The equation to solve is: $$(x^3y^3+x^2y^2+xy+1)ydx+(x^3y^3-x^2y^2-xy+1)xdy=0$$

I tried putting $xy=t$ but that just gave me this: $$\frac{t^3-t^2-t+1}{t^3+t^2+t+1}dt=\frac{dx}{x}$$

I suppose there must be some clever factoring involved somewhere but I can't see it so can someone guide me on how to advance or perhaps suggest an alternate method?

  • 6
    $\begingroup$ $\frac{t^3-t^2-t+1}{t^3+t^2+t+1}=\frac{(t+1)(t-1)^2}{(t+1)(t^2+1)}=1-\frac{2t}{t^2+1}$. $\endgroup$
    – Feng
    Sep 6, 2019 at 4:54

2 Answers 2


The given equation is of the form $f_1(xy)ydx+f_2(xy)xdy=0.$ Here, $M=(x^3y^3+x^2y^2+xy+1)y$ & $N=(x^3y^3-x^2y^2-xy+1)x$ $$\therefore I.F.=\frac1{Mx-Ny}\\ =\frac1{2x^2y^2(xy+1)}$$ $\text{Multiplying the I.F. with the equation, we get}$ \begin{align}\\ &{\begin{aligned}\\ \frac{(x^3y^3+x^2y^2+xy+1)y}{2x^2y^2(xy+1)}dx &+\frac{(x^3y^3-x^2y^2-xy+1)x}{2x^2y^2(xy+1)}dy=0 \end{aligned}}\\ &{\begin{aligned}\\ \implies\frac{(x^2y^2+1)(xy+1)}{x^2y(xy+1)}dx &+\frac{(x^2y^2-1)(xy-1)}{xy^2(xy+1)}dy=0 \end{aligned}}\\ &{\begin{aligned}\\ \implies\frac{x^2y^2+1}{x^2y}dx &+\frac{(x^2y^2-1)(xy-1)}{xy^2(xy+1)}dy=0 \end{aligned}}\\ &{\begin{aligned}\\ \implies ydx+\frac{dx}{x^2y}+&\frac{(x^2y^2-1)(xy-1)}{xy^2(xy+1)}dy=0 \end{aligned}}\\ \end{align} $\therefore\text{the solution is}$ \begin{align} &{\begin{aligned}\\ \int_{(\text{y const})}Mdx+ &\int(\text{terms of N not containing x}) dy=c \end{aligned}}\\ &{\implies \int ydx+\int\frac{dx}{x^2y}=c}\\ &{\implies xy-\frac{1}{xy}=c}\\ \end{align}


Grouping terms, we get

$$(x^3y^3+1)(ydx+xdy) - (x^2y^2+xy)(xdy-ydx) = 0$$

Now substitute $t=xy$ and $s=\frac{y}{x}$

$$\implies dt = ydx+xdy \hspace{20 pt} x^2 ds = \frac{t}{s}ds = xdy - ydx$$

turning the differential equation into

$$s(t^3+1)dt - t^2(t+1)ds = 0$$

which is now separable and yields

$$\int \frac{ds}{s} = \int \frac{t^3+1}{t^2(t+1)}dt = \int 1 - \frac{1}{t} + \frac{1}{t^2}dt$$

$$\implies \log|s| = t - \log|t| - \frac{1}{t} + C$$

Substituting back in for $x$ and $y$

$$\log (y^2) - xy + \frac{1}{xy} = C$$

or rearranging terms we can have

$$y^2\exp\left(\frac{1}{xy} - xy \right) = C > 0$$


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