- (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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Question
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
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Q Search for textbooks, step-by-step explanations to homework questions, .
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< 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 →
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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 →
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