# Using seperation of variables?

• Apr 19th 2010, 10:18 AM
punkstart
Using seperation of variables?
What is the procedure or what are the steps to be carried out here:
$\displaystyle \frac{X\prime}{X}=\frac{1}{4}(\frac{Y}{Y\prime})=\ lambda$
The answer is given as $\displaystyle X=e^{\lambda\alpha}$ and $\displaystyle Y=e^{\frac{\beta}{4\lambda}}$
I don't know how to get to the answer, i have done seperation of variables, but not when i have a derivative as a denominator!
• Apr 19th 2010, 11:01 AM
AllanCuz
Quote:

Originally Posted by punkstart
What is the procedure or what are the steps to be carried out here:
$\displaystyle \frac{X\prime}{X}=\frac{1}{4}(\frac{Y}{Y\prime})=\ lambda$
The answer is given as $\displaystyle X=e^{\lambda\alpha}$ and $\displaystyle Y=e^{\frac{\beta}{4\lambda}}$
I don't know how to get to the answer, i have done seperation of variables, but not when i have a derivative as a denominator!

This produces 2 equations

$\displaystyle X^{ \prime } - \lambda X = 0$

and

$\displaystyle \lambda Y^{ \prime } - \frac{Y}{4} = 0$

This can become

$\displaystyle Y^ { \prime } - \frac{ Y }{ \lambda 4} = 0$

To solve both,

let $\displaystyle X=e^{rx}$

let $\displaystyle Y=e^{qy}$

Therefore the first equation becomes

$\displaystyle re^{rx} - \lambda e^{rx} = 0$

$\displaystyle r= \lambda$

Hence,

$\displaystyle X=C_1 e^{ \lambda x }$

Of course the co-efficient is just a constant to represent the general solution. Which can be re-written in the form that they gave you.

The same follows for the Y equation!