<s>[INST] initial value problem $y'=-y^2, y(1)= \sqrt{2}$ hey I cant seem to solve this at all, not sure how you do it with no x values 
$$y'=-y^2, y(1)= \sqrt{2}$$
 [/INST] We have $y'=-y^2 \iff \frac{y'}{-y^2}=1 \iff \frac{d}{dx}(\frac{1}{y})=\frac{dx}{dx} \iff \frac{1}{y}=x+c$
$y(1)=\sqrt{2} \iff c=\frac{\sqrt{2}-2}{2}$
So $1/y=\frac{2x+\sqrt{2}-2}{2}\iff y=\frac{2}{2x+\sqrt{2}-2}$
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