (T) (F) The function y = cx is a solution to the differential equation xy = y for every real value of c. (b) (T) (F) The general solution to a 1st order linear nonhomogeneous differential equation is the product of the general solution to the corresponding homogeneous equation and a particular solution to the nonhomogeneous equation. (c) (T) (F) The equation (y x 3)dx + (yx + 4)dy = 0 is exact. (d) (T) (F) For two matrices A, B, we always have AB = BA. Solutions: (a) TRUE - since when we plug in y = cx into the ODE, we get a true statement xy = x(cx) = xc = cx = y (b) FALSE - Our theorem states the general solution is the SUM... not the product! (c) TRUE - test for exactness (P y = Q x ) P = y x 3 P y = yx and Q = yx + 4 Q x = yx = P y (d) FALSE - Consider matrix A with size 3 4 and matrix B with size 4 3. Then AB is a 3 3 but BA is a 4 4, which means AB BA.
3 3. Given a linear system Ax = b as follows x + 6y + z = 7 x + y z = 1 5x + 8y 4z = (a) Apply row operations to simplify the augmented matrix to row echelon form (DO NOT use a calculator, you must show all of your work!). (b) Write the solution of the linear system in a parametric form. Solutions: (a) The augmented matrix is: To make things easier, start by switching row 1 and row : Then use the operations R-R1, and R3-5R1, to create new rows and 3: Add R3+R, for a new row 3: (b) Now we are ready to back solve. Row 3 says 4z = 16, thus z = 4. Row says y + 3(4) = 9 y = 3/. And row 1 says x + ( 3/) 4 x = 6 The parametric solution looks like x 6 y = 3/ z 4
4 Practice Problems - DO NOT TURN IN 1. Let x(t) be a solution to the initial value problem tx (t) + x(t) = 4t, x(1) = 3. Show that x(t) > (t + 1/t ) for all t for which x(t) is defined. Solution: First we need to show that x(t) is a unique solution. Putting the ODE in normal form x (t) = (/t)x(t) + 4t = F (x, t), we see that F (x, t) is continuous for all x, and for t 0. Thus F (x, t) is continuous on the rectangle R(y, t) = {0 < t < 3, 0 < x < 5}, which the point (1, 3) inside R. Thus by the existance theorem, x(t) exists. To show uniqueness, consider F = /t. This is also continuous on the R defined above x (which still contains our point), thus by the uniqueness theorem, x(t) is also unique. We know a unique solution cannot cross another solution. So, to prove x(t) > (t + 1/t ), we need to show that t + 1/t is also a solution. We can do this by plugging in the equation into the ODE, in replace of x(t): tx (t) + x(t) = t ( t + 1/t ) + ( t + 1/t ) = t ( t /t 3) + t + /t = t /t + t + /t = 4t Thus since (t + 1/t ) satisfies the ODE, it is also a solution. Since x(t) is unique, it cannot cross (t + 1/t ). So we need to see where x(t) lies. To do this we check one point. When t = 1, we know x(1) = 3 by our initial condition. When t = 1, for (t + 1/t ), we have (1) + 1 =. Since < 3, this means (t + 1/t ) must always be less than x(t), so x(t) > (t + 1/t ). Find the largest interval where the solution can be defined for the following problems (a) ty + y = t 3, y( 1) = 3 (b) ty + y = t 3, y(1) = 3 (c) (t 3)y + (ln t)y = t, y(1) = Soltuion: (a) In normal form the ODE looks like y = (1/t)y + t. Thus the largest interval is R = { < t < 0, < y < } (b) In normal form the ODE looks like y = (1/t)y + t. Thus the largest interval is R = {0 < t <, < y < } (c) In normal form the ODE looks like y = ln t y + t. Thus the largest interval is t 3 t 3 R = {0 < t < 3, < y < }
