1. ## question about separable diff equation

Hi,

I have the following separable differential equation

$\displaystyle \frac{dy}{dx} = 1-y^2$

$\displaystyle y^2dy = 1dx$

$\displaystyle \frac{y^3}{3} = x$

$\displaystyle y = (3x)^{1/3}$

I know this is wrong, but i don't understand why. Why can't you solve it like this?

/Jones

2. Take a look at the equation:

$\displaystyle \frac{dy}{dx} = 1 - y^2$

In order to move the $\displaystyle y^2$ to the other side you have to add it, not divide it:

$\displaystyle y^2 + \frac{dy}{dx} = 1$

This isn't something you can separate.... However, if you take what is on the right hand side of the original equation and divide ALL of it throughout on both sides, you will get something interesting:

$\displaystyle \frac{1}{1 - y^2}\frac{dy}{dx} = 1$

Now, by the property of the differential you get:

$\displaystyle \frac{1}{1 - y^2}~dy = ~dx$

This is what you need to integrate.

3. Originally Posted by Aryth
Take a look at the equation:

$\displaystyle \frac{dy}{dx} = 1 - y^2$

In order to move the $\displaystyle y^2$ to the other side you have to add it, not divide it:

$\displaystyle y^2 + \frac{dy}{dx} = 1$

This isn't something you can separate.... However, if you take what is on the right hand side of the original equation and divide ALL of it throughout on both sides, you will get something interesting:

$\displaystyle \frac{1}{1 - y^2}\frac{dy}{dx} = 1$

Now, by the property of the differential you get:

$\displaystyle \frac{1}{1 - y^2}~dy = ~dx$

This is what you need to integrate.
you can integrate the left side either with partial fractions or trig sub.

4. Like this?

$\displaystyle \int \frac{1}{1-y^2}dy = \int 1 dx$

= $\displaystyle ln|1-y^2| = x$

5. Originally Posted by Jones
Like this?

$\displaystyle \int \frac{1}{1-y^2}dy = \int 1 dx$

= $\displaystyle ln|1-y^2| = x$
NO! You can't just ignore the fact that $\displaystyle 1- y^2$ is itself a function of x. The derivative of $\displaystyle ln|1- y^2|$ is $\displaystyle \frac{1}{1- y^2}$ times the derivative of $\displaystyle 1- y^2$ which is -2y.

Instead use "partial fraction" to write $\displaystyle \frac{1}{1- y^2}= \frac{1}{(1- y)(1+ y)}= \frac{A}{1- y}+ \frac{B}{1+ y}$.

Quite frankly, if you are studying differential equations, you are going to have to be a lot better at both algebra and calculus!

6. Originally Posted by HallsofIvy
YES! YES! NO!
Quite frankly, if you are studying differential equations, you are going to have to be a lot better at both algebra and calculus!

you're probably right