· y(x + y)dx + (x + 2y 1)ay (determination of integrating factor)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Can you solve this using the flow given for determination of

Transcribed Image Text:Determination of Integrating Factor
We have already mentioned in the previous lesson the aid of integrating factor in solving a
first-order linear differential equation and that the same idea sometimes works for a nonexact
differential equation
M(x, y)dx + N(x, y)dy = 0
(1).
We mean to say that it is sometimes possible to determine an integrating factor u(x, y)
such that after multiplying it to a given nonexact equation (1), the left side of the resulting
equation
u(x, y) M(x, y)dx + u(x, y)N(x, y)dy = 0 (2)
becomes an exact differential.
To determine such integrating factor u(x, y), recall the criterion for exactness. Equation (2)
is exact if and only if
or U
U
Ə
- (UM) =
U
ду
Solving for u, u(u) = ef
ƏM
Əy
ƏM ƏN
dy дх
+ M
ƏN
əx
du ƏN
dy ?x
du
əx
du
U
=
a
da (UN)
= U + N
= N
If u satisfies (3), then u is an integrating factor for (1). Note that M, N,
ƏN
are
dy
əx
known functions of r and y. To determine such function u, from (3), we have to solve for
a partial differential equation, but were not yet ready to do that. Thus, we first make an
assumption that u is a function of only one variable.
Let us first assume that u depends on a alone. Then, =
becomes
Ju du
?х dx
ƏM
Ndu
dr
[ du = 12 (M
u
1 (ƏM ƏN
N dy Ər
du
əx
- M
ƏM ƏN
Əy ?х
ON
N ду дх
Solving for u, we have u(x) =
Similarly, if u depends on y alone, then we will have,
du
ду
It is still not easy to determine u even after taking the integral of both sides of (4) if its
right side depends on both x and y. However, if after some algebraic simplications are made,
1 (OM ON
the expression
turns out to be dependent alone on the variable y, then (4)
is now a first-order ordinary differential equation and we can finally solved for u in (4) by
separation of variables. That is,
ду Әх
ƏM ƏN
L (OM_W) dr
dx
du
1 /OM
ƏN
M ду əx
dy
du
dy
dx
(3)
dx
=-M
1
ƏN
-- (-2) dy
U
M
Əy
Əx
ƏM
, and
Ju
and = 0. Hence, (3)
Əy

Transcribed Image Text:5. y(x + y)dx + (x+2y-1)dy = 0 (determination of integrating factor)
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