Differential Equation (not sure what order?)

[Saladsamurai]

Homework Statement

I don't think it's a typo since no one 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?

 

[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.

 

[Saladsamurai]

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

 

[Saladsamurai]

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 ..

 

[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.

 

[Saladsamurai]

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

Now note the solution y=k is not a particular case of those solutions however, I don't think it envelopes the general solution so y=k is not a singular solution but I'm not sure.

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 ...

 

[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.