Solve Homogenous Diff EQ: x\sqrt{x^2 - y^2}\,dx = (x+y)(y\,dx - x\,dy)

[Saladsamurai]

Homework Statement

I was given:

x\sqrt{x^2 - y^2}\,dx = (x+y)(y\,dx - x\,dy)\qquad(1)

\Rightarrow x\sqrt{x^2 - y^2}\,dx = xy\,dx - x^2\,dy +y^2\,dx - xy\,dy

Dividing by dx we have

x\sqrt{x^2 - y^2} = xy - x^2\frac{dy}{dx} +y^2 - xy\frac{dy}{dx}

\Rightarrow x\sqrt{x^2 - y^2} -xy - y^2 = -(x^2 +xy)\frac{dy}{dx}\qquad(2)

The Attempt at a Solution

Letting y = ux --> dy/dx = u + x(du/dx) and putting into (2) we have

x\sqrt{x^2(1 - u^2)} - x^2u - u^2x^2 = -(x^2 +x^2u)(u +x\frac{du}{dx})\qquad(3)

Dividing by x^2 gives:

\sqrt{1 - u^2} - u - u^2 = -(1 - u)(\frac{u}{x^2} + \frac{du}{x\,dx})\qquad(4)

Here is where I get stuck ... supposed;y this is separable, but I cannot see it. Did I mess up somewhere? Or am I correct so far and cannot see the next step?

Thanks!

 

[fzero]

Use y = ux in equation (1). You can put it in the form M(x) dx = N(u) du.

 

[Saladsamurai]

Use y = ux in equation (1). You can put it in the form M(x) dx = N(u) du.

Hi fzero Isn't that what I did? I got it to (4) and can not see how to separate u and x.

 

[tiny-tim]

Hi Saladsamurai!

Why are you making this so complicated?

(ydx - xdy)/x² = d(y/x) = du …

carry on from there ​

 

[Saladsamurai]

Hi Saladsamurai!

Why are you making this so complicated?

(ydx - xdy)/x² = d(y/x) = du …

carry on from there ​

Sorry tiny-tim! I (a) do not understand what you are getting at and (b) don't understand why it matters what form my equation takes? I have done it twice like this and arrive at the same (4). Either way, it should be separable.

I don't understand what "(ydx - xdy)/x² = d(y/x) = du …"
means. What are you doing?

 

[tiny-tim]

i'm using the quotient rule for y/x (or the product rule for y * 1/x)

 

[Saladsamurai]

i'm using the quotient rule for y/x (or the product rule for y * 1/x)

Yeah, but I don't understand your approach. Where is that supposed to go? The approach I have always been taught is that if an EQ is homogenous you take 'u*x' and stick it wherever you see a 'y' and hence you must also replace 'dy' with an appropriate substitution. This is what I did. I simplified (or complicated ) (1) to (2), then replaced all of my y's with (ux)'s and replaced dy/dx with u+x(du/dx) and got (4). I see nothing wrong with what I did. Could you tell me why you think this would not work (forgetting for the moment that there might be an easier way).

Can anyone tell me why my approach is failing? I am kind of desperate here ... I thought I knew how to do these, but now I am worried ... big exam tomorrow.

 

[tiny-tim]

The approach I have always been taught is that if an EQ is homogenous you take 'u*x' and stick it wherever you see a 'y' and hence you must also replace 'dy' with an appropriate substitution. This is what I did. I simplified (or complicated ) (1) to (2), then replaced all of my y's with (ux)'s and replaced dy/dx with u+x(du/dx) and got (4). I see nothing wrong with what I did.

Yes, that is the correct approach (except you should have divided by x² at the very start).

Unfortunately, you got the signs wrong in your line following (1), and then you complicated everything by switching some of the RHS to the LHS (if you'd left the RHS as it was, things would have cancelled).

 

[Saladsamurai]

Yes, that is the correct approach (except you should have divided by x² at the very start).

