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

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

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

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

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

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

Mr.Math
 2 years ago
Best ResponseYou've already chosen the best response.1Putting \(\large u=\frac{dy}{dx}\) 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.

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