hi
i want to solve
$\displaystyle (x-1)dx-(3x-2y-5)dy=0$
But I could not classify it
any help?
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 $\displaystyle (x_{0},y_{0}).$ You want to do the substitution $\displaystyle t=x-x_{0},$ and $\displaystyle z=y-y_{0}.$ The resulting DE will be of a form that you know, I think. What do you get?
@ 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 $\displaystyle \displaystyle (x - 1)\frac{dx}{dy} - 3x = -2y - 5$
which is a much more recognizable form than the original.
-Dan
Uh. D*mn. I guess this one doesn't work out that well.
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 $\displaystyle x_{0},y_{0}$ on the other side of the equation. Equivalently, my substitution is $\displaystyle t+x_{0}=x,$ and $\displaystyle z+y_{0}=y.$