Question

show that if an equation m(x y)dx + n(x y)dy=0 has an integration factor f(x) and also an integrating factor g(y) , then the equation is separable .

Answer 1

The integrating factors are
\[f(x)=\exp\left(\int\frac{M_y-N_x}{N}~dx\right)\]
and
\[g(y)=\exp\left(\int\frac{N_x-M_y}{M}~dy\right)\]
So given the equation
\[M~dx+N~dy=0\]
and multiplying both sides by either integrating factor, you have
\[\exp\left(\int\frac{M_y-N_x}{N}~dx\right)(M~dx+N~dy)=0\]
or\[\exp\left(\int\frac{N_x-M_y}{M}~dy\right)(M~dx+N~dy)=0\]
which tells you
\[\begin{align*}\exp\left(\int\frac{M_y-N_x}{N}~dx\right)&=\exp\left(\int\frac{N_x-M_y}{M}~dy\right)\\\\
\int\frac{M_y-N_x}{N}~dx&=\int\frac{N_x-M_y}{M}~dy
\end{align*}\]
Now, how to show that the original equation is separable...

Answer 2

A Warm Welcome to Opens Study

Answer 3

\[\begin{align*}M(x,y)~dx+N(x,y)~dy&=0\\
N(x,y)~dy&=-M(x,y)~dx
\end{align*}\]
For this equation to be separable, \(N\) and \(M\) must be functions of only \(y\) and \(x\), respectively. This means you should have \(N_x=M_y=0\), which means the equation in my previous post would be an identity.

The problem with this reasoning is that I'm working backwards, so you have to establish that the following indeed holds:
\[\int\frac{M_y-N_x}{N}~dx=\int\frac{N_x-M_y}{M}~dy\]
My guess is that there's some theorem out there that says if an equation has two integrating factors like this, then they must be equivalent, or something to that effect.