Step 1 Given nonlinear differential equation; bxn-k Xn+1 = Axn + Bxn-k + Cxp-1 + Dxn-o + [dxn-k-exn-1 .(1)
Step 1 Given nonlinear differential equation; bxn-k Xn+1 = Axn + Bxn-k + Cxp-1 + Dxn-o + [dxn-k-exn-1 .(1)
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 determine red
![Step 1
Given nonlinear differential equation;
bxn-k
Xn+1 = Axn + Bxn-k + Cxn-1 + Dxn-o +
[dxn-k-exn-1]
..... ....
Step 2
Considering equation (1) for some N 2 0,
bxN-k
XN+1 =
= AxN + BXN-k + CxN-1 + DxN-o +
[dxN-k-exN-1
. .(a)
for 2 < XN < 1 and b # e,
given initial condition,
XN-I+1, •.. ,
...., XN-1, XN E 2, 1|
b
S XN-+1,…. , XN-k+1, •·• ,
XN-1, XN < 1
d
applying initial condition,
sa(1) + B(1) + c(1) + >(1) +
(a) = XN+1 < A
bxN-k
XN+1 <A +B + C + D +
[dxN-k-exN-1]
which is named as equation (46) and can be written
as
XN+1 < A +B + C + D +
b(1)
[dxN-k-exN-1]
b
XN+1 <A +B + C + D+
[dxN-k-exN-1]
..(b)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F5fc002be-1341-40b7-a5cf-7819a16e5415%2Fqp3bz5s_processed.png&w=3840&q=75)
Transcribed Image Text:Step 1
Given nonlinear differential equation;
bxn-k
Xn+1 = Axn + Bxn-k + Cxn-1 + Dxn-o +
[dxn-k-exn-1]
..... ....
Step 2
Considering equation (1) for some N 2 0,
bxN-k
XN+1 =
= AxN + BXN-k + CxN-1 + DxN-o +
[dxN-k-exN-1
. .(a)
for 2 < XN < 1 and b # e,
given initial condition,
XN-I+1, •.. ,
...., XN-1, XN E 2, 1|
b
S XN-+1,…. , XN-k+1, •·• ,
XN-1, XN < 1
d
applying initial condition,
sa(1) + B(1) + c(1) + >(1) +
(a) = XN+1 < A
bxN-k
XN+1 <A +B + C + D +
[dxN-k-exN-1]
which is named as equation (46) and can be written
as
XN+1 < A +B + C + D +
b(1)
[dxN-k-exN-1]
b
XN+1 <A +B + C + D+
[dxN-k-exN-1]
..(b)
![< XN-I+1,…
d
XN-k+1,
XN-1, XN < 1
•.. ,
applying initial condition,
+ B(1) + c(1) + D(1)
(а) —>
XN+1 <A(1)
+ F
bxN-k
XN+1 <A + B +C + D +
[dxN-k-exN-1
v-1]
which is named as equation (46) and can be written
as
b(1)
XN+1 <A + B + C +D +
[dxy-k-exn-1]
XN+1 <A + B+ C + D +
[dxy-k-exN-1]
..(b)
Also from initial condition,
< XN-k < 1
* < XN-1 <1
d
d
b< dxn-k <d
XN-1 < 1
= dxN-k 2 b
→ -XN-1 2 -1
» dxN-k
exN-1 2 b – e(1)
(using
→ dxN-k
еxN-1 > b —е
1
dxN-k-exN-1
b-e
above inequalities)
substituting in (b),
b
XN+1 <A + B + C + D +
b-e
which is named as equation (47).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F5fc002be-1341-40b7-a5cf-7819a16e5415%2F0lbeceh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:< XN-I+1,…
d
XN-k+1,
XN-1, XN < 1
•.. ,
applying initial condition,
+ B(1) + c(1) + D(1)
(а) —>
XN+1 <A(1)
+ F
bxN-k
XN+1 <A + B +C + D +
[dxN-k-exN-1
v-1]
which is named as equation (46) and can be written
as
b(1)
XN+1 <A + B + C +D +
[dxy-k-exn-1]
XN+1 <A + B+ C + D +
[dxy-k-exN-1]
..(b)
Also from initial condition,
< XN-k < 1
* < XN-1 <1
d
d
b< dxn-k <d
XN-1 < 1
= dxN-k 2 b
→ -XN-1 2 -1
» dxN-k
exN-1 2 b – e(1)
(using
→ dxN-k
еxN-1 > b —е
1
dxN-k-exN-1
b-e
above inequalities)
substituting in (b),
b
XN+1 <A + B + C + D +
b-e
which is named as equation (47).
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