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UnkleRhaukus

  • 5 years ago

\[y''-yy'=0\]

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  1. UnkleRhaukus
    • 5 years ago
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    the equation is x-absent let \[y'=p\]\[y''=pp'\]

  2. UnkleRhaukus
    • 5 years ago
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    \[pp'-yp=0\]\[p'-y=0\]\[dp/dx=y\]

  3. UnkleRhaukus
    • 5 years ago
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    ?

  4. anonymous
    • 5 years ago
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    whats your question?

  5. UnkleRhaukus
    • 5 years ago
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    what do i do next i am tring to solve for y

  6. UnkleRhaukus
    • 5 years ago
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    y=y(x)

  7. anonymous
    • 5 years ago
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    is p a constant?

  8. UnkleRhaukus
    • 5 years ago
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    p=dy/dx

  9. anonymous
    • 5 years ago
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    y'=p =>y''=0

  10. anonymous
    • 5 years ago
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    if p is constant

  11. UnkleRhaukus
    • 5 years ago
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    y''=pp'

  12. anonymous
    • 5 years ago
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    p must be a const just check it out!

  13. UnkleRhaukus
    • 5 years ago
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    that is just the trivial solution

  14. anonymous
    • 5 years ago
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    k

  15. UnkleRhaukus
    • 5 years ago
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    the other solution has tangent of something in it

  16. anonymous
    • 5 years ago
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    y'=p y''=p'

  17. anonymous
    • 5 years ago
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    ur solving for y right

  18. anonymous
    • 5 years ago
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    ?

  19. UnkleRhaukus
    • 5 years ago
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    i am solving for y(x) yes. y'=p however y''=pp'

  20. anonymous
    • 5 years ago
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    then you should integrate this equation y'=p => y(x)=px+c

  21. UnkleRhaukus
    • 5 years ago
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    y(x)=y'x+c

  22. anonymous
    • 5 years ago
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    err p is a function of x dhashni

  23. UnkleRhaukus
    • 5 years ago
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    I am tring to show that the solution is \[y=c\] or \[y=atan({ax \over 2}+b)\]

  24. anonymous
    • 5 years ago
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    rather \[y = a \tan(\frac{ax}{2}+b)\]

  25. UnkleRhaukus
    • 5 years ago
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    um...

  26. UnkleRhaukus
    • 5 years ago
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    i dont know

  27. anonymous
    • 5 years ago
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    aha - it goes tan -> sec^2 -> tan*sec^2 as you differentiate

  28. anonymous
    • 5 years ago
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    wait why do you think that y'' = pp' rather y'' = yy'

  29. anonymous
    • 5 years ago
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    \[y'' - yy' = 0\]\[y' - \frac{1}{2}y^2 = c\]\[y' = \frac{1}{2}y^2 + c\]\[\frac{y'}{\frac{1}{2}y^2 + c} = 1\]\[\frac{2y'}{y^2 + 2c} = 1\]\[\sqrt{\frac{2}{c}}\arctan(\frac{y}{\sqrt{2c}})=x+c_2\]\[y = \sqrt{2c} \tan(\sqrt{\frac{c}{2}}(x+c_2))\]

  30. anonymous
    • 5 years ago
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    the two hard steps are where you integrate both sides with respect to x

  31. UnkleRhaukus
    • 5 years ago
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    hey thankyou very much

  32. anonymous
    • 5 years ago
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    nps :)

  33. UnkleRhaukus
    • 5 years ago
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    i can not see what you did to \[y''-yy'=0\]to get \[y'-{1 \over 2}y^2=c\]

  34. anonymous
    • 5 years ago
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    that's the step where i integrated. the integral of y'' is y'+c and the integral of 2yy' is y^2+c (chain rule)

  35. anonymous
    • 5 years ago
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    \[\int (y''-yy') dx = \int 0 \space dx\] \[\int \frac{d}{dx}(y') dx - \frac{1}{2} \int 2y\frac{dy}{dx} dx = 0 + C_3\] \[ y' + C_1 - \frac{1}{2}y^2 - C_2 = C_3 \] \[ y' - \frac{1}{2}y^2 = (-C_1+C_2+C_3) \] \[ y' - \frac{1}{2}y^2 = c \]

  36. UnkleRhaukus
    • 5 years ago
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    that is it thank you Broken Fixer

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