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Quasi-homogeneous 1st Order ODEs

Wondering if anyone could explain this example to me, please?

I'm looking at a worked example in our notes and the qu is:

Solve dy/dx = (x + y + 3)/(x − y − 5).

The solution starts:

Put x = x0 + X, y = y0 + Y
We require x0 + y0 + 3 = 0 and x0 − y0 − 5 = 0 which has solution x0 = 1 and y0 = −4.
Then we have dY/dX = (X + Y)/(X − Y).

I understand all of that.

Then, it says:

Put Y = uX to get X*du/dX + u = (1 + u)/(1 − u).

I understand where the RHS came from... but how did they get the LHS of that?

And what happens to the 1 and -4 we found earlier?

Any help would be much appreciated. Thank you :smile:
differentiated uX with respect to X? (just a guess, i shouldnt really be looking at undergraduate)
Reply 2
Avatar for DPLSK
DPLSK
14
funkiichiicka
I understand where the RHS came from... but how did they get the LHS of that?
The LHS is the product rule for differentiation. It is necessary to find dYdX=d(uX)dX\dfrac{\text{d}Y}{\text{d}X}= \dfrac{\text{d}(uX)}{\text{d}X}.

funkiichiicka
What happened to the 1 and -4 we found earlier?
It was used to give us dYdX=X+YX−Y\dfrac{\text{d}Y}{\text{d}X}= \dfrac{X+Y}{X-Y}.

I hope that helps.

Darren
(edited 12 years ago)
Reply 3
Avatar for DFranklin
DFranklin
18
Yes, Y = uX, so ddXY=ddX(uX)\dfrac{d}{dX} Y = \dfrac{d}{dX}(uX).
Original post by DPLSK
The LHS is the product rule for differentiation. It is necessary to find dYdX=d(uX)dX\dfrac{\text{d}Y}{\text{d}X}= \dfrac{\text{d}(uX)}{\text{d}X}.

It was used to give us dYdX=X+YX−Y\dfrac{\text{d}Y}{\text{d}X}= \dfrac{X+Y}{X-Y}.

I hope that helps.

Darren


Thank you.

I thought it was something to do with product rule, but I can't figure out how exactly it was used :frown:

Please can you explain :confused:
Reply 5
Avatar for DPLSK
DPLSK
14
funkiichiicka
Thank you.

I thought it was something to do with product rule, but I can't figure out how exactly it was used :frown:

Please can you explain :confused:


You're welcome.

We know the product rule for differentiation is the following.

ddx(u⋅v)=u⋅dvdx+v⋅dudx\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}

Replace v with X and the required result follows.

I hope that helps.

Darren

P.S.: The centre dot denotes product. It isn't necessary here, but some people choose to use it.
(edited 12 years ago)
Reply 6
Avatar for around
around
13
Original post by funkiichiicka

Thank you.

I thought it was something to do with product rule, but I can't figure out how exactly it was used :frown:

Please can you explain :confused:


uXuX is the function u(X)u(X) multiplied by the function XX. We can just product rule it:

d(uX)dX=dudX+dXdX\dfrac{d(uX)}{dX} = \dfrac{du}{dX} + \dfrac{dX}{dX}
Original post by DPLSK
You're welcome.

We know the product rule for differentiation is the following.

ddx(u⋅v)=u⋅dvdx+v⋅dudx\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}

Replace v with X and the required result follows.

I hope that helps.

Darren

P.S.: The centre dot denotes product. It isn't necessary here, but some people choose to use it.


Aah!!! Got it! Thank you so much!

Thank you for the help. Much appreciated. :biggrin:

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