# Differential equation question

I was going through the following question: Solve the following differential equation $(1+xy)ydx + (1-xy)xdy=0$.

I have been taught to solve differential equations of the following type:

1.Homogeneous

2.Variable separable

3.Linear differential equation

But, I'm unable to find to which category the given differential equation belongs.Please help. I would like some hints towards solving the question and please don't post the answer.

• Is your expression equal to something ? – Claude Leibovici Dec 4 '13 at 12:11
• One of the $dy$ terms should be a $dx$. Then presumably this is an exact differential equation. – hunter Dec 4 '13 at 12:11
• Sorry for the typo...I was just getting mad trying to solve this question. – Rajath Radhakrishnan Dec 4 '13 at 12:12
• Good luck ! It is just awful to me ! – Claude Leibovici Dec 4 '13 at 12:15
• hmm, it's not quite exact after all. – hunter Dec 4 '13 at 12:18

Set $\xi = xy$, $\eta = \frac{x}{y}$. Then \begin{align} \mathrm{d}\xi &= y\,\mathrm{d}x + x\,\mathrm{d}y & \mathrm{d}\eta &= \frac{y\,\mathrm{d}x - x\,\mathrm{d}y}{y^2} \end{align} And therefore $$\mathrm{d}\xi + \frac{\xi^2}{\eta}\mathrm{d}\eta = 0$$ I suppose, you can take it from there.
$$(1+xy)ydx+(1-xy)xdy=0$$ The given equation is of the form $$f_1(xy)ydx+f_2(xy)xdy=0$$ .Here, $$M=(1+xy)y$$ $$N=(1-xy)x$$ $$\therefore I.F.=\frac1{Mx-Ny}=\frac1{2x^2y^2}$$ $$\therefore\text{the solution is}$$ $$\int_{(\text {y const})}M\times(I.F.)dx+\int (\text{terms of N not containing x})\times(I.F.)dy=c$$ I think you can take help from this..