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spyros

  • 4 years ago

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

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  1. spyros
    • 4 years ago
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    \[y \prime \prime=\frac{1+y \prime ^2}{2y}\]

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

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

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

  5. spyros
    • 4 years ago
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    y=y(x)

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

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

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

  9. Mr.Math
    • 4 years ago
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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.

  10. spyros
    • 4 years ago
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    will do

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

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