Unfortunately, you got the signs wrong in your line following (1), and then you complicated everything by switching some of the RHS to the LHS (if you'd left the RHS as it was, things would have cancelled).

I must be going senile. Where do you see a sign error?

Here is the OP with renumbered EQs

x\sqrt{x^2 - y^2}\,dx = (x+y)(y\,dx - x\,dy)\qquad(1)

\Rightarrow x\sqrt{x^2 - y^2}\,dx = xy\,dx - x^2\,dy +y^2\,dx - xy\,dy\qquad(2)

Dividing by dx we have

x\sqrt{x^2 - y^2} = xy - x^2\frac{dy}{dx} +y^2 - xy\frac{dy}{dx}\qquad(3)

\Rightarrow x\sqrt{x^2 - y^2} -xy - y^2 = -(x^2 +xy)\frac{dy}{dx}\qquad(4)

The Attempt at a Solution

Letting y = ux --> dy/dx = u + x(du/dx) and putting into (4) we have

x\sqrt{x^2(1 - u^2)} - x^2u - u^2x^2 = -(x^2 +x^2u)(u +x\frac{du}{dx})\qquad(5)

Dividing by x^2 gives:

\sqrt{1 - u^2} - u - u^2 = -(1 - u)(\frac{u}{x^2} + \frac{du}{x\,dx})\qquad(6)

 

[tiny-tim]

sorry, the sign mistake isn't until line (4)

 

[Saladsamurai]

sorry, the sign mistake isn't until line (4)

tiny-tim, I don't mean to sound fresh ... but where is this "sign error" ?! Can you be more explicit? I see no sign error. Perhaps you are mistaken? EDIT: I found an error, but it is not in sign. If you look at line (6) when I divided the very last term by x^2, I distributed 1/x^2 to both factors, which is clearly not correct. The corrected (6) should be

\sqrt{1 - u^2} - u - u^2 = -(1 - u)(u + x\frac{du}{\,dx})\qquad(6*)

but again, I am not sure that this is separable ...

 

[tiny-tim]

oh, you've renumbered the lines, i didn't see that …

your old line (4) is your new line (6)

EDIT: and (6*)

 

[Saladsamurai]

oh, you've renumbered the lines, i didn't see that …

your old line (4) is your new line (6)

EDIT: and (6*)

I thought I was about to go nutty Ok. Taking into account the sign error AND the fact that I "double divided" the last term in (6) by x^2, the newly corrected line (6) should be:

<br /> \sqrt{1 - u^2} - u - u^2 = -(1 + u)(u + x\frac{du}{\,dx})\qquad(6**)<br />

which is separable. I did not have the sign error on paper, and it was not what caused the eqaution to be inseparable, it was crappy algebra skills.

Thanks for your patience tiny-tim! I'll post back with the full solution soon and then I want to try it your crazy d(y/x/) way. I think I now see where you were going with that. Had I divided by x^2 right off the bat, I would have had exactly the expression for du, assuming that I was using y = ux --> u = y/x. However, having never seen that 'trick' before, I never would have thought of it on my own. I definitely am going to try it since it seems like it could be a lifesaver on a test (but only if you know that it's there).

 

[Saladsamurai]

That trick is so good!

 

[fzero]

You've still made things a bit complicated:

<br /> x\sqrt{x^2 - y^2}\,dx = (x+y)(y\,dx - x\,dy)\qquad(1)<br />

is just

<br /> x^2\sqrt{1 - u^2}\,dx = x (1+u) du<br />

or

x\, dx = \sqrt{ \frac{1+u}{1-u} } du.

 

[Saladsamurai]

Hmmm... I don't know. I got the same answer twice with different approaches and it doen't look like that fzero. Either way, I'll take it!

 

[tiny-tim]

Hi Saladsamurai!

Your dx/x solution looks correct to me

(and you've corrected the mistake in sign I made in my post #4)