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: 9781119096023

Author: Kreyszig

Publisher: WILEY

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Solve the initial value problem: y' = x3 + y³ 3 , y(1) = 2 xy² y(x) = Hint: Notice that the equation on the right is homogeneous and see Homework exercise 23 in section 1.2 of our textbook to review techniques for solving homogeneous equations. Note that we've been given an intial value of the form y(a) = b where a > 0, so this only determines a solution corresponding to the right half of the graph of In(x), i.e., the part of the graph corresponding to positive values of x. Therefore, we should write In(x) instead of ln(|x|), since the left half of the graph is not determined by the initial condition given.

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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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