# Homework Help: Non linear, non exact first order DE

1. Sep 16, 2011

### fluidistic

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Not really sure.

3. 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?

2. Sep 16, 2011

### jackmell

Ok, that one is linear in the two variables (x and y). There is a section in any DE text book 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.

3. Sep 16, 2011

### 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. :)

4. Sep 16, 2011

### 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. :)

5. Sep 16, 2011

### icystrike

It is not very simplified yet.

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

6. Sep 16, 2011

### fluidistic

Hmm I see but how can this help me?

7. Sep 17, 2011

### 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.

8. Sep 17, 2011

### 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.

9. Sep 17, 2011

### LCKurtz

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

10. Sep 17, 2011

### fluidistic

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?