I just spent more time than I care to admit pushing through the algebra of this question and, although I arrived at the answer, I'm curious to know if there is some simplification that could be done earlier on that would reduce the amount of algebra involved or otherwise improve the method (without introducing entirely new concepts; i.e. sticking to the general framework used here)?

The question: "Find a 1-parameter family of solutions of the differential equation $3xy + y^2 + (x^2+xy)\frac{dy}{dx} = 0$ using the integrating factor $\mu(x,y) = \frac{1}{xy(2x+y)}$"

The work:

Multiply in the integrating factor:

$\frac{3xy + y^2}{xy(2x+y)} + \frac{x^2+xy}{xy(2x+y)}\frac{dy}{dx} = 0$

Normally, I would expect some nice simplification in one of the expressions, but I could not find anything substantial (at least not for both terms) and so after simplifying the first term a bit:

$\frac{3xy + y^2}{xy(2x+y)} = \frac{1}{x} + \frac{1}{2x+y}$

I integrated w.r.t. $x$ to obtain:

$\int \frac{1}{x} + \frac{1}{2x+y} \,dx = \ln x + \frac{1}{2}\ln(2x+y) + k(y)$

Differentiating this expression w.r.t. $y$ should yield an expression equivalent to the coefficient on $\frac{dy}{dx}$ in the first line, above, i.e.:

$\frac{d}{dy}\big(\ln x + \ln(2x+y) + k(y)\big) = \frac{1}{2(2x+y)} + k'(y) = \frac{x+y}{y(2x+y)}$

Isolating for $k'(y)$:

$k'(y) = \frac{x+y}{y(2x+y)} - \frac{1}{2(2x+y)} = \frac{1}{2y}$

therefore $k(y) = \frac{1}{2}\ln y + C$ and an implicit solution is given by $\phi(x,y) = \ln x + \frac{1}{2}\ln(2x+y) + \frac{1}{2}\ln y = c$.

Any insight would be greatly appreciated - usually by the end of such a question I would have an idea of how to construct similar questions, but in this case I do not understand where the complexity is coming from / how it is being created (or from whence it arises).


1 Answer 1


Your approach seems right in itself (although I believe there's a minor mistake). I think the easiest way to see this is to write both terms in partial fractions:

\begin{align} \left(\frac 1x + \frac{1}{2x + y}\right)dx + \frac 12\left(\frac{1}{y} + \frac{1}{2x + y}\right)dy & = 0. \end{align} It is easy that this equation is exact. You can now move things around, like this: \begin{align} \frac 12\left(2\frac{dx}x + \frac{dy}y + \frac{d(2x + y)}{2x + y}\right) & = 0. \end{align} This can now be integrated term by term, yielding $$ 2 \log x + \log y + \log(2x + y) = C. $$

  • $\begingroup$ The partial fractions move was what I was missing. I had never seen it applied to a multivariable context. I'm interested to see how it would apply to higher-order factors in more than one variable. If you happen to know a good source on the topic, I'd be very interested. Thank you kindly for your time and excellent answer! $\endgroup$
    – Rax Adaam
    Feb 7, 2016 at 17:18

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