Is This Nonlinear O.D.E. Solvable?

[asdf1]

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

 

[TD]

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

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

 

[asdf1]

how'd you think of making that substitution?

 

[TD]

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.

 

[asdf1]

wow! thank you very much! :)

 

[TD]

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:

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

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

 

[asdf1]

ok~ great! :)

 

[HallsofIvy]

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.

 

[asdf1]

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]

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]

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.

 

[asdf1]

thank you very much for explaining! :)

 

[asdf1]

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?

 

[TD]

If

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

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

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

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

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

 

[asdf1]

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

 

[TD]

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):

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

 

[asdf1]

ok, thank you very much! :)