1
$\begingroup$

I'm getting ready for my differential equations course, and I read that if you have an equation of the form

$$y' + p(x)y=f(x)$$

then this can be solved by multiplying the entire equation by an integrating factor $u(x)=e^{\int p(s)ds}$ so that it becomes

$$(uy)'=u(x)f(x)$$

then $y(x) = \frac{1}{u(x)} \int u(x)f(x) dx$.

On the other hand, if you have an equation of the form

$$M(x,y)dx+N(x,y)dy = 0$$ then first you check if the equation is exact (check if $M_y=N_x$). If the equation is not exact then you multiply $M$ and $N$ by an integrating factor $u(x,y)$ and force $[u(x,y)M(x,y)]_y=[u(x,y)N(x,y)]_x$. Then you go on to solve it.

My question is, are these two equations really the same? When we use an integrating factor in the first case it is already assumed to be of a certain form..... is it just a special case of the equation of the second type?

I suppose they could be equivalent because we could rewrite $y'+p(x)y=f(x)$ as $M(x)dx + Ndy=0$ where $M(x,y)=M(x)=[p(x)y(x)-f(x)]$ and $N(x,y)=1$ but then $M_y=0=N_x$ however we still need to manipulate the first equation by an integrating factor.

$\endgroup$
  • $\begingroup$ Check again, with $M(x,y)=p(x)y-f(x)$ you get $M_y=p$. $\endgroup$ – LutzL Aug 20 '16 at 16:25
  • $\begingroup$ DOH! I can't believe I missed this... and I am glad to see there is actually a connection. $\endgroup$ – Clclstdnt Aug 20 '16 at 16:30
1
$\begingroup$

In your last test for exactness, $M_y = p(x)$ not zero. Then to test whether there exists an integrating factor of the form $u(x)$, we look at $(M_y-N_x)/N =p(x)$, and since this is a function of $x$ alone, we solve $u^{\prime} = p(x) u$ to the the integrating factor. And, this is exactly (pun intended) the same equation we solve to get the integrating factor if we treat it as a linear equation. So the answer to your question is "yes", they are the same.

$\endgroup$
  • $\begingroup$ Thanks! This makes sense. I see that if $\xi(x)=\frac{M_y-N_x}{N}$ is a function of only $x$ then the I.F. is $\mu(x) = \exp[\int \xi(x) dx]$ and if $\psi(y)=\frac{M_y-N_x}{-M}$ is a function of only $y$ then the I.F. is $\mu(y) = \exp[\int \psi(y) dy]$. Where is a good source for a proof of this? Also what can you say if neither of these fractions are a function of only one variable? No solution? What then? $\endgroup$ – Clclstdnt Aug 20 '16 at 18:34
  • $\begingroup$ I think most DE books have the derivation of these conditions. Boyce and Diprima certainly has it. Basically, you assume that there exists an integrating factor of some form or another (a function of x alone, a function of y alone, a function of xy, a function of x/y, etc) and then plug it into the equation and see what is forced. That results in the "test" and also gives you the equation to solve to get the integrating factor. $\endgroup$ – B. Goddard Aug 20 '16 at 20:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.