# One more o.d.e. question

for this O.D.E.
y= (1-2y-4x)/(1+y+2x)
how do you treat it?

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TD
Homework Helper
$$y' = \frac{{1 - 2y - 4x}}{{1 + y + 2x}} = \frac{{ - 2\left( { - {\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}} + y + 2x} \right)}}{{1 + y + 2x}}$$

Now try the substitution $$u = y + 2x$$, it should become separable.

how'd you think of making that substitution?

TD
Homework Helper
It's a typical strategy for ODE's of the form:

$$y' = \frac{{ax + by + c}}{{a'x + b'y + c'}}$$

Consider the nominator and denominator as 2 lines.

Case 1: they intersect -> substitute so you move the intersection to the origine to lose the constant term, making it homogenous

Case 2: they're parallel -> there is a k that you can factor out (like in this case) to get the same coefficients in x and y, substitue for that expression.

wow!!! thank you very much!!! :)

TD
Homework Helper
No problem In a bit more detail, in case 1 (intersecting lines), search the point of intersection, I'll call it $\left( {x_0 ,y_0 } \right)$

Then do the substitution:

$$\left\{ \begin{array}{l} x = u + x_0 \\ y = v + y_0 \\ \end{array} \right$$

The DE will become homogenous. The other case (parallel) is the one that applies to your problem and it will make the DE separable.

ok~ great!!! :)

HallsofIvy
Homework Helper
The way I would have done y= (1-2y-4x)/(1+y+2x) is to rewrite it as

(1+ y+ 2x)dy+ (4x+ 2y-1)dx= 0. This is an "exact" equation.

then (1+y+2x)dy=-(4x+ 2y-1)dx
but then when you integrate both sides,
say on the left side, what do you do with "2x"? is it still possible to integrate that?

HallsofIvy
Homework Helper
No, I did not intend that you try to integrate the two parts separately- not only is there a "2x" in the "dy" term but there is a "2y" in the "dx" term.

I said this was an exact equation. That means that the "differential form"
(1+ y+ 2x)dy+ (4x+ 2y-1)dx (and I want them on the same side of the equation!) IS an exact differential- that is, there exist some function F(x,y) so that
dF= (1+ y+ 2x)dy+ (4x+ 2y- 1)dx. That, in turn, means that
$$\frac{\partial F}{\partial x}= 4x+ 2y- 1$$ and
$$\frac{\partial F}{\partial y}= 1+ y+ 2x$$.
Since the "partial derivative of F with respect to x" treats y as if it were a constant, an "anti-partial-derivative" of that first would give F(x,y)=2x2+ 2xy- x+ g(y). Notice that, since we are treating y as a constant here, the "constant of integration could be any function of y- that's the "g(y)".

Now differentiate that with respect to y:
$$2x+ g'(y)= \frac{\partial F}{\partial y}= 1+ y+ 2x$$.
Fortunately for us, the "2x" terms on each side cancel (that's because this was an "exact differential") leaving g'(y)= 1+ y so that g(y)= y+ (1/2)y2+ C.
Putting that into the previous function, F(x,y)= 2x2+ 2xy- x+ y+ (1/2)y2+ C.

The original equation was "dF= 0" so F(x,y)= constant. Including the "C" above in that constant, 2x2+ 2xy- x+ y+ (1/2)y2= C is the general solution to the differential equation.

TD
Homework Helper
Of course, I forgot to check whether it was exact If it's not though, the method I described above works for any of these types in general.

thank you very much for explaining!!! :)

hmm... here's my work:
suppose v= y+2x
=> v= y+2
so y= (1-2v)/(1+v)
=> v= 3/(1+v)
=> (1+v)dv = 3dx
=> v+(v^2)/2 = 2x+c
=> y+2x +(y^2 +4xy + 4x^2)/2 = 3x +c
however, the correct answer should be 2y - 2x + (y+2x)^2=c
does anybody know where my calculations went wrong?

Last edited:
TD
Homework Helper
If

$$y' = \frac{{1 - 2y - 4x}}{{1 + y + 2x}} = \frac{{ - 2\left( { - {\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}} + y + 2x} \right)}}{{1 + y + 2x}}$$

Then using $u = y + 2x \Leftrightarrow y = u - 2x \Leftrightarrow y' = u' - 2$ gives

$$u' - 2 = \frac{{ - 2\left( { - {\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}} + u} \right)}}{{1 + u}} \Leftrightarrow u' = \frac{{1 - 2u}}{{1 + u}} + 2 = \frac{3}{{u + 1}}$$

I believe you forgot that -2 after substituting y'.

But as HallsofIvy said, you can solve this as an exact DE too.

sorry, i mistyped~
@@a
the result is still the same as yours, though...
so there's still some place that went wrong?

TD
Homework Helper
Your answer seems correct, it's just not in the same form I think.
Solved for y, both solutions give (apart from a factor 2 for the constant, but that's still a constant):

$$y = \pm \sqrt {6x + c + 1} - 2x - 1$$

ok, thank you very much!!! :)