# how to solve this differential equation $\frac{dy}{dx} = \frac{1+xy}{x(1-xy)}$ by substitution?

I've tried with this differential equation $\displaystyle \frac{dy}{dx} = \frac{1+xy}{x(1-xy)}$ , put $u=xy$ then $\displaystyle\frac{du}{dx}=x\frac{dy}{dx}+y$ So,

It will be after editing $\displaystyle \frac{du}{dx} = \frac{1+u}{(1-u)} + \frac{u}{x}$

I am stuck here, keeping in mind only these three techniques are allowed: substitution, separating variable, and converting it to homogeneous equation

• Are you sure of the ode. Maybe you are asked to solve $\displaystyle \frac{dy}{dx} = \frac{y(1+xy)}{x(1-xy)}$. – Mhenni Benghorbal Sep 25 '13 at 7:10
• @MhenniBenghorbal I've checked the textbook(it's in Arabic) many times, to be sure I've copy it correctly. the ODE that you mentioned is easily solvable , but I think there is a good chance to be a typing mistake. I'll ask my teacher. – SomeOne Sep 25 '13 at 11:20
• Is this a first course in differential equations? – Mhenni Benghorbal Sep 26 '13 at 5:00

Approach $1$:

$\dfrac{dy}{dx}=\dfrac{1+xy}{x(1-xy)}$

$(x-x^2y)\dfrac{dy}{dx}=xy+1$

Let $u=\dfrac{1}{x}-y$ ,

Then $y=\dfrac{1}{x}-u$

$\dfrac{dy}{dx}=-\dfrac{1}{x^2}-\dfrac{du}{dx}$

$\therefore x^2u\left(-\dfrac{1}{x^2}-\dfrac{du}{dx}\right)=x\left(\dfrac{1}{x}-u\right)+1$

$-u-x^2u\dfrac{du}{dx}=2-xu$

$x^2u\dfrac{du}{dx}=(x-1)u-2$

$u\dfrac{du}{dx}=\dfrac{(x-1)u}{x^2}-\dfrac{2}{x^2}$

This belongs to an Abel equation of the second kind.

Let $v=\dfrac{1}{x}$ ,

Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=-\dfrac{1}{x^2}\dfrac{du}{dv}$

$\therefore-\dfrac{u}{x^2}\dfrac{du}{dv}=\dfrac{(x-1)u}{x^2}-\dfrac{2}{x^2}$

$u\dfrac{du}{dv}=(1-x)u+2$

$u\dfrac{du}{dv}=\left(1-\dfrac{1}{v}\right)u+2$

Let $s=u-v$ ,

Then $\dfrac{ds}{dv}=\dfrac{du}{dv}-1$

$\therefore(s+v)\left(\dfrac{ds}{dv}+1\right)=\left(1-\dfrac{1}{v}\right)(s+v)+2$

$(s+v)\left(\dfrac{ds}{dv}+\dfrac{1}{v}\right)=2$

$(s+v)\dfrac{ds}{dv}+\dfrac{s}{v}+1=2$

$(s+v)\dfrac{ds}{dv}=1-\dfrac{s}{v}$

$(v-s)\dfrac{dv}{ds}=v(s+v)$

Approach $2$:

$\dfrac{dy}{dx}=\dfrac{1+xy}{x(1-xy)}$

$(yx+1)\dfrac{dx}{dy}=x-yx^2$

Let $u=x+\dfrac{1}{y}$ ,

Then $x=u-\dfrac{1}{y}$

$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$

$\therefore yu\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-y\left(u-\dfrac{1}{y}\right)^2$

$yu\dfrac{du}{dy}+\dfrac{u}{y}=-yu^2+3u-\dfrac{2}{y}$

$yu\dfrac{du}{dy}=-yu^2+\dfrac{(3y-1)u}{y}-\dfrac{2}{y}$

$u\dfrac{du}{dy}=-u^2+\dfrac{(3y-1)u}{y^2}-\dfrac{2}{y^2}$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $u=\dfrac{1}{v}$ ,

Then $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$

$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dy}=-\dfrac{1}{v^2}+\dfrac{3y-1}{y^2v}-\dfrac{2}{y^2}$

$\dfrac{dv}{dy}=\dfrac{v^3}{y^2}-\dfrac{(3y-1)v^2}{y^2}+v$