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- (Ax + B) - Ar + B where A = acz and B = bez. When i--a <0, the differential equation becomes X" – a²X = 0 which gives the solution X = a cosh ax + b sinh ax. Thus, the product solution is u (x. 0) = X (x) T () - (a cosh ax + b sinh ax) (c3e") -i (A cosh ax + B sinh ax) where A = acz and B = bez. When i = a > 0, the differential equation becomes X* + a²X = 0 which gives the solution X = a cos ax + b sin ax. Therefore, the product solution is u (x, () = X (x) T (4) - (a cos ax + b sin ax) (ce") -ci (A cos ax + Bsin ax) Where A- aez and B = bez- Thus, the product solution of the given partial differential equation when à = 0is Ar + B, when à = -a² < 0is dar's (A cosh ax + B sinh ax) and when à = a' > 0 is eưi (A cos ax + B sin ax). + Chapter 12.1, Problem 9E Chapter 12.1, Problem 11E →

- (Ax + B) - Ar + B where A = acz and B = bez. When i--a <0, the differential equation becomes X" – a²X = 0 which gives the solution X = a cosh ax + b sinh ax. Thus, the product solution is u (x. 0) = X (x) T () - (a cosh ax + b sinh ax) (c3e") -i (A cosh ax + B sinh ax) where A = acz and B = bez. When i = a > 0, the differential equation becomes X* + a²X = 0 which gives the solution X = a cos ax + b sin ax. Therefore, the product solution is u (x, () = X (x) T (4) - (a cos ax + b sin ax) (ce") -ci (A cos ax + Bsin ax) Where A- aez and B = bez- Thus, the product solution of the given partial differential equation when à = 0is Ar + B, when à = -a² < 0is dar's (A cosh ax + B sinh ax) and when à = a' > 0 is eưi (A cos ax + B sin ax). + Chapter 12.1, Problem 9E Chapter 12.1, Problem 11E →

Advanced Engineering Mathematics
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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I dont understand why X=acoshax+bsinhax

                                       X=acosax+bsinax

 

Can you please explain it to me. Thank you

M (no subject) - praguechan3000@ x M Inbox (18) - pchandr4@bingham x A Homework 1 6 about:blank#blocked b In Problems 1-16 use separation X a Amazon.com: yarn + A bartleby.com/solution-answer/chapter-121-problem-10e-differential-equations-with-boundary-value-problems-mindtap-course-list-9th-edition/9781337604918/in-problems-116-use-separation-of-variables-to-find-if-possible-product-solutio. B Paused = bartleby Q Search for textbooks, step-by-step explanations to homework questions, . E Ask an Expert e Bundle: Differential Equations with Bou... < Chapter 12.1, Problem 10E > = (ax + b) (C2e = e-0 (Ax + B) = Ax + B where A = acz and B = bc2. When A = -a? < 0, the differential equation becomes X" – aX = 0 which gives the solution X = a cosh ax + b sinh ax. Thus, the product solution is и (х, 1) %3D X (х) T() = (a cosh ax + b sinh ax) (cze-ikt) = eka't (A cosh ax + B sinh ax) where A = ac2 and B = bc2. When A = a? > 0, the differential equation becomes X" + a?X = 0 which gives the solution X = a cos ax + b sin ax. Therefore, the product solution is u (x, t) = X (x) T (1) = (a cos ax + b sin ax) (cze-ikt) = e-ka't (A cos ax + B sin ax) Where A = acz and B = bcz. Thus, the product solution of the given partial differential equation when 1 = 0 is Ax + B , when 2 = -a? < 0is eka'i (A cosh ax + B sinh ax) and when 1 = a? > O is e-ka'ı (A cos ax + B sin ax). E Chapter 12.1, Problem 9E Chapter 12.1, Problem 11E → Privacy - Terms image (40).png image (39).png Show all
Transcribed Image Text:M (no subject) - praguechan3000@ x M Inbox (18) - pchandr4@bingham x A Homework 1 6 about:blank#blocked b In Problems 1-16 use separation X a Amazon.com: yarn + A bartleby.com/solution-answer/chapter-121-problem-10e-differential-equations-with-boundary-value-problems-mindtap-course-list-9th-edition/9781337604918/in-problems-116-use-separation-of-variables-to-find-if-possible-product-solutio. B Paused = bartleby Q Search for textbooks, step-by-step explanations to homework questions, . E Ask an Expert e Bundle: Differential Equations with Bou... < Chapter 12.1, Problem 10E > = (ax + b) (C2e = e-0 (Ax + B) = Ax + B where A = acz and B = bc2. When A = -a? < 0, the differential equation becomes X" – aX = 0 which gives the solution X = a cosh ax + b sinh ax. Thus, the product solution is и (х, 1) %3D X (х) T() = (a cosh ax + b sinh ax) (cze-ikt) = eka't (A cosh ax + B sinh ax) where A = ac2 and B = bc2. When A = a? > 0, the differential equation becomes X" + a?X = 0 which gives the solution X = a cos ax + b sin ax. Therefore, the product solution is u (x, t) = X (x) T (1) = (a cos ax + b sin ax) (cze-ikt) = e-ka't (A cos ax + B sin ax) Where A = acz and B = bcz. Thus, the product solution of the given partial differential equation when 1 = 0 is Ax + B , when 2 = -a? < 0is eka'i (A cosh ax + B sinh ax) and when 1 = a? > O is e-ka'ı (A cos ax + B sin ax). E Chapter 12.1, Problem 9E Chapter 12.1, Problem 11E → Privacy - Terms image (40).png image (39).png Show all
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