Now use partial fractions on the left side.

[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 anyone 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

[itex]y'=\frac{y+2y}{y-sx}[/itex]

use the substitution [itex]u = \frac{y}{x}[/itex]
ie

[itex]y = ux[/itex]

Is there a reason why they chose [itex]u = \frac{y}{x}[/itex] ?




regards

 

[Hurkyl]

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?)

 

[HallsofIvy]

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 anyone 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

[itex]y'=\frac{y+2y}{y-sx}[/itex]

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

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

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

use the substitution [itex]u = \frac{y}{x}[/itex]
ie

[itex]y = ux[/itex]

Is there a reason why they chose [itex]u = \frac{y}{x}[/itex] ?




regards

 

[beetle2]

Sorry it was,

[itex] y '= \frac{y+2x}{y-2x}[/itex]

 

[beetle2]

[itex]y ' = \frac{y+2x}{y-2x}[/itex]

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


[itex]y ' = \frac{\frac{y}{x}+2}{\frac{y}{x}-2}[/itex]

use the substitution u= y/x


[itex]y ' = \frac{u+2}{u-2}[/itex]

now on the left side I still have [itex]y'[/itex]
ie

[itex]\frac{dy}{dx} = \frac{u+2}{u-2}[/itex]

How does the left hand side become a function wrt [itex]\frac{du}{dx}[/itex] ?

 

[Char. Limit]

[itex]y ' = \frac{y+2x}{y-2x}[/itex]

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


[itex]y ' = \frac{\frac{y}{x}+2}{\frac{y}{x}-2}[/itex]

use the substitution u= y/x


[itex]y ' = \frac{u+2}{u-2}[/itex]

now on the left side I still have [itex]y'[/itex]
ie

[itex]\frac{dy}{dx} = \frac{u+2}{u-2}[/itex]

How does the left hand side become a function wrt [itex]\frac{du}{dx}[/itex] ?

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

 

[beetle2]

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

 

[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
[tex]x\frac{du}{dx}+ u= \frac{u+ 2}{u- 2}[/tex]

[tex]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}[/tex]

[tex]\frac{u- 2}{2- u- u^2}du= x dx[/tex]