Question

Help please. I don't know what to do now.

Integrating Factors found by Inspection
1. x(xdy-ydx) - y^2lnxdx
2. (siny + ysinx) dx - (cosx - cosy)dy=0
3. dx/dy = (x/y)^2 + 2(x/y)
4. (2xy + 4x + 3)dx + (x^2 + 2y - 5)dy = 0
5. y^2(xdy+ydx) = (xy +2)dy
6. y(y^3-x)dx + x(y^3+x)dy = 0

any help is appreciated.

Answer 1

Well, for the first one we can do a substitution, letting y = vx and dy = vdx + xdv:

\[x(xdy-ydx) - y ^{2}\ln(x)dx\]
After substituting we have:
\[x[x(vdx+xdv)-vxdx] - v ^{2}x ^{2}\ln(x)dx\]
I'll simplify that left hand portion some:
\[x[xvdx+x ^{2}dv-vxdx] - v ^{2}x ^{2}\ln(x)dx\]
xvdx cancels inside the brackets and I can multiply in the x term.
\[x ^{3}dv - v ^{2}x ^{2}\ln(x)dx\]
I can now separate variables by moving the v^2 portion to the other side of the equation (I'm assuming this is an actual equation) and dividing out.
\[\frac{ dv }{ v ^{2} } = \frac{ \ln(x)dx }{ x }\]
You can now integrate both sides from here.

Answer 2

ohh. so you can't really apply the default formulas?

Answer 3

Well, you need a way to separate the variables. Sometimes you have problems that are in other forms like y' + p(x)y = q(x), in which there are straightforward methods for solving. But when you don't have a DE that looks like it can be put into any specific form, you need to try and separate the variables.

Answer 4

is there something in the given that I can apply the default formulas? o.o

Answer 5

Well, I guess it depends what default formulas you're referring to. To be honest, I don't think of a lot of these as formulas, but I suppose it depends on what level of DEs you're doing.

Answer 6

I mean, the default formulas for this topic. Our prof gave 7 formulas under this topic. Yet.. I can't see where and how can I use it. @_@

Answer 7

Do you have them available so I can see?

Answer 8

okay. here they are.
\[xdy + ydx = d(xy)\]
\[\frac{ ydx-xdx }{ y^2} = d( \frac{ x }{ y })\]
\[\frac{ x^2dy - 2xydx }{ y^2 } = d(\frac{ x^2 }{ y })\]
\[\frac{ x^2dy - 2xydx }{ x^2 } = d(\frac{ y^2 }{ x })\]
\[\frac{ xdy-ydx }{ x^2+y^2} = d(\tan^{-1} \frac{ x }{ y })\]
\[\frac{ xdy + ydx }{xy } = d(\ln xy)\]

Answer 9

Ah. Yeah, a couple of these are what I used for the first problem, I just use v and x and not x and y for a lot ofthem. Yes, all of these are applicable if you can get the equation into the proper form.

Answer 10

can you try the 1st problem for me? I've tried it, but I can't seem to get it. ._.

Answer 11

Using your substitutions instead of mine, correct? xD

Answer 12

yeap.

Answer 13

Alright, I'll write yours down. They expand upon what I know, so they're useful for me to hang on to :P Alright, lemme give it a try.

Answer 14

Im just writing down your formulas first.

Answer 15

okay. I'll just wait.. ._.

Answer 16

those 'formulas' are just derivatives

Answer 17

I figure they're just extra substitution guides that will help you separate variables, but what I do with y = vx and dy = vdx + xdv usually works fine. I mean, it's kinda similar, but I'm not familiar enough with using those other substitutions to be able to adapt to it on the spot x_x

Answer 18

@aldrin, yeah it is just derivatives..
@psymon, it's okay. I find it hard to adapt too. ._.

Answer 19

For me, those formulas I can see being useful once you know how to use them properly, I just don't. What I'm guessing is that if you can get any one of those derivatives onto one side of the equation, you can integrate them back into what is given, x/y, x^2/y, etc. Im taking a last look :P

Answer 20

Yeah, without seeing it done, not sure how to go about it x_x I can solve it with my method, but not with this. I need to be shown first, lol >.<