To determine The current from the given figure. Explanation Let the currents in the left and right loops be I 1 ( t ) and I 2 ( t ) . From the figure, the voltage drops in the right loop are L I ′ 2 and R ( I 2 − I 1 ) . Note that, by the sum of the voltage drop over the capacitor is zero. By the note, the equation becomes as L I ′ 2 + R ( I 2 − I 1 ) = 0 . Given that, R = 1 Ω , L = 1.25 H, C = 0.2 F with initial conditions on current as [ I 1 ( 0 ) I 2 ( 0 ) ] = [ 1 1 ] . Substitute those values in the equation L I ′ 2 + R ( I 2 − I 1 ) = 0 and solve for I ′ 1 . ( 1.25 ) I ′ 2 + ( I 2 − I 1 ) = 0 1.25 I ′ 2 = I 1 − I 2 I ′ 2 = 1 1.25 I 1 − 1 1.25 I 2 I ′ 2 = 0.8 I 1 − 0.8 I 2 From the figure, the voltage drops in the left loop are R ( I 1 − I 2 ) and 1 C ∫ I 1 d t . The sum of the voltage drop over the capacitor is zero. That is, R ( I 1 − I 2 ) + 1 C ∫ I 1 d t = 0 . Differentiate the equation R ( I 2 − I 1 ) + 1 C ∫ I 2 d t = 0 both sides with respect to t . d d t ( R ( I 1 − I 2 ) ) + d d t ( 1 C ∫ I 1 d t ) = d d t ( 0 ) R ( I ′ 1 − I ′ 2 ) + 1 C I 1 = 0 R I ′ 1 − R I ′ 2 + I 1 C = 0 Substitute R = 1 and C = 0.2 in the equation R I ′ 1 − R I ′ 2 + I 1 C = 0 . I ′ 1 − I ′ 2 + I 1 0.2 = 0 I ′ 1 = I ′ 2 − 5 I 1 Substitute the value of I ′ 2 = 0.8 I 1 − 0.8 I 2 in the equation I ′ 1 = I ′ 2 − 5 I 1 . I ′ 1 = 0.8 I 1 − 0.8 I 2 − 5 I 1 = − 4.2 I 1 − 0.8 I 2 Now, the system of differential equations in terms currents is { I ′ 1 = − 4.2 I 1 − 0.8 I 2 I ′ 2 = 0.8 I 1 − 0.8 I 2 . Solve the system of differential equation as follows. The coefficient matrix of the corresponding systems of differential equations is A = [ − 4.2 − 0.8 0.8 − 0.8 ] . Hence, a single vector equation is I ′ = [ − 4.2 − 0.8 0.8 − 0.8 ] I , where I = [ I 1 I 2 ] . Characteristic equation of the coefficient matrix A = [ − 4.2 − 0.8 0.8 − 0.8 ] is det ( A − λ I ) = 0 , where λ eigenvalue and I is 2 × 2 identity matrix . Simplify for the det ( A − λ I ) = 0 using the coefficient matrix as follows. det ( A − λ I ) = 0 det ( [ − 4.2 − 0.8 0.8 − 0.8 ] − λ [ 1 0 0 1 ] ) = 0 det ( [ − 4.2 − λ − 0.8 0.8 − 0.8 − λ ] ) = 0 ( λ + 4.2 ) ( λ + 0.8 ) + 0.64 = 0 Further simplified as, λ 2 + 5 λ + 4 = 0 ( λ + 4 ) ( λ + 1 ) = 0 λ + 4 = 0 or λ + 1 = 0 λ = − 4 or λ = − 1 Compute the eigenvector t ( 1 ) = [ t 1 t 2 ] corresponding to the eigenvalue λ = − 1 as follows

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 4, Problem 26RQ

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Chapter 4 Solutions

Advanced Engineering Mathematics

Ch. 4.1 - Prob. 1P Ch. 4.1 - Prob. 2P Ch. 4.1 - Prob. 3P Ch. 4.1 - Prob. 4P Ch. 4.1 - If you extend Example 1 by a tank T3 of the same...Ch. 4.1 - Find a “general solution” of the system in Prob....Ch. 4.1 - In Example 2 find the currents: 7. If the initial...Ch. 4.1 - Prob. 8P Ch. 4.1 - Prob. 9P Ch. 4.1 - Find a general solution of the given ODE (a) by...

