In Problems 5–14 solve the given linear system.
11.
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A First Course in Differential Equations with Modeling Applications (MindTap Course List)
- 9. Deternine if S = (7 - 4x + 4x²,6 + 2x − 3x²,20 − 6x + 5x²) is linearly independent or dependentarrow_forwardSection 2.2 2.1. Solve the following difference equations: (a) Yk+1+Yk = 2+ k, (b) Yk+1 – 2Yk k3, (c) Yk+1 – 3 (d) Yk+1 – Yk = 1/k(k+ 1), (e) Yk+1+ Yk = 1/k(k+ 1), (f) (k + 2)yk+1 – (k+1)yk = 5+ 2* – k2, (g) Yk+1+ Yk = k +2 · 3k, (h) Yk+1 Yk 0, Yk = ke*, (i) Yk+1 Bak? Yk (j) Yk+1 ayk = cos(bk), (k) Yk+1 + Yk = (-1)k, (1) - * = k. Yk+1 k+1arrow_forward4. (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_forward
- can you solve (D) (E) (F)arrow_forward3. Determine whether x = T (0 3 0 9 2) is a solution to 3x2 + 2x5 max X1,2,3,4,5 ER subject to x1 + x2x5 x2 + x3 + x5 -2x2 + x4+x5 X1, X2, X3, X4, X5 = = 1 5 5 0 (3)arrow_forwardSection 2.2 2.1. Solve the following difference equations: (a) Yk+1+ Yk = 2+ k, (b) Yk+1 – 2yk = k³, (с) ук+1 "Yk = 0, (d) Yk+1 – Yk = 1/k(k+1), (e) Yk+1+ Yk = 1/k(k+1), (f) (k+2)yk+1 – (k + 1)yk = 5 + 2k – k², (g) Yk+1+ Yk = k + 2 · 3k, (h) Yk+1 – Yk = ke“, Yk = Bak*, = cos (bk), (k) Yk+1 + Yk = (-1)*, Yk – k. ,2k (i) Ук+1 (j) Yk+1 – aYk (1) Yk+1 k+1arrow_forward
- 11 Why can't a 1 by-3 system have xp = (2, 4, 0) and xn = any multiple of (1, 1, 1)?arrow_forward8 - 12F dt 2.arrow_forwardIn case an equation is in the form y = f(ax+by+c), i.e., the RHS is a linear function of x and y. We will use the substitution = ax + by + c to find an implicit general solution. The right hand side of the following first order problem y = (4x − 3y + 1) 5/6 +, y(0) = 0 is a function of a linear combination of x and y, i.e., y = f(ax +by+c). To solve this problem we use the substitution v= ax + by + c which transforms the equation into a separable equation. We obtain the following separable equation in the variables x and v: U' = Solving this equation an implicit general solution in terms of x, u can be written in the form x+ Transforming back to the variables x and y the above equation becomes x+ = C. y = = C. Next using the initial condition y(0) = 0 we find C = 6 Then, after a little algebra, we can write the unique explicit solution of the initial value problem asarrow_forward
- Linear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage Learning