ODE Integrating factor question

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?

• you don't differentiate, you just assume that there is some function of $y$ that get's factored from exact equation. – Santosh Linkha Sep 26 '13 at 16:35
• 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. – Throwaway Sep 26 '13 at 16:49
• can be found here in detail. – Santosh Linkha Sep 26 '13 at 16:51
• $e^{\int \displaystyle \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}dy}$ is the integrating factor. This is derived in most books. See Zill and Cullen for example. – Wintermute Sep 26 '13 at 16:51
• Thank you, my book did not explain this concept well. – Throwaway Sep 26 '13 at 16:55

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 $$\frac{b'(y)}{b(y)}=\frac{N_x-M_y}{M}$$