Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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- Solve the nonhomogeneous system of differential equations 8 -2 2 8 d -2 8 2|x - - e2t| 16 dt 2 8 24 Hint: The characteristic polynomial of the coefficient matrix is -(A – 10)²(A – 4).arrow_forward10arrow_forwardConsider the second-order system of differential equations d²x dt² d²y dt2 =x+3y, The coefficient matrix 1 3 3 3x+y. wwwwww has eigenvectors Hand [1] are 4 and - 2. Select the option that gives the general solution of this system. Select one: ° 1) = 0[¹] con (24) + C₁ [₁] (²0₁₁+₂ [1¹] CA cos(2t) sin(2t) + C3 ev2t [*] = C₁ [1] cos(2t) + 0₂ sin(2t) + C3 cos(√2t) + C4 Casin(√21) = + HHH] co √21) + C [¹] sin (√2) C₁ e2t cos(√2t) C4 C3 and the corresponding eigenvalues [3] = C₁ [1] cos(2t) + C₂ e2t 2t [1] = ₂ [1] ² + ₂ [1] ² ² + ₁ [₁¹] e√²+ + C₁ [ 1¹ ] e-√₂² 2t e C4 A WEW MEXAND sin(2t) + C3 · [₁¹] ₁ √²+ + 0₁ [1¹] ₁ = √² √√2t earrow_forward
- Theorem 1. The following system of equations Un-1Vn-3 Un-3Vn-1 Un+1 = and vn+1 = (35) Vn-1(A+ Bun-1Vn–3) Un-1(C+ Dun-3Vn–1)' (1-А) has a 2-periodic solution if u_3 = u_1, v_3 = v_1, u_3v_3 = u_2v_2 = " and A = 42 C, B = D+ 0. 41 -A), where A = C,B = Proof. Let u _3 = u_1,v_3 = v_1 and u_3v_3 = u_2v_2 = D in the exact equation (34a). Then 2m-1 2m + (1-C) Σ C n-1 l=0 U4n-2 =u-2 |T 2m A2m+1 + (1– A) E A! m=0 1=0 =u-2. (36)arrow_forwardmatrix algebra qusetion need help solving it.arrow_forwardConsider the second-order system of differential equations dy d²x dt² =x-5y, The coefficient matrix 1 -5 eigenvalues are 6 and -4. Select the option that gives the general solution of this system. I dt² C₁ -5x + y. 1 -5 11 1 has eigenvectors Select one: [*] = C₁ [ ¹₁ ] cos(√õt) + C₂ [ ¹₁ ] sin(√6t) + Cs [1] cos(2t) + C₁ [1] C3 C4 sin(2t) evot + Vot C3 - [3] = a₁ [¹₁] ev [*] = C₁ [1] cos(√ēt) + C₂ H ₂[1] HHHHH [+] = C₁₂ [ ¹₁ ] ev√² + C₂ [ ¹₁ ] e-√₂ 3 [1] 2²¹ -C₂₁ [ ²₁ ] e-√² + C₂ [1] sin(√6t) + C3 [1₁] [1] Cos evot +03 e2t+CA C₂ [4] [1] and H cos(√6t) + C₂ and the corresponding cos(2t) + C4 H e2t + C4 sin(2t) -2t [¹1] e ²2 -2t e2t sin (√6t) + C3 as He" + a₁₂H₁² -2tarrow_forward
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