Question

If $x^4-xy+y^4=1$, then $y_2$ at (0,1 ) is

• A. $\frac{1}{8}$
• B. $\frac{-1}{8}$
• C. $\frac{1}{16}$
• D. $\frac{-1}{16}$

Answer 1

The correct option is D $\frac{-1}{16}$

$4x^3-y-x{y}^{\prime }+4y^3{y}^{\prime }=0$
$y^{\prime }=\frac{y-4x^3}{4y^3-x}\Rightarrow y^{\prime }{|}_{x=0,y=1}=\frac{1}{4}$
$12x^2-y^{\prime }-y^{\prime }-x{y}^{\prime \prime }+12y{y}^{\prime 2}+4y^{\prime \prime }{y}^3=0$
$\frac{-1}{2}+\frac{12}{16}+4y^{11}=0$

$y^1{|}_{x=0,y=1}=\frac{-1}{16}$