How do you find #(d^2y)/(dx^2)# given #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)#