# on first order differential equations

My question is

If $u(x,y)$ and $v(x,y)$ are two integrating factors of a diff eqn $M(x,y)dx + N(x,y)dy$, $u/v$ is not a constant. then $u(x,y) = cv(x,y)$is a solution to the differential eqn for every constant $c$. I m totally stuck :(

Another doubt i have is how to derive the singular solution for the Clairaut's equation. i tried it we have $y= px + f(p)$ diff wrt $x$ and considering $dp/dx=0$ we get $p=c$, how to solve the other part?

-
Ask one question per.. question. – Mariano Suárez-Alvarez Sep 18 '11 at 2:48
Are you given both $u$ and $v$ and tasked with showing $u/v$ "is not a constant," or are you given $v$ and tasked with showing the function defined by $u:=cv$ is also an integrating factor? These are totally different questions. Also, an integrating factor is not "a solution to the differential eqn." And precisely how did you go from Clairaut's to $dp/dx=0$ (because that's completely wrong in general)? – anon Sep 18 '11 at 2:55

The fact that $u$ is an integrating factor means that $(u M)_y = (uN)_x$, i.e. $u_x = \frac{u (M_y - N_x) + u_y M }{N}$, and similarly $v_x = \frac{v (M_y - N_x) + v_y M }{N}$. So $(u/v)_x = \frac{u_x v - u v_x}{v^2} = \frac{(u_y v - u v_y) M}{N v^2} = \frac{M}{N} (u/v)_y$. The curves $u/v = c$ satisfy the differential equation $(u/v)_x \ dx + (u/v)_y \ dy = 0$ which is a multiple of $M \ dx + N \ dy = 0$.