# Introduction to substitutions

1. Aug 9, 2010

### beetle2

Hi Guy's,
I have just started DE's and I'm having difficulty following my notes on using the substitution method. I was wondering if any one new a good link that could show me some good techniques to use when approaching these problems. Is there a good step by step process which any of you use?

An example in my notes says for

$y'=\frac{y+2y}{y-sx}$

use the substitution $u = \frac{y}{x}$
ie

$y = ux$

Is there a reason why they chose $u = \frac{y}{x}$ ?

regards

2. Aug 9, 2010

### Hurkyl

Staff Emeritus
Because it works!

It was probably thought of because y/x appears all over the right hand side, if you reorganize it. (I assume there's a typo in what you wrote?)

3. Aug 9, 2010

### HallsofIvy

I'm pretty sure that's incorrect. Wasn't it
$$y'= \frac{y+ 2x}{y- sx}$$

In any case, if you divide both numerator and denominator of either
$$\frac{3y}{y- sx}$$
or
$$\frac{y+ 2x}{y- sx}$$
by x you get
$$\frac{3\frac{y}{x}}{\frac{y}{x}- s}$$
or
$$\frac{\frac{y}{x}+ 2}{\frac{y}{x}- s}$$
and taking u= y/x those are
$$\frac{3u}{u- s}$$
and
$$\frac{u+ 2}{u- s}$$

Now you have just the one variable, "u" rather than both "x" and "y".

4. Aug 9, 2010

### beetle2

Sorry it was,

$y '= \frac{y+2x}{y-2x}$

Last edited: Aug 9, 2010
5. Aug 9, 2010

### beetle2

$y ' = \frac{y+2x}{y-2x}$

So like you said if I divide both numerator and denominator by x I get

$y ' = \frac{\frac{y}{x}+2}{\frac{y}{x}-2}$

use the substitution u= y/x

$y ' = \frac{u+2}{u-2}$

now on the left side I still have $y'$
ie

$\frac{dy}{dx} = \frac{u+2}{u-2}$

How does the left hand side become a function wrt $\frac{du}{dx}$ ?

6. Aug 9, 2010

### Char. Limit

Well, you have y=ux, so differentiate both sides wrt x.

7. Aug 9, 2010

### beetle2

So do we differentiate $dy/dx dx$
and $\frac{u+2}{u-2}dx$

8. Aug 10, 2010

### HallsofIvy

??? There is no reason to have that additional "dx".

Since u= y/x, y= xu. Now use the product rule: dy/dx= (dx/dx)u+ x(du/dx)= u+ x du/dx.

The differential equation becomes
$$x\frac{du}{dx}+ u= \frac{u+ 2}{u- 2}$$

$$x\frac{du}{dx}= \frac{u+ 2}{u- 2}- u \frac{u+ 2- u^2- 2u}{u- 2}= \frac{2- u- u^2}{u- 2}$$

$$\frac{u- 2}{2- u- u^2}du= x dx$$