Solving the DE: I Tried a Substitution and Failed

In summary: Here is a graph of the ODE:In summary, Simon tried using a substitution of the form u=ax+by+c, however was unsuccessful. He multiplied both sides by the denominator and grouped like terms. This suggested a substitution z=\dot{y} + 1 - which could be solved for z in terms of t. He then tried z=\dot y + 3 and still did not get anywhere. He lost his derivatives when doing this. Finally, he tried z=\dot{y}-1 and got x \dot{z} - (\dot{z}+4) y - \dot{z} + 6=0. This was
  • #1
namu
33
0
How do I solve the DE

[itex]\dot{y}[/itex]=[itex]\frac{t+3y-5}{t-y-1}[/itex]

I tried using a substitution of the form u=ax+by+c, however was
unsucessful. Any suggestions?
 
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  • #2
I multiplied both sides by the denominator and grouped like terms.
This suggested a substitution [itex]z=\dot{y} + 1[/itex] - which could be solved for z in terms of t.

There's probably a better way.
 
  • #3
Simon,

Thank you for the suggestion. I tried your suggestion, however do
not seem to be getting anywhere. When I group like terms, I get

[itex](\dot{y}-1)x-(\dot{y}+3)-\dot{y}+5=0[/itex]

If I use

[itex] z=\dot{y}-1 [/itex],

I get no where, plus I lose my derivatives.

If I use

[itex] \dot{z}=\dot{y}-1 [/itex],

I get

[itex] x \dot{z} - (\dot{z}+4) y - \dot{z} + 6=0 [/itex]

which is worse than before. I am still lost.
 
  • #4
Hmmm ... did I mess up? Did I misplace 2y ... I just scribbled it on my window, let's see:

[tex](t-y-1)\dot y = t + 3y -5[/tex]... expand
[tex]t \dot y - y\dot y -\dot y -t -3y +5 = 0[/tex]... group terms
[tex]t(\dot y -1) - y(\dot y +3) - \dot y +5 = 0[/tex]... yep, looks like I did!

How about [itex]z=\dot y + 3[/itex] ...

note: [itex]t(\dot y -1) = t(\dot y +3) - 4t[/itex]

dammit! I still get a stray y.
 
  • #5
First get rid of the constants on the right hand side. If you solve
t + 3y - 5 = 0
t - y - 1 = 0
you get t = 2, y = 1

So let T = t + 2, Y = t + 1
DY/DT = (T + 3Y)/(T - Y)
Then you can separate the variables by letting Y(T) = Tu(T)
 
  • #6
Thank you!
 
  • #7
Yes - thank you and my apologies to namu.

aside:
did you know that "namu" is Maori for sand-fly?
 
  • #8
Hi!

In fact this ODE is of the homogeneous kind (attached page) :
 

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  • #9
Simon Bridge said:
Yes - thank you and my apologies to namu.

aside:
did you know that "namu" is Maori for sand-fly?

No, i didn't know that. That is interesting. Namu is my name. Its often
included in buddhist chants. Also in korean, namu means tree. It is interesting
how one word has many different meanings in different cultures.
 
  • #10
Yes, I saw that once I realized the clever substitution that was pointed out. Thank you for the work.
 
  • #11
That's cool (the name)!

I can see why a tree would come up in chants quite a lot. Trees are a feature in the life of Siddartha don't they?

"Namu" can be a nickname, and they have spiritual connotations to the Maori too.
This has been a neat window between cultures, thanks.
 
  • #12
First, you need to make an "additive" substitution:
[tex]
t = \tilde{t} + \alpha, y = \tilde{y} + \beta
[/tex]
that will cancel the free terms in the numerator and denominator of the fraction. Effectively, you need to solve the following linear system:
[tex]
\alpha + 3 \beta - 5 = 0, \ \alpha - \beta - 1 = 0
[/tex]
After that, considering [itex]\tilde{y} = f(\tilde{t})[/itex], you get a homogeneous ODE:
[tex]
\dot{\tilde{y}} = \frac{\tilde{t} + 3 \tilde{y}}{\tilde{t} - \tilde{y}} = \frac{1 + 3 \frac{\tilde{y}}{\tilde{t}}}{1 - \frac{\tilde{y}}{\tilde{t}}} = f(\frac{\tilde{y}}{\tilde{t}}), f(z) = \frac{1 + 3 z}{1 - z}
[/tex]
You can separate the variables by making one more subtitution:
[tex]
z(\tilde{t}) = \frac{\tilde{y}}{\tilde{t}} \Rightarrow \tilde{y} = \tilde{t} \, z, \ \dot{\tilde{y}} = \tilde{t} \, \dot{z} + z
[/tex]
 

What is a substitution method for solving differential equations?

A substitution method is a technique used to transform a differential equation into a simpler form that can be easily solved. It involves substituting a new variable for the original variable in the equation.

Why did my substitution method fail to solve the differential equation?

There could be several reasons why a substitution method failed to solve a differential equation. It could be due to an error in the substitution itself, the equation may not be separable, or the substitution may not be applicable to the given equation.

Is the substitution method the only way to solve differential equations?

No, the substitution method is just one of the many techniques used to solve differential equations. Other methods include separation of variables, integrating factors, and power series methods.

Can I use any substitution for solving differential equations?

No, not all substitutions will work for all differential equations. The substitution used must be applicable to the equation and must result in a simpler form that can be easily solved.

What are some tips for using substitution method to solve differential equations?

Some tips for using the substitution method include carefully choosing the substitution, checking for errors in the substitution, and practicing with various types of equations to become familiar with the technique.

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