# ODE Homogeneous Eqn - What did I do wrong this time?

1. May 26, 2009

### hahutzy

Hey guys, I've been stuck on this problem for a good hour... I have no idea how to finish it up.

1. The problem statement, all variables and given/known data

Solve: dy/dx = 4y-3x / (2x-y)
Use homogeneous equations method.

2. Relevant equations
Answer : |y - x| = c|y + 3x|5 (also y = -3x)

3. The attempt at a solution
dy/dx = 4y-3x / (2x-y)
dy/dx = 4y-3x / (2x-y) * [(1/x) / (1/x)]
dy/dx = (4y/x)-3 / (2-(y/x))

let v = y/x
y = v(x) x
dy/dx = x dv/dx + v

x dv/dx + v = 4v-3 / (2-v)
x dv/dx = [4v-3 / (2-v)] - v
x dv/dx = 4v-3 -[2v-v2] / (2-v)
x dv/dx = 2v-3+v2] / (2-v)
dx/x = [(-v + 2) / (v^2 + 2v -3)] dv
dx/x = [(-5/4)(1/(v+3)) + (1/4)(1/(v-1))] dv (partial fractions)

Integrating both sides:

ln |x| = (1/4) ln |v-1| - (5/4) ln |v+3| + c
ln |x| = (1/4) [ln |v-1| - 5 ln |v+3| + c]
|x| = (1/4) exp {[ln |v-1| - 5 ln |v+3| + c]}
|x| = c |v-1| |v+3|-5
c|v+3|5 = |v-1|/|x|
|(y/x)-1| / |x| = c|(y/x)+3|5

Answer from book : |y - x| = c|y + 3x|5 (also y = -3x)

Much appreciated.

Last edited: May 26, 2009
2. May 26, 2009

### djeitnstine

c|v+3|^5 = |v-1|/|x|

What happened to your 1/4!! it disappeared

rework it, you should have
c|v+3|^-5 = |v-1|/|x|^4

rearrange and substitute

|x|^4 = |y/x-1|/c|y/x+3|^-5

factor out a 1/x^-5 from the right

|x|^4 = |y/x-1| / (c|y+3x|^-5 * 1/x^-5)

Multiply both sides

c 1/x = |y/x-1| / (|y+3x|^-5 )

c 1/x*(|y+3x|^-5 ) = |y/x-1|

facor out 1/x from the right and everything should be clear

- srry it was just quick and dirty but bear with me =]

Edit! O man sorry OP, that ^5 was ^-5, recheck your partial fractions, now it all works out

Last edited: May 27, 2009
3. May 26, 2009

### hahutzy

Thanks! But just making sure I'm getting this.

Code (Text):
Multiply both sides

cx = |y/x-1| / (|y+3x|^5 )

cx*(|y+3x|^5 ) = |y/x-1|
Should be

Code (Text):
Multiply both sides

c(1/|x|) = |y/x-1| / (|y+3x|^5 )

c*(|y+3x|^5 ) = |y/x-1|*|x|

c*(|y+3x|^5 ) = |y-x|

Right?

Also, is there any easy way to remember that (y/x + 3)^5 = (y + 3x)^5 * (1/x)^5?

4. May 26, 2009

### djeitnstine

Yes you're right (check my edit - my net is giving me probs) umm I don't know what to say, you just have to train yourself to spot it I guess.

5. May 27, 2009

### HallsofIvy

Staff Emeritus
I don't think it is a matter of "remembering". Just add the fractions:
y/x+ 3= y/x+ 3x/x= (y+ 3x)/x.

6. May 30, 2009

### hahutzy

I understand that, however, what I can't wrap my head around is doing the same thing if the term is being raised to the power of something. Namely, ((y/x)+3)N where N is integer > 1

Let's say I had N=2, ((y/x)+3)2

Expanding gives y2/x2 + 6(y/x) + 9

y2/x2 + 6(y/x)(x/x) + 9(x2/x2)

[y2 + 6(yx) + 9x2] (1/x2)

(y+3x)2 (1/x)2

.'. ((y/x)+3)2 = (y+3x)2 (1/x)2

Now, since N=1 is true (intuitive), N=2 is true (proven), and you guys claim N=5 is also true, does that mean ((y/x)+3)N = (y+3x)N (1/x)N for 1..2..3..N by induction?

If the above is true, am I overthinking this all? Is it mathematically correct to do this:

((y/x)+3)N
((y/x)+3(x/x))N
((y+3x)/x))N
(y+3x)N (1/x)N

7. May 30, 2009

### Cyosis

You are definitely over thinking this. What you've done is correct. Remember $a^n/b^n=(a/b)^n, a^nb^n=(ab)^n$. It is just basic arithmetic.

8. May 30, 2009

### djeitnstine

I concur, definitely over thought. Simple algebra.