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Napervillian
 3 years ago
solve this differential equation:
y'+(4/x)y = 2
initial condition: y(1)=4
Napervillian
 3 years ago
solve this differential equation: y'+(4/x)y = 2 initial condition: y(1)=4

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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1this is linear do you know how to use an integrating factor?

Napervillian
 3 years ago
Best ResponseYou've already chosen the best response.0yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.1how far did you get?

Napervillian
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1if 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

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

Napervillian
 3 years ago
Best ResponseYou've already chosen the best response.0from this: yx^4=(2/5)x^5+c do I go to y=(2/5)x + c or y=(2/5)x+cx^4?

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yes, that is correct

Napervillian
 3 years ago
Best ResponseYou've already chosen the best response.0which one? the second one?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1oh 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)

Napervillian
 3 years ago
Best ResponseYou've already chosen the best response.0gotcha, thank you! I appreciate that. makes life a whole lot easier. lol

Napervillian
 3 years ago
Best ResponseYou've already chosen the best response.0from there applying initial conditions is a piece of cake!

Napervillian
 3 years ago
Best ResponseYou've already chosen the best response.0thanks for your help man(or woman), I really appreciate it.
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