solve this differential equation: y'+(4/x)y = 2 initial condition: y(1)=-4

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solve this differential equation: y'+(4/x)y = 2 initial condition: y(1)=-4

Mathematics
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this is linear do you know how to use an integrating factor?
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
how far did you get?

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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
also, how do I handle the absolute value that gets kicked out of a logarithmic function?
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
yes, but am I solving y=(2/5)x+c or am i solving y=(2/5)x+cx^-4?
y=(2/5)x+cx^-4?
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?
which can also be written as y=(2/5)x+c/(x^4)
yes, that is correct
which one? the second one?
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)
gotcha, thank you! I appreciate that. makes life a whole lot easier. lol
from there applying initial conditions is a piece of cake!
thanks for your help man(or woman), I really appreciate it.
no problem

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