Non linear, non exact first order DE

[fluidistic]

Homework Statement

I must solve the following DE: x+y+1+(2x+2y-1)y'=0.
I can't write the DE under the form y'+P(x)y=Q(x) so I can't use the integrating factor method. I checked out of the DE is exact, and it's not.

Homework Equations

Not really sure.

The Attempt at a Solution

I tried a z-substitution but I fell over a non separable DE.
Let z=2x+2y+1 \Rightarrow z'=2+2y' \Rightarrow y'=\frac{z'-2}{2}.
The original DE then turns out to be worth \frac{z}{2}+\frac{1}{2}+\frac{z}{2} (z'-2)=0 \Rightarrow - \frac{z}{2} + \frac{zz'}{2}+\frac{1}{2}=0. Thus -z+zz'=- \frac{1}{2} \Rightarrow -z+z \frac{dz}{dx}=-\frac{1}{2} which isn't separable. Hmm maybe I could use the integrating factor method on this DE and solve for z? Hmm no either, I can't put the DE under the right form.

By the way I have a general question on DE's. The z-substitution I have made would work if and only if the DE was exact?

 

[jackmell]

Ok, that one is linear in the two variables (x and y). There is a section in any DE textbook that covers that. Know what, I'd have to review it too and by the time I did that you could do so also. Best though if you do it so you can learn better that way.

 

[jackmell]

Ok, I got it. You with me on this? Checked it yet? It's like so easy. Just let w=x+y. You got it now I'm sure but still check out the text. :)

 

[fluidistic]

Ok I'm with you.
I wish I was understanding what you understand!
I've let w=x+y, reached that the original DE is worth \frac{w+1}{2w-1}+w'=1. Which is still not separable nor linear in my opinion. I'd like more help. :)

 

[icystrike]

It is not very simplified yet.

1-\frac{w+1}{2w-1}

 

[fluidistic]

It is not very simplified yet.

1-\frac{w+1}{2w-1}

Hmm I see but how can this help me?

 

[jackmell]

Let's start at the beginning:
(x+y+1)dx+(2(x+y)-1)dy=0
and letting w=x+y so that dw=dx+dy:
(w+1)(dw-dy)+(2w-1)dy=0
and then we can separate variables, solve for w in terms of y, then back-substitute w=x+y to get an implicit expression for y in terms of x. You can do that I bet.

 

[fluidistic]

I'm still stuck. How do you separate the variables here?
I have something like f(w)(dw-dy)+dy=0. But I want something of the form f(w)dw+f(y)dy=0. I don't know how to reach this.

 

[LCKurtz]

I'm still stuck. How do you separate the variables here?
I have something like f(w)(dw-dy)+dy=0. But I want something of the form f(w)dw+f(y)dy=0. I don't know how to reach this.

Collect the variables on dw and dy: (...)dw + (...)dy and you will see how to separate the variables.

 

[fluidistic]

Collect the variables on dw and dy: (...)dw + (...)dy and you will see how to separate the variables.

Ok thanks!
I reach y=-x-y-3 \ln (|2-x-y|)+C. Is that the "implicit expression of y"? I bet there's no way to get explicitly y(x), although I guess the implicit function theorem could demonstrate this but I'm not really sure.

By the way, how is this method of the substitution called? What is the method we've used to solve the DE, in other words?