Question

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

Answer 1

\[y \prime \prime=\frac{1+y \prime ^2}{2y}\]

Answer 2

hint : \[y\prime=u(x)\]

Answer 3

maybe putting it like this\[y = 1+u ^{2}/2u'\]

Answer 4

its expressed in terms of u and its drivative, so guess thats it

Answer 5

y=y(x)

Answer 6

we were told to substitute y' by u(x) solve for u(x) then for y(x)

Answer 7

\[\frac{du(x)}{dx}=\frac{1+u(x)^2}{2y(x)} ,y(x)=\int\limits_{}^{}{u(x)}\]

Answer 8

i think you need to make u a function of y
u(y) = y'

then y'' = u*(du/dy)

Answer 9

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.

Answer 10

will do

Answer 11

rearrange it so it looks like
\[\frac{2u*du}{1+u^{2}} = \frac{dy}{y}\]