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 left loop are L I ′ 1 and R ( I 1 − I 2 ) . Note that, the voltage (the electromotive force EMF) impressed on a closed loop is equal to the sum of the voltage drops across all the other elements of the loop. By the note, the equation becomes as L I ′ 1 + R ( I 1 − I 2 ) = E . Given that, R = 2.5 Ω , L = 1 H, C = 0.04 F, and E ( t ) = 169 sin t V with initial conditions on current as [ I 1 ( 0 ) I 2 ( 0 ) ] = [ 0 0 ] . Substitute those values in the equation L I ′ 1 + R ( I 1 − I 2 ) = E and solve for I ′ 1 . ( 1 ) I ′ 1 + 2.5 ( I 1 − I 2 ) = 169 sin t I ′ 1 + 2.5 ( I 1 − I 2 ) = 169 sin t I ′ 1 = − 2.5 I 1 + 2.5 I 2 + 169 sin t From the figure, the voltage drops in the right loop are R ( I 2 − I 1 ) and 1 C ∫ I 2 d t . The sum of the voltage drop over the capacitor is zero. That is, R ( I 2 − I 1 ) + 1 C ∫ I 2 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 2 − I 1 ) ) + d d t ( 1 C ∫ I 2 d t ) = d d t ( 0 ) R ( I ′ 2 − I ′ 1 ) + 1 C I 2 = 0 R I ′ 2 − R I ′ 1 + I 2 C = 0 Substitute R = 2.5 and C = 0.04 in the equation R I ′ 2 − R I ′ 1 + I 2 C = 0 . 2.5 I ′ 2 − 2.5 I ′ 1 + I 2 0.04 = 0 2.5 I ′ 2 = 2.5 I ′ 1 − 25 I 2 I ′ 2 = I ′ 1 − 10 I 2 Substitute the value of I ′ 1 = − 2.5 I 1 + 2.5 I 2 + 169 sin t in the equation I ′ 2 = I ′ 1 − 10 I 2 . I ′ 2 = − 2.5 I 1 + 2.5 I 2 + 169 sin t − 10 I 2 I ′ 2 = − 2.5 I 1 − 7.5 I 2 + 169 sin t Now, the system of differential equations in terms currents is { I ′ 1 = − 2.5 I 1 + 2.5 I 2 + 169 sin t I ′ 2 = − 2.5 I 1 − 7.5 I 2 + 169 sin t . Solve the system of differential equation as follows. The coefficient matrix of the corresponding systems of differential equations is A = [ − 2.5 2.5 − 2.5 − 7.5 ] . Note that, a single vector equation is I ′ = A I + g ( t ) , where A is coefficient matrix and g ( t ) = [ 169 sin t 169 sin t ] . Hence, a single vector equation is I ′ = [ − 2.5 2.5 − 2.5 − 7.5 ] I + [ 169 sin t 169 sin t ] , where I = [ I 1 I 2 ] . Characteristic equation of the coefficient matrix A = [ − 2.5 2.5 − 2.5 − 7.5 ] 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 ( [ − 2.5 2.5 − 2.5 − 7.5 ] − λ [ 1 0 0 1 ] ) = 0 det ( [ − 2.5 − λ 2.5 − 2.5 − 7.5 − λ ] ) = 0 ( λ + 2.5 ) ( λ + 7.5 ) + 6.25 = 0 Further simplified as, λ 2 + 10 λ + 25 = 0 ( λ + 5 ) 2 = 0 λ + 5 = 0 λ = − 5 Compute the eigenvector t ( 1 ) = [ t 1 t 2 ] corresponding to the eigenvalue λ = − 5 as follows. A t − ( − 5 ) t = 0 [ − 2.5 + 5 2.5 − 2.5 − 7.5 + 5 ] [ t 1 t 2 ] = [ 0 0 ] [ 2.5 t 1 + 2.5 t 2 − 2.5 t 1 − 2.5 t 2 ] = [ 0 0 ] 2.5 t 1 + 2.5 t 2 = 0 and − 2.5 t 1 − 2.5 t 2 = 0 t 1 − 1 = t 2 1 Hence, the possible vector corresponding to the eigenvalue λ = − 5 is t ( 1 ) = [ − 1 1 ] . Compute the eigenvector t ( 2 ) = [ t 1 t 2 ] such that ( A − λ I ) t ( 2 ) = t ( 1 ) as follows. ( A + 5 I ) t ( 2 ) = t ( 1 ) ( [ − 2.5 2.5 − 2.5 − 7.5 ] − 5 [ 1 0 0 1 ] ) [ t 1 t 2 ] = [ − 1 1 ] [ − 2.5 + 5 2.5 − 2.5 − 7.5 + 5 ] [ t 1 t 2 ] = [ − 1 1 ] [ 2.5 t 1 + 2.5 t 2 − 2.5 t 1 − 2.5 t 2 ] = [ − 1 1 ] Compute the eigenvector t ( 2 ) = [ t 1 t 2 ] corresponding to the eigenvalue λ = − 1 as follows. A t − ( − 1 ) t = 0 [ 1 + 1 4 1 1 + 1 ] [ t 1 t 2 ] = [ 0 0 ] [ 2 t 1 + 4 t 2 t 1 + 2 t 2 ] = [ 0 0 ] 2 t 1 + 4 t 2 = 0 and t 1 + 2 t 2 = 0 t 1 2 = t 2 − 1 Further simplified as, 2.5 t 1 + 2.5 t 2 = − 1 and − 2.5 t 1 − 2.5 t 2 = 1 t 1 = 2.5 t 2 − 1 2.5 t 1 = 2.5 t 2 − 0.4 Hence, the possible vector corresponding to the eigenvalue λ = − 5 is t ( 2 ) = [ − 0.4 0 ] . Thus, the homogenous solution of the system of differential equation is I h = c 1 e − 5 t [ − 1 1 ] + c 2 ( t [ − 1 1 ] + [ − 0.4 0 ] ) e − 5 t . Now obtain the particular solution of the system of differential equation using undetermined coefficient as follows. Let the particular solution of the differential equation be of the form I p = [ a 1 a 2 ] cos t + [ b 1 b 2 ] sin t Differentiate the solution with respect to t . I ′ p = [ a 1 a 2 ] d d t ( cos t ) + [ b 1 b 2 ] d d t ( sin t ) = − [ a 1 a 2 ] sin t + [ a 2 b 2 ] cos t Substitute I p = [ a 1 a 2 ] cos t + [ b 1 b 2 ] sin t in the equation I ′ = [ − 2.5 2.5 − 2.5 − 7.5 ] I + [ 169 sin t 169 sin t ] . I ′ p = [ − 2.5 2.5 − 2.5 − 7.5 ] ( [ a 1 a 2 ] cos t + [ b 1 b 2 ] sin t ) + [ 169 sin t 169 sin t ] = [ − 2.5 2

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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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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