anonymous 4 years ago y''=(1+y'^2)/2y

1. anonymous

$y \prime \prime=\frac{1+y \prime ^2}{2y}$

2. anonymous

hint : $y\prime=u(x)$

3. anonymous

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

4. anonymous

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

5. anonymous

y=y(x)

6. anonymous

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

7. anonymous

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

8. anonymous

i think you need to make u a function of y u(y) = y' then y'' = u*(du/dy)

9. Mr.Math

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. anonymous

will do

11. anonymous

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