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I Integrating factor of (y+1)dx+(4x-y)dy=0

  1. Sep 17, 2016 #1
    I tried to put it in standard form as (dx/dy)+4x(1/(1+y))=y/(y+1). I get that the integrating factor is (y+1) but i am not sure if i am doing it right or what am I suppose to do next? I get (y+1) because the integrating of 1/(y+1) is ln(y+1) and since it has e, then ln cancels and i am left with (c(y+1)) as the integrating factor
     
  2. jcsd
  3. Sep 17, 2016 #2
    oh no I get integrating factor of (y+1)^4. that should be right. but i am confused about the next part?
     
  4. Sep 17, 2016 #3
    now I get that (x)(1+y)^4=y(1+y)^4/4 + (1+y)^5/5. now I am really stuck!! the answer should be 20x=4y-1+c(1+y)^-4. but i am not sure how to simplify to that
     
  5. Sep 17, 2016 #4

    Charles Link

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    Let's change y to t. On the right side you get an integral ## \int (t)(t+1)^3 \, dt ##.(I agree with your integrating factor of ## (y+1)^4 ##.) If you do it by parts using ## (1/4) \int t \, d(t+1)^4 ## , you do get the answer they give: ## x(t+1)^4=(1/4)t(t+1)^4-(1/4) \int (t+1)^4 \, dt =(1/4)t(t+1)^4-(1/20)(t+1)^5 +C ##. Finally, just divide out the ## (t+1)^4 ## and multiply by 20.
     
    Last edited: Sep 17, 2016
  6. Sep 17, 2016 #5
    but I have to do integration by parts on the right side right? So i get t(t+1)^4/4 - (t+1)^5/5=t(t+1)^4. I find a common multiple on the left side and I get a 20, and I divide (t+1)^4 on the right side and the left side and I get, 20t=5t-4(t+1)+c(t+1)^-4. which is 20t=t-4+c(t+1)^-4. but I am not sure what am i doing wrong. I now it has to do with some sign error but I can't find it.
     
  7. Sep 17, 2016 #6

    Charles Link

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    The "4" should not be there in front of (t+1). See my post #4 which I edited after you saw it. It's 20x= 5t-(t+1)+c(t+1)^{-4}. (There's a 1/4 factor in the second term of your first line.)
     
  8. Sep 17, 2016 #7
    oooooohhh my God :'[ thank you for helping meee!! engineering is not so for me I am changing majors!! lol
     
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