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

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

 

[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

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

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".

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

y = ux

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




regards

 

[beetle2]

Sorry it was,

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

 

[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} ?

 

[Char. Limit]

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} ?

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

 

[beetle2]

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

 

[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