• Support PF! Buy your school textbooks, materials and every day products Here!

One more o.d.e. question

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

Answers and Replies

TD
Homework Helper
1,020
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.
 
734
0
how'd you think of making that substitution?
 
TD
Homework Helper
1,020
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.
 
734
0
wow!!! thank you very much!!! :)
 
TD
Homework Helper
1,020
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.
 
734
0
ok~ great!!! :)
 
HallsofIvy
Science Advisor
Homework Helper
41,736
894
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.
 
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?
 
HallsofIvy
Science Advisor
Homework Helper
41,736
894
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.
 
TD
Homework Helper
1,020
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.
 
734
0
thank you very much for explaining!!! :)
 
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:
TD
Homework Helper
1,020
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.
 
734
0
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
1,020
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]
 
734
0
ok, thank you very much!!! :)
 

Related Threads for: One more o.d.e. question

  • Last Post
Replies
1
Views
982
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
3
Views
838
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
3
Views
1K
Top