0
$\begingroup$

Show that $M(x,y) + N(x,y) \displaystyle \frac{dy}{dx}$ = 0 has an integrating factor that is a function of y alone provided

$ \displaystyle \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}$ is a function of y alone. What first order linear (or seperable) equation do you solve to find this integrating factor?

$\endgroup$
  • $\begingroup$ you don't differentiate, you just assume that there is some function of $y$ that get's factored from exact equation. $\endgroup$ – Santosh Linkha Sep 26 '13 at 16:35
  • $\begingroup$ I don't really understand what you mean by that, also i'm a bit confused because I thought for any integrating factor for a exact transformation is e^R(x)dx, thus, always resulting in a function with x. $\endgroup$ – Throwaway Sep 26 '13 at 16:49
  • $\begingroup$ can be found here in detail. $\endgroup$ – Santosh Linkha Sep 26 '13 at 16:51
  • $\begingroup$ $e^{\int \displaystyle \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}dy}$ is the integrating factor. This is derived in most books. See Zill and Cullen for example. $\endgroup$ – Wintermute Sep 26 '13 at 16:51
  • $\begingroup$ Thank you, my book did not explain this concept well. $\endgroup$ – Throwaway Sep 26 '13 at 16:55
0
$\begingroup$

If the differential equation $$ b(y ) M(x,y) \,dx + b(y) N(x,y) \,dy = $$ is exact, then $$ \partial_x \; b(y)N(x,y)=\partial_y\, b(y)M(x,y) $$ Then $$ b(y)N_x=b'(y)M+bM_y $$ Or equivantly $$ \frac{b'(y)}{b(y)}=\frac{N_x-M_y}{M} $$

$\endgroup$

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.