# First order DE

1. Sep 13, 2011

### fluidistic

1. The problem statement, all variables and given/known data
Calculate the general solution of the following DE reducible to a homogeneous one: x+y-2+(x-y+4)y'=0.

2. Relevant equations
Not sure.

3. The attempt at a solution
My idea is first to write the DE into the homogeneous form and then solve it via any method that work.
I've read on the Internet that a homogeneous DE is of the form y'(x)=ay(x) where a is a constant.
What I've done is $\frac{dy}{dx}=\frac{2-x-y}{x-y+4}$ for x-y different from 4. So it seems I could write $\frac{2-x-y}{x-y+4} =\alpha y(x)$ but I do not see how it's possible.
I don't really know how to go further.

2. Sep 13, 2011

### HallsofIvy

Staff Emeritus
There are different meaning for "homogeous" for d.e.s of first order and higher order. A first order equation, of the form dy/dx= f(x,y)/g(x,y) is "homogeneous" if replacing both x and in f(x,y)/g(x,y) by ax and ay respectively, gives the same as f(x,y)/g(x,y)- that is, the "a" cancels out. Here, f(x,y)= 2- x- y and g(x,y)= x-y+ 4 and f(ax,ay)= 2- ax- ay and g(ax, ay)= ax- ay+ 4 so that f(x,y)/g(x,y)= (2- ax- ay)/(ax- ay+ 4) which does not equal f(x,y)/g(x,y). This equation is not "homogeneous" in that sense.

For higher order, linear equations, a differential equation is "homogeneous" if there are NO terms that does not involve y or a derivative of y. This d.e. is not a linear equation to begin with so it is NOT homogeneous in that sense.

3. Sep 13, 2011

### fluidistic

Thank you for the information.
Hmm so the problem is wrongly stated?
I am not sure how to solve the DE. I had read the integrating factor method can solve any first order DE but I think it requires the DE to be linear which isn't the case here.
What other method(s) could I use here?

4. Sep 14, 2011

### ehild

z=x-y+4?

ehild

5. Sep 14, 2011

### ehild

The other method can be to check if the ODE is exact, that is
(x+y-2)dx+(x-y+4)dy =d(F(x,y).

ehild

6. Sep 15, 2011

### fluidistic

Hmm, I didn't reach anything useful.
Just checked out, it is exact. I reach $F(x,y)=\frac{x^2}{2}-\frac{y^2}{2}+xy-2x+4y$ and according to Boa's mathematical methods for physics, "the solution is F(x,y)=C". But obviously it isn't the solution to the DE since I should get y(x)...

7. Sep 15, 2011

### flyingpig

I hope I helped...

Write it as

$$(x + y)dx - (2x - 2y + 4)dy = 0$$

Don't check because it isn't exact, so you need an integrating factor.

8. Sep 15, 2011

### Char. Limit

Let F(x,y)=C. Then $x^2 - y^2 + 2xy + 4x + 8y = 2C_1 = C_2$. That's an implicit function (or rather, collection of functions) of x in terms of y. It can be solved using the quadratic formula, arranging the terms like so:

$$(-1) y^2 + (8+2x) y + (x^2 + 4x - C_2) = 0$$

9. Sep 15, 2011

### fluidistic

Thanks for helping too. I don't really understand why I can write the DE under this form.
The one of ehild was understandable to me and looked exact too. But from my last step in my last post, I reached y(x) but it didn't satisfy the original DE so I think I made at least 1 mistake somewhere.
I will come back tomorrow on this problem, must sleep now.

Edit: Just saw your post char.limit. That is exactly what I've done on my draft but made a mistake I guess. Will come back tomorrow on it.
By the way, could someone explain me the post of flyingpig? Looks an interesting approach I'm missing.

10. Sep 15, 2011

### Char. Limit

As I said above, you didn't make a mistake. You just neglected to go all the way with it.

11. Sep 15, 2011

### LCKurtz

:tongue2:
That is the solution:
$$\frac{x^2}{2}-\frac{y^2}{2}+xy-2x+4y = C$$

It defines y implicitly. If you must have an explicit formula, solve for y using the quadratic formula. But it will be ugly.

 Never mind, I see someone else types faster :tongue2:

12. Sep 16, 2011

### ehild

Is that useful?
z=x-y+4
y=x-z+4
y'=1-z'

Original equation: x+y-2+(x-y+4)y'

Substituting z:

x+(x-z+4)-2+(x-(x-z+4)+4)(1-z')=0

2x-z+2+z(1-z')=0

2x-z+2+z-zz'=0

2x+2-zz'=0

2x+2=zz'
can you find y(x)? Is the solution the same as yours?

Last edited: Sep 16, 2011
13. Sep 16, 2011

### fluidistic

Unfortunately I reach different answers. I had solved the quadratic yesterday and I just redid the arithmetics, fixing some error. I reach $y(x)=x+4 \pm \sqrt {2x^2+4x+C}$.
Using ehild's z-substitution, I reach $y(x)=x+4 \pm \sqrt {2x^2+16+C}$.
I can't seem to find any error in either methods; must be another arithmetics errors...

14. Sep 16, 2011

### flyingpig

I need to make a correction. I read the question wrong. Don't even take my post seriously ...

15. Sep 16, 2011

### fluidistic

I see. No problem :)
Thanks for your will of helping me though.

16. Sep 17, 2011

### ehild

I just wonder how could you get 16 instead of 4x. Next time show your derivations in detail, to spare extra work for everybody.

$2x+2=zz'$

$\int{(2x+2)dx}=\int{zdz}$

$x^2+2x+C'=\frac{1}{2}z^2$

$\mp \sqrt{2x^2+4x+C}=z$

$y=x-z+4$

$y=x+4\pm \sqrt{2x^2+4x+C}$

ehild

17. Sep 17, 2011

### flyingpig

Oh wait, this is exact...Sorry lol

18. Sep 17, 2011

### fluidistic

My bad, this is exactly what I've found using your method. I meant that using your method I reach $y=x+4\pm \sqrt{2x^2+4x+C}$ while using the other method, I find a different answer.
I start from the general solution $\frac{x^2}{2}-\frac{y^2}{2}+xy-2x+4y=C$. Multiplying everything by -2, I get $y^2+y(-2x-8)-x^2+4x+C=0$. Solving the quadratic for y(x) gives me a "+16" rather than a "+4x" compared to your result.
$y(x)= \frac{2x+8 \pm \sqrt {4x^2+16x +64 -4(-x^2+4x+C)} }{2}=x+4 \pm \frac{ \sqrt {8x^2+64+C}}{2}=x+4 \pm \sqrt {2x^2+16+C}$.
I don't see a single mistake in any step of either derivations, yet I reach a different result.