# ODE with substitution

1. Nov 21, 2009

### manenbu

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

$$y' = \frac{2x+3y-5}{x+4y}$$

2. Relevant equations

3. The attempt at a solution

First of all, I switched it to another coordinates, a and b, giving:

$$b' = \frac{2a+3b}{a+4b}$$
where $a = x-4$ and $b = y+1$.
Then using the substitution $z = \frac{b}{a}$ and some algebra I get:
$$z' = \frac{2z+2-4z^2}{a(1+4z)}$$
the integral is:
$$\frac{1}{2} \int \frac{(1+4z)dz}{2+1-2z^2} = \int \frac{da}{a}$$
solving the integral (using partial fractions) gives me:
$$5\ln{|z-1|} + \ln{|z+\frac{1}{2}|} = 6\ln{|a|}+C$$
getting rid of the logarithm, and putting back a and b inside:
$$\left(\frac{b-a}{a}\right)^5 \left(\frac{2b+a}{2a}\right)=C(a)^6$$
making it back in x and y:
$$\left(\frac{y-x+5}{x-4}\right)^5 \left(\frac{2y+x-2}{2x-8}\right)=C(x-4)^6$$

Which should be the final answer, but when I look at the answers, it's different, being:
$$(y-x+5)^5(2y+x-2)=C$$

And my question is - where did I go wrong?
I had a similar problem where I got similar results, me having denominators which didn't appear in the given answer. Am I doing something wrong in my technique, or both of the answers are equal (and I'm missing some algebra here)?

Thanks!

2. Nov 21, 2009

### tiny-tim

Hi manenbu!

No, it's the same (if you keep cancelling) …

except that their C is twice your C (which of course doesn't matter).

3. Nov 22, 2009

### manenbu

Ok thanks, but how is it the same?
I'm ok with solving the ODE but then I guess that the algebra part is a bit hard for me.
I tried getting from one form to another but no success.

4. Nov 22, 2009

### tiny-tim

Hi manenbu!

Because on the bottom left, you have (x - 4)5(2x - 8), which is 2(x- 4)6, and on the right …

ohhh … you need a minus there (i didn't notice that before) …

you missed out a "minus" on the LHS when you changed to partial fractions.

5. Nov 22, 2009

### manenbu

Oh so assuming I get the minus right that's -6 and -6 powers on both sides. Ok got it!
Thanks. :)