# Homework Help: Nonlinear DE

1. Dec 14, 2011

### namu

How do I solve the DE

$\dot{y}$=$\frac{t+3y-5}{t-y-1}$

I tried using a substitution of the form u=ax+by+c, however was
unsucessful. Any suggestions?

2. Dec 14, 2011

### Simon Bridge

I 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.

There's probably a better way.

3. Dec 14, 2011

### namu

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

$(\dot{y}-1)x-(\dot{y}+3)-\dot{y}+5=0$

If I use

$z=\dot{y}-1$,

I get no where, plus I lose my derivatives.

If I use

$\dot{z}=\dot{y}-1$,

I get

$x \dot{z} - (\dot{z}+4) y - \dot{z} + 6=0$

which is worse than before. Im still lost.

4. Dec 14, 2011

### Simon Bridge

Hmmm ... did I mess up? Did I misplace 2y ... I just scribbled it on my window, lets see:

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

How about $z=\dot y + 3$ ...

note: $t(\dot y -1) = t(\dot y +3) - 4t$

dammit! I still get a stray y.

5. Dec 14, 2011

### AlephZero

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. Dec 14, 2011

### namu

Thank you!

7. Dec 14, 2011

### Simon Bridge

Yes - thank you and my apologies to namu.

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

8. Dec 15, 2011

### JJacquelin

Hi!

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

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9. Dec 18, 2011

### namu

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. Dec 18, 2011

### namu

Yes, I saw that once I realized the clever substitution that was pointed out. Thank you for the work.

11. Dec 18, 2011

### Simon Bridge

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. Dec 18, 2011

### Dickfore

First, you need to make an "additive" substitution:
$$t = \tilde{t} + \alpha, y = \tilde{y} + \beta$$
that will cancel the free terms in the numerator and denominator of the fraction. Effectively, you need to solve the following linear system:
$$\alpha + 3 \beta - 5 = 0, \ \alpha - \beta - 1 = 0$$
After that, considering $\tilde{y} = f(\tilde{t})$, you get a homogeneous ODE:
$$\dot{\tilde{y}} = \frac{\tilde{t} + 3 \tilde{y}}{\tilde{t} - \tilde{y}} = \frac{1 + 3 \frac{\tilde{y}}{\tilde{t}}}{1 - \frac{\tilde{y}}{\tilde{t}}} = f(\frac{\tilde{y}}{\tilde{t}}), f(z) = \frac{1 + 3 z}{1 - z}$$
You can separate the variables by making one more subtitution:
$$z(\tilde{t}) = \frac{\tilde{y}}{\tilde{t}} \Rightarrow \tilde{y} = \tilde{t} \, z, \ \dot{\tilde{y}} = \tilde{t} \, \dot{z} + z$$