1. ## Solve (x-1)dx-(3x-2y-5)dy= 0.

hi

i want to solve

$(x-1)dx-(3x-2y-5)dy=0$

But I could not classify it
any help?

2. When you have linear functions multiplying the differentials like that, what you want to do is have a substitution based on the point of intersection of the two lines (if they are not parallel, as here). So, the left-hand line is x - 1 = 0, and the right-hand line is 3x - 2y - 5 = 0. Obviously, x = 1, and you can find the corresponding y. So let's call the point of intersection $(x_{0},y_{0}).$ You want to do the substitution $t=x-x_{0},$ and $z=y-y_{0}.$ The resulting DE will be of a form that you know, I think. What do you get?

3. Originally Posted by Miss
hi

i want to solve

$(x-1)dx-(3x-2y-5)dy=0$
@ Ackbeet. Interesting. I've never head of that before, but it looks like a keeper.

I had another idea. Let's turn the usual method on its head: Solve this as x(y) instead of y(x):

I get $\displaystyle (x - 1)\frac{dx}{dy} - 3x = -2y - 5$
which is a much more recognizable form than the original.

-Dan

Uh. D*mn. I guess this one doesn't work out that well.

4. @ Ackbeet:

Actually, the substitutions should be $t=x+x_0$ & $z=y+y_0$.
I know that kind of ODEs, but I forgot it.

@ topsquark:

Well, what can be done for the new form?

5. Originally Posted by Miss
@ Ackbeet:

Actually, the substitutions should be $t=x+x_0$ & $z=y+y_0$.
Well, you can certainly try that substitution, but I don't think it'll do for you what my substitution will do for you. Good luck!

I know that kind of ODEs, but I forgot it.

@ topsquark:

Well, what can be done for the new form?

6. Originally Posted by topsquark
@ Ackbeet. Interesting. I've never head of that before, but it looks like a keeper.
Yeah, I once saw someone on MHF mention that method, and then I saw it in Tenenbaum and Pollard, and so I practiced it quite a few times. A good reason to supplement Zill with T&P!

7. Originally Posted by Miss

@ topsquark:

Well, what can be done for the new form?
Apparently not much. It looks all nice, but even WolframAlpha had to think about it for a minute. (I put the link in the addendum to my first post) It looks similar to a problem I once saw, but apparently I'm remembering wrong.

-Dan

8. I found the method is explained here by the member General :

http://www.mathhelpforum.com/math-he...on-163550.html

9. Originally Posted by Miss
I found the method is explained here by the member General :

http://www.mathhelpforum.com/math-he...on-163550.html
Right; thanks for the link. If you examine his substitution closely, you'll see that it's the same as mine. He just has the $x_{0},y_{0}$ on the other side of the equation. Equivalently, my substitution is $t+x_{0}=x,$ and $z+y_{0}=y.$