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UnkleRhaukus

  • 5 years ago

\[{dy \over dx}= 4{y^2 \over x^2}+5{y \over x} +1\]

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  1. UnkleRhaukus
    • 5 years ago
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    let \[y=vx\]\[{dy \over dx}=x{dv \over dx}+v\]

  2. UnkleRhaukus
    • 5 years ago
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    yeah \[y=y(x)\]

  3. UnkleRhaukus
    • 5 years ago
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    \[x{dv \over dx}+v= 4v^2+5v+1\]

  4. UnkleRhaukus
    • 5 years ago
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    \[{dv \over dx}={4v^2+4v+1 \over x}\] \[\int{dv \over 4v^2+4v+1}= \int{dx\over x}\]

  5. UnkleRhaukus
    • 5 years ago
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    am i doing this right?

  6. anonymous
    • 5 years ago
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    yes

  7. UnkleRhaukus
    • 5 years ago
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    how am i supposed to integrate the LHS

  8. TuringTest
    • 5 years ago
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    complete the square, it will be an arctan thing..

  9. UnkleRhaukus
    • 5 years ago
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    maybe i shouldn't have misused the v

  10. anonymous
    • 5 years ago
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    completing the square 4v^2+4v+1 a^2 +2ab+b^2 2(2)(1)=4 so b must be 1 b^2=1 this is perfect square (2v+1)(2v +1)=4v^2+4v+1

  11. TuringTest
    • 5 years ago
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    ...so partial fractions works too then I guess

  12. UnkleRhaukus
    • 5 years ago
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    \[\int{dv \over (2v+1)^2}=\int {dx \over x}\]\[{-1 \over 4v+2}=ln(x) +c_1\]\[4v+2 = ln(1/x)+c_2\]

  13. TuringTest
    • 5 years ago
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    haha, oh yeah.... I was overcomplicating it

  14. UnkleRhaukus
    • 5 years ago
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    \[y=x/4(c_3-lnx)\]

  15. UnkleRhaukus
    • 5 years ago
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    oh wait, the answer in my book has arctans and lns

  16. UnkleRhaukus
    • 5 years ago
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    i must have put a negative in the wrong stop somewhere

  17. UnkleRhaukus
    • 5 years ago
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    the answer in the back of my book is \[tan^{-1}({y+3 \over x+2})+{1 \over 2}ln(x^2+y^24x+6y+13)=c\]

  18. UnkleRhaukus
    • 5 years ago
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    someone put me on the right track,

  19. UnkleRhaukus
    • 5 years ago
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    where did i go wrong

  20. UnkleRhaukus
    • 5 years ago
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    .. somebody?!

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