# Integrating Factor and the Exact Equation $(y-\frac{1}{x})dx+\frac{1}{y}dy=0$.

$$(y - \frac{1}{x})dx + \frac{1}{y}dy = 0 \,\,\, , \,\,\, y(\sqrt[]{2}) = \sqrt[]{2}$$

So I need to find the integrating factor. $\frac{M_y - N_x}{N} = \frac{1 - 0}{0}$ which we cannot have so that leaves us with only $\frac{N_x - M_y}{M} = -1$. This last condition says if the result is dependent on $y$, then the integrating factor is $u(y) = exp(\int \frac{N_x - M_y}{M})$, but our result was a constant. It relies neither on $x$ or $y$. What now?

You have some mistakes, you should divide over $N$, but you divide over $N_y$. this is the correct one $$\frac{M_y - N_x}{N} = y$$ Which is not a function of $x$. So we rewrite the equation as $$y - \frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 0 \\ \frac{dy}{dx} - \frac{1}{x}y = y^2$$
which is a Bernoulli DE

• Ah, yes of course. Still though, we needed for $\frac{M_y - N_x}{N}$ to be dependent on $x$, not $y$. Feb 9, 2014 at 23:48
• Ah! Yes, a Bernoulli DE! I've got it now, thanks. Feb 9, 2014 at 23:53

We can get an exact form $$\left(y - \frac{1}{x} \right)dx + \frac{1}{y}dy = 0 \implies ydx-xdy=xy^2 dx$$ $$\frac{ydx-xdy}{y^2}=xdx \implies d\left(\frac{x}{y} \right)=xdx$$

On integration we get

$$\frac{x}{y} =\frac{x^2}{2} +k$$

$$k$$ is some constant

Now using $$y(\sqrt{2})=\sqrt{2}$$ we get $$k=0$$

Therefore the specific solution is $$y=\frac{2}{x}$$