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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 current-voltage 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 current-voltage 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 current-voltage 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 ....
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