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spyros Group Title

y''=(1+y'^2)/2y

  • 2 years ago
  • 2 years ago

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  1. spyros Group Title
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    \[y \prime \prime=\frac{1+y \prime ^2}{2y}\]

    • 2 years ago
  2. spyros Group Title
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    hint : \[y\prime=u(x)\]

    • 2 years ago
  3. myko Group Title
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    maybe putting it like this\[y = 1+u ^{2}/2u'\]

    • 2 years ago
  4. myko Group Title
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    its expressed in terms of u and its drivative, so guess thats it

    • 2 years ago
  5. spyros Group Title
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    y=y(x)

    • 2 years ago
  6. spyros Group Title
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    we were told to substitute y' by u(x) solve for u(x) then for y(x)

    • 2 years ago
  7. spyros Group Title
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    \[\frac{du(x)}{dx}=\frac{1+u(x)^2}{2y(x)} ,y(x)=\int\limits_{}^{}{u(x)}\]

    • 2 years ago
  8. dumbcow Group Title
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    i think you need to make u a function of y u(y) = y' then y'' = u*(du/dy)

    • 2 years ago
  9. Mr.Math Group Title
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    Putting \(\large u=\frac{dy}{dx}\) gives \(\large \frac{d^2y}{dx^2}=\frac{du}{dx}=\frac{dy}{dx}\frac{du}{dy}=u\frac{du}{dy}.\) Substitute that in your DE.

    • 2 years ago
  10. spyros Group Title
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    will do

    • 2 years ago
  11. dumbcow Group Title
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    rearrange it so it looks like \[\frac{2u*du}{1+u^{2}} = \frac{dy}{y}\]

    • 2 years ago
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