Prove the principle of superposition: if y₁ and y2 are solutions of the homogeneous second-orderlinear differential equationL[y] =y" + p(x)y' + g(x)y=0on some interval I, then y = c₁y₁ + C2y2 is also a solution of L[y] = 0 on I for any c₁, c₂ € R.

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Answered: Prove the principle of superposition:…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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You can verify that the differential equation: 2t°y" - 4t(t + 1)y + 4(t +1)y= 0, t> 0 has solutions y1 = 2t and y2 = 3t exp (2t). a. Compute the Wronskian W between y1 and y2. W (t) = | 12-e2t b. The solutions y1 and y2 form a fundamental set of solutions because there is a point to where W(to) + 0: W # 0. Note: It simply wants you to find some number to to put into the first answer blank and plug into the Wronskian to get something nonzero, which goes into the second answer blank.
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Find the second linearly solution of the D.E (2x +1) y" – 4(x+1) y' + 4y = 0, given that : y,(x) = e²*
Consider the following partial differential equation: y(e +1)u, – e"(y² + 1)²u, = 0, y >0 1. By using the separation method with u(r, y) = f(r)+ g(y) to solve (F), we obtain that : (F) 1+ e" df (y² + 1)² dg a) da fip df b) 1+ er dr (y? + 1)? dg dy dg et %3D 1+ e" df c) et dr (y? +1)2 dy d) None of the above. 2. A solution u(x, y) of (F) is ( B is an arbitrary constant): a) u(r, y) = À (In(1+ e*) – In(y² +1)) + B. 1 b) u(x, y) = (In(1 + e")- + B. 2(y² + 1) 1 c) u(r, y) = A + B. 2(y? + 1). d) None af the above.
! Find a second solution y2(r) of the differential equation 22y" + 6xy' + 6y = 0, r>0, by the method of reduction of order given that yı(x) = is a solution of the equation. Compute the Wronskian and show that y1 and y2 are linearly independent.
1. Consider the following partial differential equation: du (T,y) -3 (r, y) + 2u(r, y) = 0....(E). Let a and b be two real numbers. For which values of a and b, the function u(x, y) = f(y)ear + g(y)eb" is a solution of (E), where f(y), g(y) are two arbitrary functions. a) a = 2 and b = 3 b) a = 1 and b= 2 c) a = -2 and b = -1 d) None of the above. 2. Using the characteristic method, the solution of the following partial differential equation: |ru + (y+1)uy = u |u(x,0) = f(x), %3D where f is a given function. a) u(r, y) = (y + 1)f ( b) u(x, y) = (y² + 1)f( ). c) u(r, y) = f() d) None of the above. 3. The partial differential equation uz.Urr+ Urr.Uyy – u -r = 0 is a) Second order quasi-linear non-homogenous equation. b) Second order quasi-linear homogenous equation. c) Second order homogenous linear equation. d) None of the above.
Solve the linear system of equations using Differential Operator method.. show details... 2x'+y'+x+5y3D4t x'+y'+2x+2y3D2 LTMIEditor
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Prove the principle of superposition: if y₁ and y2 are solutions of the homogeneous second-order linear differential equation L[y] =y" + p(x)y' + g(x)y=0 on some interval I, then y = c₁y₁ + C2y2 is also a solution of L[y] = 0 on I for any c₁, c₂ € R.