# First order DE

Gold Member

## Homework Statement

Calculate the general solution of the following DE reducible to a homogeneous one: x+y-2+(x-y+4)y'=0.

Not sure.

## 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.

HallsofIvy
Homework Helper
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.

Gold Member
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?

ehild
Homework Helper

z=x-y+4?

ehild

ehild
Homework Helper
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

Gold Member

z=x-y+4?

ehild
Hmm, I didn't reach anything useful.
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
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)...

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.

Char. Limit
Gold Member
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)...

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:

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

Gold Member
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.

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.

Char. Limit
Gold Member
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.

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

LCKurtz
Homework Helper
Gold Member
:tongue2:
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)...

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:

ehild
Homework Helper

z=x-y+4?

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:
Gold Member
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...

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

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

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

ehild
Homework Helper
Using ehild's z-substitution, I reach $y(x)=x+4 \pm \sqrt {2x^2+16+C}$.

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

Oh wait, this is exact...Sorry lol

Gold Member
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

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.