2. Solve utt = c²uxx, u(x, 0) = log(1 + x²), u₁(x, 0) = 4+x.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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[Second Order Equations] How do you solve this question?
The second picture is d'Alembert formula
Transcribed Image Text:2. Solve utt = c²uxx, u(x, 0) = log(1 + x²), u₁(x, 0) = 4+x.
![This simplifies to
u(x, t) =
= 1 [6(x +
z[0(x + ct) + 6(x − ct)] +
x+ct
1/2 fitte (s) ds.
-
2c x-ct
(8)
This is the solution formula for the initial-value problem, due to
d'Alembert in 1746. Assuming to have a continuous second derivative
to have a continuous first derivative (C¹), we
(written € C²) and
see from (8) that u itself has continuous second partial derivatives in x and t
(u € C²). Then (8) is a bona fide solution of (1) and (5). You may check this
directly by differentiation and by setting t = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7aab369-35b1-4304-9027-e26df89c8efc%2F3c710caf-45da-47b6-8b98-f5b734f31599%2Fmus63ra_processed.png&w=3840&q=75)
Transcribed Image Text:This simplifies to
u(x, t) =
= 1 [6(x +
z[0(x + ct) + 6(x − ct)] +
x+ct
1/2 fitte (s) ds.
-
2c x-ct
(8)
This is the solution formula for the initial-value problem, due to
d'Alembert in 1746. Assuming to have a continuous second derivative
to have a continuous first derivative (C¹), we
(written € C²) and
see from (8) that u itself has continuous second partial derivatives in x and t
(u € C²). Then (8) is a bona fide solution of (1) and (5). You may check this
directly by differentiation and by setting t = 0.
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