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anonymous

  • 5 years ago

Solve (x-y)dx+xdy=0

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  1. anonymous
    • 5 years ago
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    y=ux dy=udu+xdx dx=1 so dy=udu+xdx I try to substitute. Is it ok if I leave out the dx?

  2. anonymous
    • 5 years ago
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    y=ux dy=udu+xdx dx=1 so dy=udu+x

  3. anonymous
    • 5 years ago
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    You have to integrate no?

  4. anonymous
    • 5 years ago
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    do I?

  5. anonymous
    • 5 years ago
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    What topic of the course is this from?

  6. anonymous
    • 5 years ago
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    homogeneous substitution

  7. myininaya
    • 5 years ago
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    i got y=-xlnx+xC

  8. anonymous
    • 5 years ago
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    (x-ux)dx+x(udu+x)=0 xdx-uxdx+xudu+x^2=0 work correct so far?

  9. myininaya
    • 5 years ago
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    where the C above is a constant

  10. anonymous
    • 5 years ago
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    tell me myininaya if my work is correct

  11. myininaya
    • 5 years ago
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  12. myininaya
    • 5 years ago
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    i dont know the way you are doing it im sorry

  13. anonymous
    • 5 years ago
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    k. ty.

  14. myininaya
    • 5 years ago
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    but i posted above how i got my answer if that helps

  15. anonymous
    • 5 years ago
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    please visit site orlandoicc.org

  16. anonymous
    • 5 years ago
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    why do you multiply by v?

  17. myininaya
    • 5 years ago
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    i multiply by v (i choose this v such that we have vy'+v'y=(yv)')

  18. myininaya
    • 5 years ago
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    if we can write the differential equation in this form y'+p(x)y=q(x) then we can multiply by v such that we have vy'+vpy=vq but we want this v so that v'=vp so we can write the next step as (vy)'=vq

  19. myininaya
    • 5 years ago
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    v'=pv dv/dx=pv 1/v dv=p dx lnv=p dx so we have v=e^(int p dx)

  20. myininaya
    • 5 years ago
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    so we multiply whatever that v is on both sides of the equation then we can write the (vy)'=vq integrate both sides vy=int(vq)+C solve for y y=1/v*int(vq) +C/v

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