In Problems 1–8 use the method of undetermined coefficients to solve the given nonhomogeneous system.
1.
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- 4. (S.10). Use Gaussian elimination with backward substitution to solve the following linear system: 2.r1 + 12 – 13 = 5, 1 + 12 – 3r3 = -9, -I1 + 12 +2r3 = 9;arrow_forward3. 2хydx - (3xу + 2y?)dy %3D0 o (x - 2y)*(2x +y) = c (х — у)"(х + у) %3 с (х + 2y) (2х- у)* %3 с (x – 2y)* = c(2x + y)arrow_forward4. Solve the system dt -1 with a1 (0) = 1 and 2(0) = -1.arrow_forward
- 6. (2x +3y = 0 /2x x+2y =-1 9. (1 1 -X+-y 5 1, -x+y%3D10 4arrow_forward13 Solve the following linear system of DE; x' = Añ. 9x15x2 + 3x3 4x2 + 3x3 O 13arrow_forward4. Use the Gauss-Siedel method to approximate the solution of the following system of linear equations. (Hint: you can stop iteration when you get very close results in three decimal places.) 5х1 — 2х2 + 3хз = -1 — 3х1 + 9х2 + хз — 2 2x1 – x2 – 7x3 = 3arrow_forward
- a " OMANTEL If E (X) = -2 and V(X) = 6 () Е(5X — 2) 3 5 E(X) — Е(2) — (5x (-2)) — 2 %3D -12 (iї) V(5X — 2) %3D 52 V(X) — V(2) 3 (25 х 6) — 0 %3D150 (iї) V(-5X — 2) %3D (-5)?V(X) — V(2) 3D (25 х 6) —0 %3D 150 (iv) E (X²) =?arrow_forward.The system x′=3(x+y−13x3−k),y′=−13(x+0.8y−0.7)x′=3(x+y−13x3−k),y′=−13(x+0.8y−0.7) is a special case of the Fitzhugh–Nagumo16 equations, which model the transmission of neural impulses along an axon. The parameter k is the external stimulus. a.Show that the system has one critical point regardless of the value of k.arrow_forwardUse (1) in Section 8.4 X = eAtc (1) to find the general solution of the given system. 1 X' = 0. X(t) =arrow_forward
- 1. Solve x' = Ax + b, if A 4 -3 and b 2t -1 -1 -2 2 -18e3t 72e3t 54e3t 2. Solve x' = Ax + b, if A : 2 4 -1 and b 3 3. Find the general solution to the system d 1 2e2t x+ dt -1 4 e3t for 0 2arrow_forward3. Simple pulley system gives the equations X1 = T - g 2x2 = T – 2g X1 + x2 = 0 (a) Determine X1, X2 and T if g = 10 (b) Verify your solutions using Gaussian eliminationarrow_forwardThis is the first part of a two-part problem. Let P = cos(6t) y(t) = |- (sin(6t)) | -6 sin(6t) , Y2(t) = -6 cos(6t) a. Show that y1 (t) is a solution to the system y = Py by evaluating derivatives and the matrix product (t) -6 0 Enter your answers in terms of the variable t b. Show that y2 (t) is a solution to the system y = Pj by evaluating derivatives and the matrix product y2(t) Enter your answers in terms of the variable t.arrow_forward
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