The second image is a reference. Answer the following items on a clean sheet of paper. Upload photos of your answer sheets showing your solutions.D² ƒ – Df = 0 wmf (0) = Df(0) = 12. Solve the initial value problem,with Solve the initial-value problem:Df – D²f = 0; f(0) = Df (0) = D²f(0) = 3Solution:We apply L to both sides of our ODEL[D®f] – L[D²f] = L[0] = 0%3DBy Theorem 1.2,L[D³f] = s³L[f] – s²f(0) – sDf(0) – D²f(0)= s³L[f] – 3s² – 3s – 3L[D²f] = s?L[f] - sf(0) – Df (0)= s?L[f]– 3s – 3%DSubstituting into the transformed ODE,(s°L[f] – 3s? – 3s – 3) – (s²L[f] – 3s – 3) = 0s³ L[f] – s²L[f] – 3s² = 03s2L[f] =3s3 - s2S- 1f(x) = L-1S -= 3e*

Answered: Image /qna-images/answer/584709f8-341c-40b7-a033-c99905f56572.jpg

Answered: Answer the following items on a clean…

Answer the following items on a clean sheet of paper. Upload photos of your answer sheets showing your solutions. D² f – Df = 0 mm f(0) = Df(0) = 1 2. Solve the initial value problem, with

Algebra: Structure And Method, Book 1

(REV)00th Edition

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Chapter3: Solving Equation And Problems

Section3.5: Equations With The Variable On Both Sides

Problem 16P

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