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Chapter 6 Solutions
DIFFERENTIAL EQUATIONS W/WILEYPLUS
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- Find two linearly independent solutions of y" + 1xy = 0 of the form 1+ a3x³ + α6x6 +. Y₂ = x + b₁x¹ +b7x7 +· Y1 Enter the first few coefficients: a3 а 6 b4 = b7 || || ||arrow_forwardObtain 2 linearly independent solutions valid near the origin for x>0 2ху" + 5(1 — 2х)у — 5у 3 0.arrow_forwardFind two linearly independent solutions of 2x²y" - xy + (−4x + 1)y = 0, x > 0 of the form Y₁ = x¹(1+ a₁x + a₂x² + aşx³ + ...) Y2 = : x*²(1+b₁x+b₂x² + b3x³ + ...) where T₁ > T2. Enter 71 = a1 = a2 = az = r2 = b₁ = b₂ = b3 Inarrow_forward
- Find two linearly independent solutions of 2x2y"- xy' + (-4x + 1)y = 0, x > 0 of the form Y1 = r" (1+ a1r+ azx² + a3r³ +) Y2 = r" (1+ bjx + b,x² + bzx³ + ...) where ri> T2 Enter Tiㅋ 1 a1 ㅋ-2 a2 ㅋ |6/5 a3 = T2 ヨ 1/2 b, = 4a0 b2 = by 1arrow_forward3 = [ ¹₂ ], b = [ ³ ], c = [ ¹₁ ], d = [¯^¹'], and v = [¯2²], 5. Given vectors a = (a) Find all real numbers x₁, x2, x3, x4, such that x₁a + x₂b + x3c + x4d = v (b) Write [at least] one sentence about what you have done above using the following mathematical term: linear combination. (c) Write [at least] one sentence about what you have done above using the following mathematical term: span.arrow_forwardof the form y₁ = (1+₁+ a₂²+az³ + ...) 3₂ = x2(1+b₁x + b₂x² + b₂x³ + ...) where T₁ > T2- Enter Find two linearly independent solutions of 2x²y" - xy + (2x+1)y=0, z>0 T1 1 01 = a₂= 03 = T2= 1/2 b₁ = b₂ = b3 =arrow_forward
- Q1. If the components of a contravariant vector in (x') coordinate system are (3,4). Find its components in (x') coordinate system, where x' = 7x – 5x² x2 = -5x + 4x?arrow_forwardVerify that (AB)" = BTAT. 1 -2 1 4 -1 А — and B = 2 0 3 1 STEP 1: Find (AB)'. (AB)™ = STEP 2: Find B'A'. BTAT = STEP 3: Are the results from Step 1 and Step 2 equivalent? Yes O Noarrow_forwardFind two linearly independent solutions of 2x²y - xy +(-3x+1)y=0, x > 0 of the form = x¹(1+x+q₂x² + x² + ...) 3₂ = x¹(1 + b₂x + b₂x² + b₂x² + ...) where r > 7₂.arrow_forward
- Find the general solution of u' = u – 2v – 2w v' = -2u + v + 2w w' = 2u – 2v –- 3warrow_forwardwrite the general solution to x+y+z+w=1 x+y-z-w=1 as a vector in the form of y or w I've tried this so many times and can't get it right, please explain how you got it.arrow_forwardUse Newton's method with the starting point (1,2)T to find a solution of the system +y 0 2+y = 1. The first iteration produces the vector (A) (2,2)T (B) (2,1) (C) (1,1)" (D) (1,2)" (E) none of the above O (A) O (B) O (C) O (D) O (E)arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage