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Napervillian

  • 3 years ago

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

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  1. TuringTest
    • 3 years ago
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    this is linear do you know how to use an integrating factor?

  2. Napervillian
    • 3 years ago
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    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
    • 3 years ago
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    how far did you get?

  4. Napervillian
    • 3 years ago
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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

  5. Napervillian
    • 3 years ago
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    also, how do I handle the absolute value that gets kicked out of a logarithmic function?

  6. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    yes, but am I solving y=(2/5)x+c or am i solving y=(2/5)x+cx^-4?

  8. TuringTest
    • 3 years ago
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    y=(2/5)x+cx^-4?

  9. Napervillian
    • 3 years ago
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    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
    • 3 years ago
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    which can also be written as y=(2/5)x+c/(x^4)

  11. TuringTest
    • 3 years ago
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    yes, that is correct

  12. Napervillian
    • 3 years ago
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    which one? the second one?

  13. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    gotcha, thank you! I appreciate that. makes life a whole lot easier. lol

  15. Napervillian
    • 3 years ago
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    from there applying initial conditions is a piece of cake!

  16. Napervillian
    • 3 years ago
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    thanks for your help man(or woman), I really appreciate it.

  17. TuringTest
    • 3 years ago
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    no problem

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