I dont understand why X=acoshax+bsinhax X=acosax+bsinax Can you please explain it to me. Thank you M (no subject) - praguechan3000@ xM Inbox (18) - pchandr4@bingham xA Homework 16 about:blank#blockedb In Problems 1-16 use separation XaAmazon.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= bartlebyQ Search for textbooks, step-by-step explanations to homework questions, .E Ask an Experte Bundle: Differential Equations with Bou...< Chapter 12.1, Problem 10E >= (ax + b) (C2e= e-0 (Ax + B)= Ax + Bwhere A = acz and B = bc2.When A = -a? < 0, the differential equation becomes X" – aX = 0which gives the solution X = a cosh ax + b sinh ax.Thus, the product solution isи (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 = 0which gives the solution X = a cos ax + b sin ax.Therefore, the product solution isu (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 equationwhen 1 = 0 is Ax + B , when 2 = -a? < 0iseka'i (A cosh ax + B sinh ax) and when 1 = a? > O ise-ka'ı (A cos ax + B sin ax).E Chapter 12.1, Problem 9EChapter 12.1, Problem 11E →Privacy - Termsimage (40).pngimage (39).pngShow all

By using relationship between trigonometric, hyperbolic and exponential functions you can solve the…

- (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

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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