# Thread: Separation of Variables for PDEs

1. ## Separation of Variables for PDEs

Couple problems:

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

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

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

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

3. 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.

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

5. 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).

6. Can I ask a question of the first one

Originally Posted by brisbane
Couple problems:

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

Where we are searching for $\displaystyle u(x,y)=X(x)Y(y)$
Did you really mean $\displaystyle yX'Y'+XY=0$ or $\displaystyle yX'Y+XY'=0$?

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

8. 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 $\displaystyle y X' Y' + X Y = 0$

then

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

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

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

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

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

9. Originally Posted by Danny
If $\displaystyle y X' Y' + X Y = 0$

then

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

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

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

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

Then multiply $\displaystyle X Y$ and combine your constants to a single constant.
You can also let $\displaystyle y = e^{ry}$ and sub into the equation to find r. Whatever you find best!