How do you find $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/(dx^2)$ given $x^{2} + x y - y^{2} = 1$ $x^2+xy-y^2=1$ ?

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Answer 1 · 2016-10-28T11:02:47

$y''(x) = pm10/(5x^2-4)^(3/2)$

Calling

$f(x,y(x))=x^2+ x y(x)-y(x)^2-1=0$

$(df)/dx = 2x+y(x)-2y(x)y'(x)=0$

solving for $y'(x)$

$y'(x)=-(2x+y(x))/(x-2y(x))$

computing now

$d/dx((df)/dx) =2 + 2 y'(x) - 2 y'(x)^2 + x y''(x) - 2 y(x) y''(x)=0$ solving for $y''(x)$

$y''(x)=(10 (x^2 + x y(x) - y(x)^2))/(x - 2 y(x))^3$ after substituting $y'(x)$

Finally substituting $y(x) = 1/2 (x pm sqrt[5 x^2-4])$ we obtain

$y''(x) = pm10/(5x^2-4)^(3/2)$

How do you find (d^2y)/(dx^2)d2ydx2(d^2y)/(dx^2) given x^2+xy-y^2=1x2+xy−y2=1x^2+xy-y^2=1?