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spyros
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\[y \prime \prime=\frac{1+y \prime ^2}{2y}\]
spyros
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hint : \[y\prime=u(x)\]
myko
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maybe putting it like this\[y = 1+u ^{2}/2u'\]
myko
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its expressed in terms of u and its drivative, so guess thats it
spyros
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y=y(x)
spyros
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we were told to substitute y' by u(x) solve for u(x) then for y(x)
spyros
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\[\frac{du(x)}{dx}=\frac{1+u(x)^2}{2y(x)} ,y(x)=\int\limits_{}^{}{u(x)}\]
dumbcow
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i think you need to make u a function of y
u(y) = y'
then y'' = u*(du/dy)
Mr.Math
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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.
spyros
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will do
dumbcow
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rearrange it so it looks like
\[\frac{2u*du}{1+u^{2}} = \frac{dy}{y}\]