Identify the integrating factor for $x > 0$ for the first order linear equation $xw'+ w = 0$

I'm really confused how to identify the integrating factor for $xw'+ w = 0$ because it appears to be solvable by separation of variables, without the need to find $\mu(t)$.

$$x\frac{dw}{dx}+w=0$$ $$x\frac{dw}{dx}=-w$$ $$\frac{x}{dx}=\frac{-w}{dw}$$ $$\frac{dx}{x}=\frac{dw}{-w}$$

When integrating:

$$\int\frac{1}{x}\,dx=\int\frac{1}{-w}\,dw$$ $$\ln(x)=-\ln(w)+\ln(a)$$ $$\ln(x)=\ln\left(\frac{a}{w}\right)$$

Again, I don't see how an integrating factor would come into play with this separable, first order homogenous linear equation. What would it be?

Any help would be greatly appreciated!

  • $\begingroup$ I'm not familiar with the Exact Approach, but I looked at this. The partial for $M$ would be $0$ and the partial for $N$ would be $\frac{-1}{x^2}$ and $N$ is $\frac{1}{x}$. So the entire expression would become $\frac{-1}{x}$, which sounds reasonable given the constraint in the question, but isn't correct. $\endgroup$ Oct 4, 2016 at 1:44
  • $\begingroup$ Not sure if I got lucky, but I figured it out. $\frac{1}{x}$ happened to be correct, I just had to integrate and exponentiate. So $\int\frac{1}{x}$ is $\ln(x)$, therefore, when exponentiating, $e^{\ln(x)} = x$, which was the correct answer. I appreciate the guidance! $\endgroup$ Oct 4, 2016 at 1:53

1 Answer 1


You said it already, but just to re-iterate in my own words:

The equation can be written as $$x\frac{dw}{dx}+\frac{w}{x}=0$$

This makes the integrating factor $$=e^{\int\frac{1}{x}\,dx}$$

which is then simply $$e^{\ln{x}}=x$$

So you multiply throughout by $x$.


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