Ch. 4.1 - Find a general solution of the given ODE (a) by...Ch. 4.1 - Find a general solution of the given ODE (a) by...Ch. 4.1 - Find a general solution of the given ODE (a) by...Ch. 4.1 - Prob. 14P Ch. 4.3 - 1–9 GENERAL SOLUTION Find a real general solution...Ch. 4.3 - 1–9 GENERAL SOLUTION Find a real general solution...Ch. 4.3 - 1–9 GENERAL SOLUTION Find a real general solution...Ch. 4.3 - 1–9 GENERAL SOLUTION Find a real general solution...Ch. 4.3 - 1–9 GENERAL SOLUTION Find a real general solution...Ch. 4.3 - 1–9 GENERAL SOLUTION Find a real general solution...Ch. 4.3 - 1–9 GENERAL SOLUTION Find a real general solution...Ch. 4.3 - Find a real general solution of the following...Ch. 4.3 - Prob. 9P Ch. 4.3 - Solve the following initial value problems. Ch. 4.3 - 10–15 IVPs Solve the following initial value...Ch. 4.3 - Prob. 12P Ch. 4.3 - Solve the following initial value problems. Ch. 4.3 - Solve the following initial value problems. Ch. 4.3 - Solve the following initial value problems. Ch. 4.3 - Prob. 16P Ch. 4.3 - Prob. 17P Ch. 4.3 - Prob. 18P Ch. 4.3 - Prob. 19P Ch. 4.4 - Prob. 1P Ch. 4.4 - Prob. 2P Ch. 4.4 - Prob. 3P Ch. 4.4 - Prob. 4P Ch. 4.4 - Prob. 5P Ch. 4.4 - Prob. 6P Ch. 4.4 - Prob. 7P Ch. 4.4 - Prob. 8P Ch. 4.4 - Prob. 9P Ch. 4.4 - Prob. 10P Ch. 4.4 - Prob. 11P Ch. 4.4 - Prob. 12P Ch. 4.4 - Prob. 13P Ch. 4.4 - Prob. 14P Ch. 4.4 - Prob. 15P Ch. 4.4 - Prob. 16P Ch. 4.4 - Prob. 17P Ch. 4.5 - Prob. 1P Ch. 4.5 - Prob. 2P Ch. 4.5 - Prob. 4P Ch. 4.5 - Prob. 5P Ch. 4.5 - Prob. 6P Ch. 4.5 - Prob. 7P Ch. 4.5 - Prob. 8P Ch. 4.5 - Prob. 9P Ch. 4.5 - Prob. 10P Ch. 4.5 - Prob. 11P Ch. 4.5 - Prob. 12P Ch. 4.5 - Prob. 13P Ch. 4.6 - Prob. 1P Ch. 4.6 - Prob. 2P Ch. 4.6 - Prob. 3P Ch. 4.6 - Prob. 4P Ch. 4.6 - Prob. 5P Ch. 4.6 - Prob. 6P Ch. 4.6 - Prob. 7P Ch. 4.6 - Prob. 9P Ch. 4.6 - Prob. 10P Ch. 4.6 - Prob. 11P Ch. 4.6 - Prob. 12P Ch. 4.6 - Prob. 13P Ch. 4.6 - Prob. 14P Ch. 4.6 - Prob. 15P Ch. 4.6 - Prob. 16P Ch. 4.6 - Prob. 17P Ch. 4.6 - Prob. 19P Ch. 4 - Prob. 1RQ Ch. 4 - Prob. 2RQ Ch. 4 - How can you transform an ODE into a system of...Ch. 4 - What are qualitative methods for systems? Why are...Ch. 4 - Prob. 5RQ Ch. 4 - Prob. 6RQ Ch. 4 - What are eigenvalues? What role did they play in...Ch. 4 - Prob. 8RQ Ch. 4 - Prob. 9RQ Ch. 4 - Prob. 10RQ Ch. 4 - Find a general solution. Determine the kind and...Ch. 4 - Find a general solution. Determine the kind and...Ch. 4 - Find a general solution. Determine the kind and...Ch. 4 - Find a general solution. Determine the kind and...Ch. 4 - Prob. 15RQ Ch. 4 - Prob. 16RQ Ch. 4 - Prob. 17RQ Ch. 4 - Prob. 18RQ Ch. 4 - Prob. 19RQ Ch. 4 - Prob. 20RQ Ch. 4 - Prob. 21RQ Ch. 4 - Prob. 22RQ Ch. 4 - Prob. 23RQ Ch. 4 - Prob. 24RQ Ch. 4 - Prob. 25RQ Ch. 4 - Network. Find the currents in Fig. 103 when R = 1...Ch. 4 - Prob. 27RQ Ch. 4 - Prob. 28RQ Ch. 4 - Find the location and kind of all critical points...Ch. 4 - Find the location and kind of all critical points...

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