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anonymous

  • 5 years ago

I need to find the partivular solution to the differential equation dy/dx=(1+y)/x Given the initial conditon f(-1)=1

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  1. bahrom7893
    • 5 years ago
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    I love diff equations! =) So first of all see if its separable

  2. bahrom7893
    • 5 years ago
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    it is!

  3. bahrom7893
    • 5 years ago
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    So separate it: \[dy/dx = (1+y)/x\]

  4. bahrom7893
    • 5 years ago
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    Sorry, I keep crashing on here but i will always reply: So multiply both sides by dx and divide both sides by 1+y

  5. bahrom7893
    • 5 years ago
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    You will have: \[dy/(1+y) = dx/x\]

  6. anonymous
    • 5 years ago
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    can't you think of it as dy/(1+y) you're in good hands later.

  7. bahrom7893
    • 5 years ago
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    Integrate both sides: Left side will be just Ln|1+y| and the right side will be Ln|x|

  8. bahrom7893
    • 5 years ago
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    So: Ln|1+y| = Ln|x| + C

  9. bahrom7893
    • 5 years ago
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    Raise e to both sides to get rid of Ln: \[e^{Ln|1+y|} = e^{Ln|x|+C}\]

  10. bahrom7893
    • 5 years ago
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    Rewrite and simplify: \[e^{Ln|1+y|} = e^{Ln|x|}*e^C\] \[1+y = Kx\]

  11. bahrom7893
    • 5 years ago
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    (e to some constant is another constant so I just let that other constant be K) Now apply initial conditions: f(-1)=1: 1+1 = -1K; 2 = -K; K = -2

  12. bahrom7893
    • 5 years ago
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    So your answer is: 1+y = -2x, or: y = -2x - 1 <= Final answer

  13. bahrom7893
    • 5 years ago
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    Im taking a differential equations course! IT IS FUN!! P.S.: Please click on become a fan if I helped, I really want to get to the next level!! Thanks =)

  14. anonymous
    • 5 years ago
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    why do you multiply by e^c

  15. bahrom7893
    • 5 years ago
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    there's a property: \[A^{B+C} = A^B * A^C\], so in our case: \[e^{Ln|x|+C} = e^{\ln|x|} * e^C\]

  16. bahrom7893
    • 5 years ago
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    and e to some constant is another constant so I said let that constant be K. Any other questions?

  17. anonymous
    • 5 years ago
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    no and thanks for your help

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