How do you find (d^2y)/(dx^2)d2ydx2 given x^2+xy-y^2=1x2+xyy2=1?

1 Answer
Oct 28, 2016

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

Explanation:

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)