Question

need help with the attachment

Answer 1

2+4y''=0

Answer 2

no i don't think so

Answer 3

you have
\[x^2+2y^2=2\] so take the first derivative and get
\[2x+4yy'=0\] solve for
\[y'\] to get
\[y'=-\frac{x}{2y}\] and then you have to do it again

Answer 4

this time with the quotient rule

Answer 5

\[y''=-\frac{1}{2}\frac{y-xy'}{y^2}\]

Answer 6

and for the last step replace
\[y'\] by
\[-\frac{x}{2y}\]and do some algebra to get rid of the compound fractions

Answer 7

I got 2y^(2)+x^(2)/2y^(3)