Explanation Consider the given systems of differential equation as { y ′ 1 = y 1 + 4 y 2 − 2 cos t y ′ 2 = y 1 + y 2 − cos t + sin t . The coefficient matrix of the corresponding systems of differential equations is A = [ 1 4 1 1 ] . Note that, a single vector equation is y ′ = A y + g ( t ) , where A is coefficient matrix and g ( t ) = [ − 2 cos t − cos t + sin t ] . Hence, a single vector equation is y ′ = [ 1 4 1 1 ] y + [ − 2 cos t − cos t + sin t ] , where y = [ y 1 y 2 ] . Characteristic equation of the coefficient matrix A = [ 1 1 4 1 ] is det ( A − λ I ) = 0 , where λ is the 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 ( [ 1 4 1 1 ] − λ [ 1 0 0 1 ] ) = 0 det ( [ 1 − λ 4 1 1 − λ ] ) = 0 ( λ − 1 ) 2 − 4 = 0 Further simplified as, ( λ − 1 ) 2 = 4 λ − 1 = ± 2 λ = 1 ± 2 λ = 3 or λ = − 1 Compute the eigenvector t ( 1 ) = [ t 1 t 2 ] corresponding to the eigenvalue λ = 3 as follows. A t − ( 3 ) t = 0 [ 1 − 3 4 1 1 − 3 ] [ 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 Hence, the possible vector corresponding to the eigenvalue λ = 3 is t ( 1 ) = [ 2 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 Hence, the possible vector corresponding to the eigenvalue λ = − 1 is t ( 2 ) = [ 2 − 1 ] . Thus, the homogenous solution of the system of differential equation is y h = c 1 e 3 t [ 2 1 ] + c 2 e − t [ 2 − 1 ] . 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 y p = [ a 1 b 1 ] cos t + [ a 2 b 2 ] sin t . Differentiate the solution with respect to t . y ′ p = [ a 1 b 1 ] d d t ( cos t ) + [ a 2 b 2 ] d d t ( sin t ) = − [ a 1 b 1 ] sin t + [ a 2 b 2 ] cos t Substitute y p = [ a 1 b 1 ] cos t + [ a 2 b 2 ] sin t in the equation y ′ = [ 1 4 1 1 ] y + [ − 2 cos t − cos t + sin t ] . y ′ p = [ 1 4 1 1 ] ( [ a 1 b 1 ] cos t + [ a 2 b 2 ] sin t ) + [ − 2 cos t − cos t + sin t ] = [ 1 4 1 1 ] [ a 1 b 1 ] cos t + [ 1 4 1 1 ] [ a 2 b 2 ] sin t + [ − 2 cos t − cos t + sin t ] = [ a 1 + 4 b 1 a 1 + b 1 ] cos t + [ a 2 + 4 b 2 a 2 + b 2 ] sin t + [ − 2 cos t − cos t + sin t ] = [ a 1 + 4 b 1 − 2 a 1 + b 1 − 1 ] cos t + [ a 2 + 4 b 2 a 2 + b 2 + 1 ] sin t Compare the coefficient sin t on both sides and obtain the system equation as below

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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