Finding an exact solution using an integration factor

[julian.irwin]

When you have a DiffEQ in the form Mdx + Ndy = 0 and you want an exact solution, sometimes you need to multiply by an integration factor p(x,y). The book tells me that the integration factor will either be p =(x^m)(y^n), p = f(x) or p = f(y) and then it gives me the procedure used to solve for all of these cases. Very straightforward.

But how besides insight or guess and check can I decide which type of integration factor to use?

Are there any tricks, rules or patterns?

Let me know if I should clarify further.

 

[rock.freak667]

Given M(x,y)dx+N(x,y)dy =0 is not exact, then an integrating factor can be found as follows:

1) if (∂M/∂y - ∂N/∂x)/N = f(x) then e^{∫f(x) dx} is an integrating factor.

2) If (∂M/∂y - ∂N/∂x)/M = -g(y) then e^{∫g(y) dy} is an integrating factor.

 

[LCKurtz]

When you have a DiffEQ in the form Mdx + Ndy = 0 and you want an exact solution, sometimes you need to multiply by an integration factor p(x,y). The book tells me that the integration factor will either be p =(x^m)(y^n), p = f(x) or p = f(y) and then it gives me the procedure used to solve for all of these cases. Very straightforward.

But how besides insight or guess and check can I decide which type of integration factor to use?

Are there any tricks, rules or patterns?

Let me know if I should clarify further.

What you mean is that sometimes there will be an integrating factor of the form f(x), f(y), or x^myⁿ, and if there is, the methods you have learned will find them. Then you can solve the DE, presuming you can actually perform the integrations. Those are a couple of the patterns that you are looking for.

Unfortunately, there is no general rule to find an integrating factor any more than there is a general method to solve a first order DE because, in principle, finding an integrating factor and solving the DE are equivalent. That is to say, these equations always have in integrating factor if they have a solution in the first place.

Look at it this way. Suppose you are given the DE

M(x,y) + N(x,y)y' = 0 (equivalent to the Mdx + Ndy = 0 form)

Now suppose f(x,y) = C defines y implicitly as a solution of the DE. Then by implicit differentiation you have

y' = -\frac{f_x}{f_y}

Substituting this in the DE gives

M(x,y) - N(x,y) \frac{f_x}{f_y} = 0
M(x,y)f_y=N(x,y)f_x
\frac{f_y}{N(x,y)}=\frac{f_x}{M(x,y)}

Since these are equal we can multiply the equation M(x,y) + N(x,y)y' = 0 by this integrating factor, using the left side to multiply the N and the right side to multiply the M, giving:

\frac{f_x}{M(x,y)}M(x,y)+\frac{f_y}{N(x,y)}N(x,y)y' = 0
f_x + f_y y' = 0

This is an exact derivative which immediately gives f(x,y) = C.

Of course, you wouldn't have known this integrating factor without knowing the solution in the first place. But in principle, it is always there. Sometimes, with special methods as you have seen, you can get lucky and find an integrating factor without knowing the solution ahead of time.