- #1

asdf1

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for this O.D.E.

y`= (1-2y-4x)/(1+y+2x)

how do you treat it?

y`= (1-2y-4x)/(1+y+2x)

how do you treat it?

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- Thread starter asdf1
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- #1

asdf1

- 734

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for this O.D.E.

y`= (1-2y-4x)/(1+y+2x)

how do you treat it?

y`= (1-2y-4x)/(1+y+2x)

how do you treat it?

- #2

TD

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\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

asdf1

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how'd you think of making that substitution?

- #4

TD

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[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

asdf1

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wow! thank you very much! :)

- #6

TD

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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

asdf1

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ok~ great! :)

- #8

HallsofIvy

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

- #9

asdf1

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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

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I

(1+ y+ 2x)dy+ (4x+ 2y-1)dx (and I

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)=2x

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

Putting that into the previous function, F(x,y)= 2x

The original equation was "dF= 0" so F(x,y)= constant. Including the "C" above in that constant, 2x

- #11

TD

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If it's not though, the method I described above works for any of these types in general.

- #12

asdf1

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thank you very much for explaining! :)

- #13

asdf1

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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?

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

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[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

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

- #15

asdf1

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@@a

the result is still the same as yours, though...

so there's still some place that went wrong?

- #16

TD

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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

asdf1

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ok, thank you very much! :)

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