anonymous
  • anonymous
solve this differential equation: y'+(4/x)y = 2 initial condition: y(1)=-4
Mathematics
jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
this is linear do you know how to use an integrating factor?
anonymous
  • anonymous
yes, i have solved it but it has been a long time and I can't remember how to handle the constant c that I get out of my integral
TuringTest
  • TuringTest
how far did you get?

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anonymous
  • anonymous
I have an answer, but when I get to this part: yx^4=(2/5)x^5+c I'm not sure if I also apply my x^-4 to the c or leave it as a constant
anonymous
  • anonymous
also, how do I handle the absolute value that gets kicked out of a logarithmic function?
TuringTest
  • TuringTest
if you want the "explicit" solution, solve for y y(1)=-4 means you can plug in x=1 and y=-4, which will allow you to solve for c
anonymous
  • anonymous
yes, but am I solving y=(2/5)x+c or am i solving y=(2/5)x+cx^-4?
TuringTest
  • TuringTest
y=(2/5)x+cx^-4?
anonymous
  • anonymous
from this: yx^4=(2/5)x^5+c do I go to y=(2/5)x + c or y=(2/5)x+cx^-4?
anonymous
  • anonymous
which can also be written as y=(2/5)x+c/(x^4)
TuringTest
  • TuringTest
yes, that is correct
anonymous
  • anonymous
which one? the second one?
TuringTest
  • TuringTest
oh I see, I didn't mean to copy the '?' the second one is right y=(2/5)x+cx^-4=(2/5)x+c/(x^4)
anonymous
  • anonymous
gotcha, thank you! I appreciate that. makes life a whole lot easier. lol
anonymous
  • anonymous
from there applying initial conditions is a piece of cake!
anonymous
  • anonymous
thanks for your help man(or woman), I really appreciate it.
TuringTest
  • TuringTest
no problem

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