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

• hahutzy
In summary, the problem is that the student is stuck and does not know how to solve it. They have been trying to use homogeneous equations, but have not been able to get anywhere.f

#### hahutzy

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

## Homework Statement

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

## Homework Equations

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

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

Code:
Multiply both sides

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

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

Should be

Code:
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?

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.

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

Code:
Multiply both sides

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

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

Should be

Code:
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?

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

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

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

If the above is true, am I over thinking 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

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.

I concur, definitely over thought. Simple algebra.