# Homework Help: Differential Equ - sub

1. Sep 14, 2013

### freezer

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

$yy' + x = \sqrt{x^2 + y^2}$

2. Relevant equations

y = vx
y' = v'x + v

3. The attempt at a solution

$y' + \frac{x}{y} = \frac{\sqrt{x^2 + y^2}}{y}$

$(v'x+v) + v =\frac{ \sqrt{x^2 + (vx)^2}}{vx}$

Last edited: Sep 14, 2013
2. Sep 14, 2013

### epenguin

Stuck? No idea what to do with vx - vx ? Not noticed the whole thing has a factor x?

3. Sep 14, 2013

### freezer

Sorry, i typed the problem wrong the first time.

4. Sep 14, 2013

### epenguin

Are you still stuck? It seems to me this is fairly easily expressed in terms of y and a variable x/y.

5. Sep 14, 2013

### freezer

I am sure it is easy one i figure out what i am trying to accomplish with this substition, I am just not certain where it is i am heading for other than d/dx(y') = and something i can integrate on the right side.

6. Sep 14, 2013

### freezer

$v' = \frac{ \sqrt{x^2 + (vx)^2}}{vx^2} - \frac{v}{x}$

7. Sep 14, 2013

### epenguin

The way I look at it if you write it like this

$y' = -\frac{x}{y} + \frac{\sqrt{x^2 + y^2}}{y}$

you see the RHS is just a function of x/y. So let this be one of your variables and keep y as the other.

8. Sep 14, 2013

### freezer

Okay, I am a bit dense. I hopefully walking through this will get me going.

v = y/x, x/y = 1/v

$v'x + v = -\frac{1}{v} + \frac{\sqrt{x^2 + (vx)^2}}{vx}$

9. Sep 14, 2013

### freezer

After talking with another student, they found out this uses v = x^2 + y^2

$v' = 2x + 2yy' = 2(x+yy')$

$\frac{1}{2} \int \frac{1}{\sqrt{v}} = \int dx$

$\sqrt{v} = x + c$

$v = (x+c)^2$

$x^2 + y^2 = (x+c)^2$

$y^2 = -x^2 + x^2 + 2xc + C$

$y = +/- \sqrt{2xc + C}$

Last edited: Sep 14, 2013
10. Sep 14, 2013

### freezer

The answer in the book is:

$x - \sqrt{x^2 + y^2} = C$

And that is not what I got :-(

11. Sep 14, 2013

### epenguin

Yes (9) was well spotted, when you see yy' be looking out for (y2)' .

My thinking was that your d.e. can be written

$$y' = - \frac{x}{y} +\sqrt{\left(\frac{x}{y}\right)^2 + 1}$$

and the RHS is purely a function of (x/y) which we could call u, which looked simple. But that is overlooking that then you have to change y' into dy/du - in fact the LHS becomes

Edit something complicated and I think there is no or less than no advantage in this approach which seemed initially attractive after all.

Last edited: Sep 15, 2013