When à = 0 the differential equation becomes X" = 0 which gives the solution X = ax + b.

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Chapter2: Second-order Linear Odes
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I dont understand how X=ax+b. 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 >
The given partial differential equation is k = du, k > 0.
Let the solution be u (x, t) = X (x) T (t).
Now, differentiating u partially with respect to x and t gives
u = X"T and du = XT'
Substituting these values in the given partial differential equation
becomes
kX"T = XT'
k = 4
* = 5 =-
where i 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 (t)
= (ax + b) (cze¬dk)
= e-0* (Ax + B)
= Ax + B
where A = ac2 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.
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Transcribed Image Text:b In Problems 1-16 use separation X M Inbox (38,533) - seabhaskar1150 x M Hw 2 - pchandr4@binghamton.e × 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 > The given partial differential equation is k = du, k > 0. Let the solution be u (x, t) = X (x) T (t). Now, differentiating u partially with respect to x and t gives u = X"T and du = XT' Substituting these values in the given partial differential equation becomes kX"T = XT' k = 4 * = 5 =- where i 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 (t) = (ax + b) (cze¬dk) = e-0* (Ax + B) = Ax + B where A = ac2 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. Privacy · Terms
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