- #1

- 103

- 0

## Homework Statement

[tex]y' = \frac{2x+3y-5}{x+4y}[/tex]

## Homework Equations

## The Attempt at a Solution

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

[tex]b' = \frac{2a+3b}{a+4b}[/tex]

where [itex]a = x-4[/itex] and [itex]b = y+1[/itex].

Then using the substitution [itex]z = \frac{b}{a}[/itex] and some algebra I get:

[tex]z' = \frac{2z+2-4z^2}{a(1+4z)}[/tex]

the integral is:

[tex]\frac{1}{2} \int \frac{(1+4z)dz}{2+1-2z^2} = \int \frac{da}{a}[/tex]

solving the integral (using partial fractions) gives me:

[tex]5\ln{|z-1|} + \ln{|z+\frac{1}{2}|} = 6\ln{|a|}+C[/tex]

getting rid of the logarithm, and putting back a and b inside:

[tex]\left(\frac{b-a}{a}\right)^5 \left(\frac{2b+a}{2a}\right)=C(a)^6[/tex]

making it back in x and y:

[tex]\left(\frac{y-x+5}{x-4}\right)^5 \left(\frac{2y+x-2}{2x-8}\right)=C(x-4)^6[/tex]

Which should be the final answer, but when I look at the answers, it's different, being:

[tex](y-x+5)^5(2y+x-2)=C[/tex]

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!