Cannot figure out how to solve this differential equation

1. Feb 10, 2008

cross652

dy/dx= 1/(x^2-xy)

I have tried all the basic steps in an attempt to make it take the form of:
separable
exact
homogenous
bernoulli
linear

I have tried substitution with:
y=ux
u=x-y
u=x^2-y

I cannot for the life of me figure this out, thank you for any help.

2. Feb 10, 2008

arildno

Well, you might try do invert it, and solve instead:
$$\frac{dx}{dy}=x^{2}-xy$$
Isn't that sort of Bernoulli-like? Don't remember all those tricks any longer..

EDIT:

If I have done this correctly, we should get something like:
$$x(y)=\frac{e^{-\frac{y^{2}{2}}}{\int{e}^{-\frac{u^{2}}{2}}du$$

Last edited: Feb 10, 2008
3. Feb 10, 2008

coomast

Arildno, there seems to be something wrong with the formula, the latex image failed to generate. I think you forgot a \frac.

Anyway, this is the right approach to solving the equation, don't forget the integration constant :-)

Some more explanation and intermediate steps might be useful. The equation will become after setting z=x and t=y (in order to avoid mistakes)

$$\frac{dz}{dt}+tz-z^2=0$$

This is a Riccati type equation, which is in it's general form:

$$\frac{dz}{dt}+Q(t)z+R(t)z^2=P(t)$$

and can be transformed into a second order linear differential equation by using the substitution:

$$z(t)=\frac{1}{R(t)u}\frac{du(t)}{dt}$$

In our case this is:

$$z(t)=\frac{-1}{u}\frac{du(t)}{dt}$$

Differentiating this gives (leaving out the independent variable notation):

$$\frac{dz}{dt}=\frac{1}{u^2}\left(\frac{du}{dt}\right)^2 -\frac{1}{u}\frac{d^2u}{dt^2}$$

Substituting this in the orginal equation gives:

$$\frac{d^2u}{dt^2}=-t\frac{du}{dt}$$

Which has the following solution:

$$\frac{du}{dt}=A\cdot e^{-\frac{t^2}{2}}$$

And thus:

$$u=A\int e^{-\frac{t^2}{2}}dt+B$$

Using the transformation:

$$z(t)=-\frac{1}{u}\frac{du}{dt}=-\frac{A\cdot e^{-\frac{t^2}{2}}} {A\int e^{-\frac{t^2}{2}}dt+B}$$

Which is after dividing with A and setting B/A=C:

$$z(t)=-\frac{e^{-\frac{t^2}{2}}} {\int e^{-\frac{t^2}{2}}dt+C}$$

[Edit 2]

It is indeed the solution, substituting back in the original equation shows this.

There might be easier ways to solve this, but I can't think of any.

Last edited: Feb 10, 2008
4. Feb 10, 2008

Rainbow Child

Another way to deal with equation is the following.
Change the independent variable $x=e^t$, thus $y'=e^{-t}\,\frac{d\,y}{d\,t}$ in order to have for the inital ODE

$$\frac{d\,y}{d\,t}=\frac{1}{e^t-y}\Rightarrow \frac{d\,t}{d\,y}=e^t-y$$

Now define $t(y)=u(y)-\frac{1}{2}\,y^2$ in order to get rid of $y$ and make the ODE separable, i.e.

$$\frac{d\,u}{d\,y}=e^{u-\frac{1}{2}\,y^2}\Rightarrow e^{-u}\,d\,u=e^{-\frac{1}{2}\,y^2}\,d\,y\Rightarrow -e^{-u}=\int e^{-\frac{1}{2}\,y^2}\,d\,y+C\Rightarrow -e^{t+\frac{1}{2}\,y^2}=\int e^{-\frac{1}{2}\,y^2}\,d\,y+C\Rightarrow -\frac{1}{x}\,e^{\frac{1}{2}\,y^2}=\int e^{-\frac{1}{2}\,y^2}\,d\,y+C$$

yielding to

$$x=-\frac{e^{-\frac{1}{2}\,y^2}}{\int e^{-\frac{1}{2}\,y^2}\,d\,y+C}$$

i.e. coomast's solution.

5. Apr 13, 2009

arildno

Okay, an old Latex error finally rectified..