## Differential Equations question

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branjb
Posts: 5
Joined: Mon Jul 21, 2008 11:09 pm UTC

### Differential Equations question

Hello,

I have a differential equation of:

xy' = y + sqrt(x^2 + y^2)

The answer I have and the answer in the back of my textbook are very similar, but not the same, and I am trying to figure out the falacy(if any) in my solution, but can't find it.

I divided the entire equation by x, and then with the sqrt term I multiplied the top and bottom by 1/x leaving me with:

y' = y/x + sqrt(1 + (y/x)^2)

For my u substitution I used u = y/x, and when switched around gives y' = u + xu'

So after substituting...

u + xu' = u + sqrt( 1 + u^2 )

Equation looks nice at this point, u's cancel out, leaving me with:

x du/dx = sqrt ( 1 + u^2 ) or du / sqrt ( 1 + u^2 ) = dx / x

Integrating both sides, it turns out to be:

ln | sqrt( u^2 + 1 ) + u | = ln | x | + c

raise both sides to e to get rid of the ln's, and you get (dropping the absolute values at this point for simplicity):

sqrt( u^2 + 1 ) + u = x + c

Substitute back in:

sqrt ( (y/x)^2 + 1 ) + y/x = x + c

Multiply by x..

x * sqrt ( (y/x)^2 + 1 ) + y = x * (x + c )

Convert the sqrt back to x^2 + y^2 to drop the extra x..

y = x * ( x + c ) * sqrt ( y^2 + x^2 )

Now, the answer to the solution I have is:

y + sqrt( x^2 + y^2) = Cx^2

I'm assuming that I made some kind of algebriac error somewhere, but I can't find it. Sorry for oversimplifying steps, I didn't want to leave anything out. Thanks for any help.
Last edited by branjb on Tue Jul 22, 2008 2:10 am UTC, edited 1 time in total.

lgonick
Posts: 17
Joined: Sun Jun 22, 2008 2:48 pm UTC

### Re: Differential Equations question

I don't follow your very first step. Maybe it's because you haven't included enough parentheses? If the original eqn is

[imath]xy' = (y/x) + [sqrt(x^2 + y^2)]/x[/imath]

then isn't

[imath]y' = y/x^2 + sqrt[(1/x^2) + (y/x^2)^2][/imath]

?

ConMan
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### Re: Differential Equations question

branjb wrote:ln | sqrt( u^2 + 1 ) + u | = ln | x | + c

raise both sides to e to get rid of the ln's, and you get (dropping the absolute values at this point for simplicity):

sqrt( u^2 + 1 ) + u = x + c

e^(a + b) != e^a + e^b

Haven't checked where else there might be a mistake.

Quick way to check the two solutions - see which one actually solves the DE (differentiate and substitute, which one works?)
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ConMan
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### Re: Differential Equations question

lgonick wrote:I don't follow your very first step. Maybe it's because you haven't included enough parentheses? If the original eqn is

[imath]xy' = (y/x) + [sqrt(x^2 + y^2)]/x[/imath]

then isn't

[imath]y' = y/x^2 + sqrt[(1/x^2) + (y/x^2)^2][/imath]

?

No, his reasoning there is fine:

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

Edit: No, wait ... I see what's happening there. There is a missing x in there.
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branjb
Posts: 5
Joined: Mon Jul 21, 2008 11:09 pm UTC

### Re: Differential Equations question

My mistake on posting the original equation.. I got ahead of myself and just copied from my notes. I already edited it, but it should be:

xy' = y + sqrt(x^2 + y^2) / x

the y/x was just getting ahead of myself and dividing everything by x.

branjb
Posts: 5
Joined: Mon Jul 21, 2008 11:09 pm UTC

### Re: Differential Equations question

ConMan wrote:
branjb wrote:ln | sqrt( u^2 + 1 ) + u | = ln | x | + c

raise both sides to e to get rid of the ln's, and you get (dropping the absolute values at this point for simplicity):

sqrt( u^2 + 1 ) + u = x + c

e^(a + b) != e^a + e^b

Haven't checked where else there might be a mistake.

Quick way to check the two solutions - see which one actually solves the DE (differentiate and substitute, which one works?)

Hm, I think this is the mistake I made?

It should have been e^(ln | x + c |) = xC ?

That would let me multiply the last x to get x^2C, which is the correct solution.

ConMan
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### Re: Differential Equations question

That looks a lot better to me.
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lgonick
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Joined: Sun Jun 22, 2008 2:48 pm UTC

### Re: Differential Equations question

And in the original equation was the square root also divided by x? I'm guessing not.

branjb
Posts: 5
Joined: Mon Jul 21, 2008 11:09 pm UTC

### Re: Differential Equations question

lgonick wrote:And in the original equation was the square root also divided by x? I'm guessing not.

You are correct. I wrote it from the book into my notes like that, so transcriped it wrong onto here.