# Homework Help: Differential Equation (not sure what order?)

1. Sep 29, 2010

1. The problem statement, all variables and given/known data

I don't think it's a typo since noone else in my class has asked about it, but the following DE was given on a handout:

$$x\frac{dy}{dx} - \frac{2}{x}\frac{dx}{dy} = 0$$

I am not sure how to deal with this one. Can I get a hint to get me started?

2. Sep 29, 2010

### jackmell

The order is the highest derivative which is one in this case. Also, with first-order derivatives, we can treat them like quotients and write:

$$xy'-\frac{2}{xy'}=0$$

Now, if you divide throughout by xy' then $xy'\ne 0$. Or if it is, then y'=0 or y=k which is one solution to the equation. To find the other one, divide by xy', take square root, keep track of plus and minus, then separate variables and integrate.

3. Sep 29, 2010

Ah.... Thank you jackmell. I think I can take it from here.

4. Sep 29, 2010

Can I just multiply through by y' to get and divide by x to get

(y')2 - 2/x2 = 0

or

(y' - √2/x)(y' + √2/x) = 0

and then solve for each factor by separation?

Hmmm.... then I will get as solution

y(x) = √2ln(x) - √2ln(x) + C1 + C2

y(x) = √2[ln(x) - ln(x)] +C1 + C2

y(x) = C1 + C2

This does not feel right ..

Last edited: Sep 29, 2010
5. Sep 29, 2010

### jackmell

You have:

$$x^2(y')^2=2$$

or:

$$xy'=\pm \sqrt{2}$$

Now just separate variables and obtain:

$$y=\pm \sqrt{2}\log(x)+c$$

Also, I'm not so sure y=k is a solution since plugging that back into the DE causes it to be singular.

6. Sep 29, 2010

Hi jackmell So I got the same thing as you. However, isn't the general solution the superposition of the individual solutions? Hence when you add them, the solution reduces to a constant, i.e. C1 + C2 = k.

I am not sure how the solution is supposed to presented ... I feel like there should be a way around the log's dropping out ...

7. Sep 29, 2010

### jackmell

You mean a higher-ordered linear differential equation is the sum of a set of linearly independent solutions. For this equation however, the solution is simply:

$$y=c\pm \sqrt{2}\log(x)$$

It's easy to back-substitute either one of those and confirm it satisfies the DE.