This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 2a Practical, Fall 2008: Solutions to Homework 5 You are bound by the honor code to not look at this file if you did not take Math 2a in the Fall of 2008 1. Rewrite the equation y 00 + 4 y 5 y = 0 as a first order linear system and write down the general solution of this system. (4pts) Solution. Introducing a new variable z = y , we may rewrite this equation as the first order linear system z = 4 z + 5 y y = z . This can be expressed in matrix form as z y = 4 5 1 z y . The characteristic polynomial of this matrix is p ( λ ) = λ 2 + 4 λ 5 = ( λ 1)( λ + 5), with roots λ = 1 , 5. An eigenvector corresponding to the eigenvalue 1 is ( 1 1 ) , and an eigenvector corresponding to the eigenvalue 5 is ( 5 1 ) . Thus the general solution to the system of equations is z ( t ) y ( t ) = c 1 5 1 e 5 t + c 2 1 1 e t . 2. (a) Consider the circuit shown in Figure 7.1.5. Use the method outlined in Problem 19 to show that the current I through the inductor and the voltage V across the capacitor satisfy the system of differential equations L dI dt = R 1 I V, C dV dt = I V R 2 . (5pts) Solution. Let I 1 , I 2 , I 3 , and I 4 represent the current through R 1 , R 2 , L , and C , respectively. Likewise, let V 1 , V 2 , V 3 , and V 4 denote the corresponding voltage drops. Using the current voltage relation along the inductor, we see that L dI dt = V 3 . Kirchhoff’s second law along the upper loop in the diagram tells us that V 1 + V 2 + V 3 = 0. Thus L dI dt = V 3 = V 1 V 2 . Using Kirchhoff’s second law again, this time along the lower loop, we see that V 2 = V 4 = V . Therefore L dI dt = V 1 V . Now applying the currentvoltage relation along R 1 , we see that V 1 = R 1 I 1 . Kirchhoff’s first law applied at the upper left node says that I 1 = I 3 = I . Thus L dI dt = V 1 V = R 1 I 1 V = R 1 I V , which is the first desired relation. Next we derive the second relation. From the currentvoltage relation for the capacitor, we have C dV dt = I 4 . Kirchhoff’s first law applied to the rightmost node in the diagram gives I 4 = I I 2 , so therefore C dV dt = I 4 = I I 2 . The currentvoltage relation also gives I 2 = V 2 R 2 . As above V 2 = V , and hence I 2 = V 2 R 2 = V R 2 . Thus C dV dt = I I 2 = I V R 2 ....
View
Full
Document
This note was uploaded on 01/08/2011 for the course MA 2a taught by Professor Makarov,n during the Fall '08 term at Caltech.
 Fall '08
 Makarov,N
 Math, Differential Equations, Statistics, Equations, Probability

Click to edit the document details