tt xt + (7 The generalized method of separation of variables can also be applie differential equations. We now show this using the example Wtt + (27)wet +(7² – 1)wrx 0,
tt xt + (7 The generalized method of separation of variables can also be applie differential equations. We now show this using the example Wtt + (27)wet +(7² – 1)wrx 0,
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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Show me the steps of deteemine red
![8.2.3
Wit + (27)Wat + (y² – 1)wæx
The generalized method of separation of variables can also be applied to partial
differential equations. We now show this using the example
Wtt + (27)wxt +(y² – 1)wrx = 0,
(8.40)
where w =
w(x, t), wt = ôw/Ət, wx = dw/dx, etc.
Assume w(x, t) to have the form
w(x, t) = X(x)T(t).
(8.41)
Substitute into equation (8.40) and divide by X(x)T(t), to obtain
X'
+ 2y
X
+ (y² – 1)
- 0,
(8.42)
X
where the primes denote differentiation with respect to the appropriate vari-
able, x or t. Now, take
T'
E= -a?, a = arbitrary constant,
(8.43)
an equation whose solution is
(T(t, a?) = A(a²)e¬«'t,
(8.44)
where A(a2) is a function of a². Substituting this result into equation (8.42)
and simplifying gives
-22)X" + (2a²)X' – a*X = 0.
(8.45)
The characteristic equation is
((1- 72)r2 + (2a²)r – a* = 0,
(8.46)
and it has the roots
a?
a2
r1
(8.47)
T2 = -
1+y'
Therefore,
(x(2, a²) = B1(a®)e(#)-
+ B2(a²)e¬()=.
1+y
X (x,
(8.48)
and w(x, t, a²) = X (x, a²)T(t, a²) is
W (x, t, a²) = A1(a²) exp
+ Azla*) exp{- [(1,)
x + a²t
(8.49)
If we now sum/integrate over a², the following result is obtained for the general
solution to equation (8.40)
w(x, t) = f[x – (1+y)t] + g[x + (1 – )t],
(8.50)
where f(z) and g(z) are arbitrary functions of z.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06b6425a-ab8a-4368-90d9-64663800e88c%2F85ca155c-c5a2-4d04-9c99-9f779b00302a%2F6hu415q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:8.2.3
Wit + (27)Wat + (y² – 1)wæx
The generalized method of separation of variables can also be applied to partial
differential equations. We now show this using the example
Wtt + (27)wxt +(y² – 1)wrx = 0,
(8.40)
where w =
w(x, t), wt = ôw/Ət, wx = dw/dx, etc.
Assume w(x, t) to have the form
w(x, t) = X(x)T(t).
(8.41)
Substitute into equation (8.40) and divide by X(x)T(t), to obtain
X'
+ 2y
X
+ (y² – 1)
- 0,
(8.42)
X
where the primes denote differentiation with respect to the appropriate vari-
able, x or t. Now, take
T'
E= -a?, a = arbitrary constant,
(8.43)
an equation whose solution is
(T(t, a?) = A(a²)e¬«'t,
(8.44)
where A(a2) is a function of a². Substituting this result into equation (8.42)
and simplifying gives
-22)X" + (2a²)X' – a*X = 0.
(8.45)
The characteristic equation is
((1- 72)r2 + (2a²)r – a* = 0,
(8.46)
and it has the roots
a?
a2
r1
(8.47)
T2 = -
1+y'
Therefore,
(x(2, a²) = B1(a®)e(#)-
+ B2(a²)e¬()=.
1+y
X (x,
(8.48)
and w(x, t, a²) = X (x, a²)T(t, a²) is
W (x, t, a²) = A1(a²) exp
+ Azla*) exp{- [(1,)
x + a²t
(8.49)
If we now sum/integrate over a², the following result is obtained for the general
solution to equation (8.40)
w(x, t) = f[x – (1+y)t] + g[x + (1 – )t],
(8.50)
where f(z) and g(z) are arbitrary functions of z.
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