Separation of Variables for PDEs

**brisbane** · May 1st 2010, 11:55 AM

Couple problems:

1. $yX'Y'+XY=0$
2. $yX'Y+xXY'=0$

Where we are searching for $u(x,y)=X(x)Y(y)$

What's throwing me off is the little x and y in these. I have no problem separating stuff like $X'Y+3XY'=0$ , for example. Thanks!

**shawsend** · May 2nd 2010, 10:01 AM

For the first one, divide by $X' Y$ and rearrange to get: $\frac{X'}{X}=-\frac{Y}{yY'}=\lambda$ Now they're separated. The second one divide by what? $xX$ , maybe $xY'$ , $xyY'$ , keep trying with different combinations until you separate the variables.

**brisbane** · May 2nd 2010, 11:51 AM

I know HOW to separate them. I don't know where to go from there because I don't know how to deal with both little x/y and big X/Y.

**shawsend** · May 2nd 2010, 01:46 PM

Lil' x and y are variables and big X and Y are functions. So take for example the first part of the first one: $-\frac{Y}{yY'}=\lambda$ rearranging I get: $Y'+\frac{1}{\lambda y} Y=0$ which you can solve for $Y(y)$ by finding an integrating factor. Same dif' for the other ones in X.

**brisbane** · May 2nd 2010, 03:27 PM

Thanks...care to walk me through it? This isn't a HW problem; I'm preparing for an exam so I need to see how it's done. I had about a 2-year break from Calculus so there's some basic stuff I forget how to do (like integrating factors).

**Danny** · May 3rd 2010, 06:01 AM

Can I ask a question of the first one

Originally Posted by brisbane

Couple problems:

1. $yX'Y'+XY=0$
2. $yX'Y+xXY'=0$

Where we are searching for $u(x,y)=X(x)Y(y)$

Did you really mean $yX'Y'+XY=0$ or $yX'Y+XY'=0$ ?

**brisbane** · May 3rd 2010, 10:23 AM

No, the problem is correct as written. Please, just a clear walkthrough? I'm pretty sure this is a straightforward problem.

**Danny** · May 3rd 2010, 11:32 AM

Originally Posted by brisbane

No, the problem is correct as written. Please, just a clear walkthrough? I'm pretty sure this is a straightforward problem.

If $y X' Y' + X Y = 0$

then

$\frac{y Y'}{Y} = - \frac{X}{X'}$

Since the LHS is only a function of $y$ and the RHS only a function of $x$ then each must be constant. Thus,

$\frac{y Y'}{Y} = - \frac{X}{X'} = \lambda$

1) $\frac{y Y'}{Y} = \lambda$ . Separate $\frac{dY}{Y} = \frac{\lambda}{y}$ so $\ln Y = \lambda \ln y + \ln c_1$ . Thus, $Y = c_1 y^{\lambda}$ .
2) $- \frac{X}{X'} = \lambda$ . Separate $\frac{dX}{X}= - \frac{dx}{\lambda}$ so $\ln X = - \frac{1}{\lambda} x + \ln c_2$ . Thus, $X = c_2 e^{-x/\lambda}$ .

Then multiply $X Y$ and combine your constants to a single constant.

**AllanCuz** · May 3rd 2010, 02:29 PM

Originally Posted by Danny

If $y X' Y' + X Y = 0$

then

$\frac{y Y'}{Y} = - \frac{X}{X'}$

Since the LHS is only a function of $y$ and the RHS only a function of $x$ then each must be constant. Thus,

$\frac{y Y'}{Y} = - \frac{X}{X'} = \lambda$

1) $\frac{y Y'}{Y} = \lambda$ . Separate $\frac{dY}{Y} = \frac{\lambda}{y}$ so $\ln Y = \lambda \ln y + \ln c_1$ . Thus, $Y = c_1 y^{\lambda}$ .
2) $- \frac{X}{X'} = \lambda$ . Separate $\frac{dX}{X}= - \frac{dx}{\lambda}$ so $\ln X = - \frac{1}{\lambda} x + \ln c_2$ . Thus, $X = c_2 e^{-x/\lambda}$ .

Then multiply $X Y$ and combine your constants to a single constant.

You can also let $y = e^{ry}$ and sub into the equation to find r. Whatever you find best!

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