The problem is partially completed, make corrections (if any) and complete. It is a Difference equation using an auxiliary equation (homogenous formula for the right-hand side) and trial and error ((nonhomogenous) for the left-hand side. there is an image a sa guide Session 7.2 Second-Order DifferenceEquationsSolving Non Homogenous Equationusing the method of "trial and error" todetermine the PI• Solving Homogenous EquationFind and solve the auxiliaryequation: ar? + br + c = 0 to getf(x) RHSf(x) = cf(x) = mt + c (linear) Y, = Pt + q (linear)Try Y,Y, = K (a constant)rootsCase 1Case 2= mt? + mt + C(quadratic)f(x) = mqtY, = Pt² + qt + r (quadratic)i.) Y, = Pq', if q is not a root of the'auxiliary equation.The roots are real andThe roots are real anddistinctequal2キり+bのroot of the auxiliary equation.Yn = Ar¡' + Br2tYn = (A + Bt)rt%3D%3Dl=P2=r 29,=r=z Y, =Pt°q", if q is a repeated3)9=1 butat2 - Ptq', is q is a root of the 12auxiliary equation.2022-03-24 Complete the following difference equation problem.Yt+2 – 4Yt+1 + 4Y; = 2(2*) + t²-Step 1: Y+2 – 4Y;+1+ 4Y; = 0r2 – 4r + 4 = 0(r – 2)(r – 2) = 0 Hence roots would ber, = 2 and rz = 2 since the roots are real and equal, we use the homogenous equationУh 3D (А + Bt)22Step 2: Particular solutionFrom the equation above, q=2 (RHS equation), where r, = 2 and rz = 2 hence we can say thatq is a root we use the formulaY, = Pt2(2) + Dt² + 2t + rY++1 = P(t + 1)²(2)*+1 + D(t + 1)² + 2(t + 1) +r= Pt? + 2Pt +P * 2(2)t + Dt2 + 2Dt + D + 2t + 2 +rY++2 = P(t + 2)²(2)*+2 + D(t + 2)2 + 2(t + 2) +r= Pt? + 4Pt + 4P * (2)2(2) + Dt2 + 4Pt + 4P + 2t + 4 +2 +rStep 3 Substitute equation in the main equation we get:Pt? + 4Pt + 4P * (2)²(2)t + Dt² + 4Pt + 4P + 2t + 4 + 2 +r– 4(Pt² + 4Pt + 4P *(2)(2) + Dt2 + 4Pt + 4P + 2t + 4 + 2 +r) + 4(Pt²(2)' + Dt² + 2t + r) = 2(2') +t²

The complete solution is given in the attached handwritten notes.

Complete the following difference equation problem. Y++2 - 4Yt+1 + 4Y = 2(2*) + t² Step 1: Y+2 – 4Y++1+ 4Y; = 0 r2 – 4r + 4 = 0 (r – 2)(r – 2) = 0 Hence roots would be r, = 2 and r2 = 2 since the roots are real and equal, we use the homogenous equation Yn = (A+ Bt)2² Step 2: Particular solution

Complete the following difference equation problem. Y++2 - 4Yt+1 + 4Y = 2(2*) + t² Step 1: Y+2 – 4Y++1+ 4Y; = 0 r2 – 4r + 4 = 0 (r – 2)(r – 2) = 0 Hence roots would be r, = 2 and r2 = 2 since the roots are real and equal, we use the homogenous equation Yn = (A+ Bt)2² Step 2: Particular solution

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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