One more o.d.e. question

  • Thread starter asdf1
  • Start date
  • #1
734
0
for this O.D.E.
y`= (1-2y-4x)/(1+y+2x)
how do you treat it?
 

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
[tex]y' = \frac{{1 - 2y - 4x}}{{1 + y + 2x}} = \frac{{ - 2\left( { - {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}} + y + 2x} \right)}}{{1 + y + 2x}}[/tex]

Now try the substitution [tex]u = y + 2x[/tex], it should become separable.
 
  • #3
734
0
how'd you think of making that substitution?
 
  • #4
TD
Homework Helper
1,022
0
It's a typical strategy for ODE's of the form:

[tex]y' = \frac{{ax + by + c}}{{a'x + b'y + c'}}[/tex]

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.
 
  • #5
734
0
wow!!! thank you very much!!! :)
 
  • #6
TD
Homework Helper
1,022
0
No problem :smile:

In a bit more detail, in case 1 (intersecting lines), search the point of intersection, I'll call it [itex]\left( {x_0 ,y_0 } \right)[/itex]

Then do the substitution:

[tex]\left\{ \begin{array}{l}
x = u + x_0 \\
y = v + y_0 \\
\end{array} \right[/tex]

The DE will become homogenous. The other case (parallel) is the one that applies to your problem and it will make the DE separable.
 
  • #7
734
0
ok~ great!!! :)
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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.
 
  • #9
734
0
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?
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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
[tex]\frac{\partial F}{\partial x}= 4x+ 2y- 1[/tex] and
[tex]\frac{\partial F}{\partial y}= 1+ y+ 2x[/tex].
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:
[tex]2x+ g'(y)= \frac{\partial F}{\partial y}= 1+ y+ 2x[/tex].
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.
 
  • #11
TD
Homework Helper
1,022
0
Of course, I forgot to check whether it was exact :blushing:

If it's not though, the method I described above works for any of these types in general.
 
  • #12
734
0
thank you very much for explaining!!! :)
 
  • #13
734
0
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:
  • #14
TD
Homework Helper
1,022
0
If

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

Then using [itex]u = y + 2x \Leftrightarrow y = u - 2x \Leftrightarrow y' = u' - 2[/itex] gives

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

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

But as HallsofIvy said, you can solve this as an exact DE too.
 
  • #15
734
0
sorry, i mistyped~
@@a
the result is still the same as yours, though...
so there's still some place that went wrong?
 
  • #16
TD
Homework Helper
1,022
0
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):

[tex]y = \pm \sqrt {6x + c + 1} - 2x - 1[/tex]
 
  • #17
734
0
ok, thank you very much!!! :)
 

Related Threads on One more o.d.e. question

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
886
  • Last Post
Replies
15
Views
2K
Replies
1
Views
6K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
5
Views
1K
Top