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spyros Group TitleBest ResponseYou've already chosen the best response.0
\[y \prime \prime=\frac{1+y \prime ^2}{2y}\]
 2 years ago

spyros Group TitleBest ResponseYou've already chosen the best response.0
hint : \[y\prime=u(x)\]
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
maybe putting it like this\[y = 1+u ^{2}/2u'\]
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
its expressed in terms of u and its drivative, so guess thats it
 2 years ago

spyros Group TitleBest ResponseYou've already chosen the best response.0
we were told to substitute y' by u(x) solve for u(x) then for y(x)
 2 years ago

spyros Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{du(x)}{dx}=\frac{1+u(x)^2}{2y(x)} ,y(x)=\int\limits_{}^{}{u(x)}\]
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.5
i think you need to make u a function of y u(y) = y' then y'' = u*(du/dy)
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.1
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

dumbcow Group TitleBest ResponseYou've already chosen the best response.5
rearrange it so it looks like \[\frac{2u*du}{1+u^{2}} = \frac{dy}{y}\]
 2 years ago
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