I don't understand why T=c2e^-lamda(k)(t). Can you please explain it to me. Thank you

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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I don't understand why T=c2e^-lamda(k)(t). Can you please explain it to me. Thank you

b In Problems 1-16 use separation X
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< Chapter 12.1, Problem 10E >
Problem 11E
Explanation of Solution
The given partial differential equation is k = , k > 0.
Let the solution be u (x, t) = X (x) T (t).
Now, differentiating u partially with respect to x and i gives
*u = X"T and = XT'
Substituting these values in the given partial differential equation
becomes
kX"T = XT'
k = 4
* = = -1
KT
where l is the separation constant.
On solving further it becomes,
X" + AX = 0 and T' + AkT = 0
On solving further it gives, T = c2e¬kt_
For first differential equation consider three cases:
When 1 = 0 the differential equation becomes X" = 0 which gives
the solution X = ax + b.
Thus, the product solution is,
u (x, t) = X (x) T (1)
= (ax + b) (cze¬dkr)
= e-0t (Ax + B)
= Ax + B
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Transcribed Image Text:b In Problems 1-16 use separation X 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-solutions-for-th... = 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 > Problem 11E Explanation of Solution The given partial differential equation is k = , k > 0. Let the solution be u (x, t) = X (x) T (t). Now, differentiating u partially with respect to x and i gives *u = X"T and = XT' Substituting these values in the given partial differential equation becomes kX"T = XT' k = 4 * = = -1 KT where l is the separation constant. On solving further it becomes, X" + AX = 0 and T' + AkT = 0 On solving further it gives, T = c2e¬kt_ For first differential equation consider three cases: When 1 = 0 the differential equation becomes X" = 0 which gives the solution X = ax + b. Thus, the product solution is, u (x, t) = X (x) T (1) = (ax + b) (cze¬dkr) = e-0t (Ax + B) = Ax + B Privacy · Terms
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