I am coursing differential equations and recently encountered with the concept of integrating factor

gnatopoditw

gnatopoditw

Answered question

2022-06-06

I am coursing differential equations and recently encountered with the concept of integrating factors. I have seen them to solve two types of ODEs: inexact and linear. A linear equation is an ODE in the form:
d y d x + p ( x ) y = q ( y )
the integrating factor ends up being:
u ( x ) = e p ( x ) d x
so that the equation comes to:
d d x ( u y ) = q ( x ) u ( x )
the equation can now be solved if q ( x ) u ( x ) d x can be computed.
An inexact equation is an equation in the form
A ( x , y ) d x + B ( x , y ) d y = 0
where
A y B x
(i.e. A d x + B d y is not an exact differential)
The integrating factor for these equations (I will call it μ for inexact equations) is a function such that
( μ A ) y = ( μ B ) x
Expanding,
μ A y + μ y A = μ B x + μ x A
I have read the Wikipedia article, which says that to solve this equation where μ = μ ( x , y ) requires partial differential equations, but if μ = μ ( x ) or μ = μ ( y ), then there is a straightforward formula for both, in terms of A and B (and their partial derivatives, respectively). But here is the important part: it says
"[...] in which case we only need to find μ with a first-order linear differential equation or a separable differential equation [...]"
Does this mean that this method can only be used for linear ODEs? In that case, I think the first method is way faster.

Answer & Explanation

aletantas1x

aletantas1x

Beginner2022-06-07Added 22 answers

No, it is referring to how to find μ, not the form of the original D.E.
Consider the equation you quote:-
μ A y + μ y A = μ B x + μ x A
If μ = μ ( x ) then this equation becomes
μ A y = μ B x + μ x A
μ ( A y B x ) = μ x A
This is the 'straightforward equation' referred to in the article.

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