ODE with substitution

  • Thread starter manenbu
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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!
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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… Which should be the final answer, but when I look at the answers, it's different …

Hi manenbu! :smile:

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

except that their C is twice your C (which of course doesn't matter). :wink:
 
  • #3
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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
tiny-tim
Science Advisor
Homework Helper
25,832
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Ok thanks, but how is it the same?

Hi manenbu! :smile:

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. :wink:
 
  • #5
103
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Oh so assuming I get the minus right that's -6 and -6 powers on both sides. Ok got it!
Thanks. :)
 

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