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

1. TuringTest

this is linear do you know how to use an integrating factor?

2. Napervillian

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

3. TuringTest

how far did you get?

4. Napervillian

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

5. Napervillian

also, how do I handle the absolute value that gets kicked out of a logarithmic function?

6. 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

7. Napervillian

yes, but am I solving y=(2/5)x+c or am i solving y=(2/5)x+cx^-4?

8. TuringTest

y=(2/5)x+cx^-4?

9. Napervillian

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?

10. Napervillian

which can also be written as y=(2/5)x+c/(x^4)

11. TuringTest

yes, that is correct

12. Napervillian

which one? the second one?

13. 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)

14. Napervillian

gotcha, thank you! I appreciate that. makes life a whole lot easier. lol

15. Napervillian

from there applying initial conditions is a piece of cake!

16. Napervillian

thanks for your help man(or woman), I really appreciate it.

17. TuringTest

no problem