Cannot figure out how to solve this differential equation

In summary, the conversation discusses an attempt to solve the equation dy/dx= 1/(x^2-xy) by transforming it into various forms such as separable, exact, homogenous, Bernoulli, and linear. The speaker also mentions trying substitution with different variables but is unable to solve the equation. Another person suggests inverting the equation and solving for x instead, which is similar to the Bernoulli method. After some discussion and calculations, they arrive at the solution x(y)=\frac{e^{-\frac{y^{2}}{2}}}{\int{e}^{-\frac{u^{2}}{2}}du, with an additional explanation and alternative approach using a transformation of variables.
  • #1
cross652
1
0
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.
 
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  • #2
Well, you might try do invert it, and solve instead:
[tex]\frac{dx}{dy}=x^{2}-xy[/tex]
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:
[tex]x(y)=\frac{e^{-\frac{y^{2}{2}}}{\int{e}^{-\frac{u^{2}}{2}}du[/tex]
 
Last edited:
  • #3
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 :-)

[Edit]

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)

[tex]\frac{dz}{dt}+tz-z^2=0[/tex]

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

[tex]\frac{dz}{dt}+Q(t)z+R(t)z^2=P(t)[/tex]

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

[tex]z(t)=\frac{1}{R(t)u}\frac{du(t)}{dt}[/tex]

In our case this is:

[tex]z(t)=\frac{-1}{u}\frac{du(t)}{dt}[/tex]

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

[tex]\frac{dz}{dt}=\frac{1}{u^2}\left(\frac{du}{dt}\right)^2 -\frac{1}{u}\frac{d^2u}{dt^2}[/tex]

Substituting this in the orginal equation gives:

[tex]\frac{d^2u}{dt^2}=-t\frac{du}{dt}[/tex]

Which has the following solution:

[tex]\frac{du}{dt}=A\cdot e^{-\frac{t^2}{2}}[/tex]

And thus:

[tex]u=A\int e^{-\frac{t^2}{2}}dt+B[/tex]

Using the transformation:

[tex]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}[/tex]

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

[tex]z(t)=-\frac{e^{-\frac{t^2}{2}}} {\int e^{-\frac{t^2}{2}}dt+C}[/tex]

[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:
  • #4
Another way to deal with equation is the following.
Change the independent variable [itex]x=e^t[/itex], thus [itex]y'=e^{-t}\,\frac{d\,y}{d\,t}[/itex] in order to have for the inital ODE

[tex]\frac{d\,y}{d\,t}=\frac{1}{e^t-y}\Rightarrow \frac{d\,t}{d\,y}=e^t-y[/tex]

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

[tex]\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[/tex]

yielding to

[tex]x=-\frac{e^{-\frac{1}{2}\,y^2}}{\int e^{-\frac{1}{2}\,y^2}\,d\,y+C}[/tex]

i.e. coomast's solution. :smile:
 
  • #5
arildno said:
Well, you might try do invert it, and solve instead:
[tex]\frac{dx}{dy}=x^{2}-xy[/tex]
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:
[tex]x(y)=\frac{e^{-\frac{y^{2}}{2}}}{\int{e}^{-\frac{u^{2}}{2}}du[/tex]

Okay, an old Latex error finally rectified..
 

Related to Cannot figure out how to solve this differential equation

1. What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It is used to model various physical phenomena in fields such as physics, engineering, and economics.

2. How do I solve a differential equation?

There are various methods for solving a differential equation, depending on its type and complexity. These include separation of variables, integrating factors, and the use of Laplace transforms. It is important to understand the problem and choose the appropriate method for solving it.

3. Why am I having trouble solving this differential equation?

Solving differential equations can be challenging as it requires a combination of mathematical skills, critical thinking, and understanding of the problem at hand. It is common to face difficulties while solving a differential equation, and it may require multiple attempts and approaches to find a solution.

4. Are there any resources available to help me solve differential equations?

Yes, there are various resources available to help with solving differential equations. These include textbooks, online tutorials, video lectures, and software programs. It is also helpful to seek guidance from a mathematics tutor or professor.

5. How can I check if my solution to a differential equation is correct?

One way to check the correctness of a solution to a differential equation is by plugging it back into the original equation and seeing if it satisfies the equation. Additionally, you can use online tools or software programs that can verify the solution for you.

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