5 3. A bacteria culture starts with 500 bacteria and after 3 hours there are 8000 bacteria. Assuming a Malthusian Model, find the number of bacteria after 4 hours. Soltuion: The solution to the Malthusian Model ODE is given by P (t) = P 0 e rt. In this example, P 0 = 5 (in hundreds), and it is given that P (3) = 80. Thus at t = 4, P (t) = 5e rt P (3) = 5e 3r = 80 e 3r = 16 r = ln(16) 3 P (4) = 5e (4/3) ln()(4) Thus after 4 hours there are roughly 0, bacteria. = 4 3 ln() 4. Suppose a species of fish in a lake is modeled by a logistic population model with relative growth rate of r = 0.3 per year and carrying capacity of K = (a) Write the differential equation describing the logistic population model for this problem. (b) Determine the equilibrium solutions for this model. (c) If 500 fish are initially introduced into the lake, find the analytic solution P (t) that models the number of fish in the lake after t years. Use it to estimate the number of fish in the lake after 5 years. (d) Estimate the time it will take for there to be 8000 fish in the lake. Soltuion: ( ) (a) The logistic ODE is given by P (t) = r 0 1 P K P. So in this case, ( P (t) = P ) P 10 (b) The equilibrium solutions are when P (t) = 0, thus when P = 0, 10 KP 0 (c) The analytic solution is given by P (t) =, thus in this case P 0 + (K P 0 )e r(t t 0) (d) Set P (t) = 10(.5).5 + (10.5)e = 5 0.3t e 0.3t Then, P (5) , so the number of fish will be approximately 5, then t 8.8 years e 0.3t = 8
6 5. Suppose you start an IRA account at age 5. You deposit $000 at the beginning and $000 each year after. The interest rate is 8% annually. How much money is in this account when you retire at 65 years old? Soltuion: The ODE looks like P (t) = 0.08P +, P (0) = This is a linear equation so we can use method of integrating factor. Then µ = e 0.08t. e 0.08t P (t) e 0.08t 0.08P = e 0.08t (e 0.08t P ) = e 0.08t (e 0.08t P ) = e 0.08t P = P = P (0) = P (t) = P (40) = e 0.08t 0.08 e 0.08t + C Ce0.08t C = C = ( ) ( e 0.08t ) e 0.08(40) Thus at the age of 65 (40 years later) you will have $637, 378.8
7 6. Given a linear system Ax = b as follows x 1 + x x 3 = 6 x 1 + 4x 4x 3 = 1 3x 1 + 6x 6x 3 = 18 (a) Apply row operations to simply the augmented matrix to the reduced row echelon form. (b) Write the solutions of the linear system in parametric form. (c) Find a basis for null(a). Solution: (a) The augmented matrix is: Use the operations R-R1, and R3-3R1, to create new rows and 3: This is already in reduced row echelon form! (b) Now we are ready to back solve. We have two free columns, so let x 3 = s and x = f. Row 1 says x 1 + f s = 6 x 1 = 6 f + s The parametric solution looks like x 1 6 x x 3 = 0 + f s 0 1 (c) Using the theorem that says x = p + v, we know the null space of A is given by span 1, 0. But for it to be a basis the vectors must also be linearly 0 1 independent. To check, consider the equation a + b a b 0 = 1 a b The only way to make this the 0 vector is if a = b = 0, thus they are linearly independent. Therefore, the basis for null(a) is span 1, 0 0 1
8 7. Is the vector (5, ) T in the span{u 1, u }, where u 1 = (1, ) T,u = (3, 0) T? If so, provide a specific linear combination that works. Solution: We want to solve the system [u 1 u ] x = (5, ) T. The augmented matrix is: [ ] 0 Use the operation R+R1: [ This have a solution. By the nd row: 6y = 8 y = 4/3. By the 1st row x + 3(4/3) = 5 x = 1. This means that [ ] [ ] [ ] (1) + (4/3) = 0 and (5, ) T is in the span{u 1, u }. ]
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