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

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TuringTestBest ResponseYou've already chosen the best response.1
this is linear do you know how to use an integrating factor?
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
how far did you get?
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
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
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
also, how do I handle the absolute value that gets kicked out of a logarithmic function?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
yes, but am I solving y=(2/5)x+c or am i solving y=(2/5)x+cx^4?
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
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?
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
which can also be written as y=(2/5)x+c/(x^4)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes, that is correct
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
which one? the second one?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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)
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
gotcha, thank you! I appreciate that. makes life a whole lot easier. lol
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
from there applying initial conditions is a piece of cake!
 one year ago

NapervillianBest ResponseYou've already chosen the best response.0
thanks for your help man(or woman), I really appreciate it.
 one year ago
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