Please find the answer to the question

Question

Find d²y/dx² implicitly in terms of x and y.

x^2y − 2x = 8

d²y/dx^{2 =?}

Please find the answer and make it clear, please.

Answer 1

y''=(1/(ln(x))²)*[(-2x^-2y)(1+1/x)+(y/x²)(ln(x)+2)]

y''=dy²/dx²

I think is same as above answer just calculated further.

Answer 2

x^2y − 2x = 8 ----> e^2yln(x) − 2x = 8 -------> d / dx (e^2yln(x) − 2x = 8) ----> ( 2y/x + 2ln(x) dy/dx ) e^2yln(x) − 2 = 0

so, solving for dy/dx, we get the following:

2e^2yln(x)y/x + 2e^2yln(x)ln(x) dy/dx = 2 -----> e^2yln(x)y/x + e^2yln(x)ln(x) dy/dx = 1

e^2yln(x)ln(x) dy/dx = 1 - e^2yln(x)y/x ------> dy/dx = [ 1 - ye^2yln(x)/x ] / [ e^2yln(x)ln(x) ]

dy/dx = [ x - ye^2yln(x) ] / [ xe^2yln(x)ln(x) ] = [ xe^-2yln(x) - y ] / [ x ln(x) ]

dy/dx = [ x*x^-2y - y ] / [ x ln(x) ] = [ x^1-2y - y ] / [ x ln(x) ] = y',

so, xln(x)y' = x^1-2y - y, and taking the derivative again, we get the following:

d²y/dx² = d/dx ( [ x^1-2y - y ] / [ x ln(x) ] ) = d/dx ( [ e^(1-2y)ln(x) - y ] / [ x ln(x) ] )

d²y/dx² = [ ( xln(x) ) d/dx(e^(1-2y)ln(x) - y) - (e^(1-2y)ln(x) - y) d/dx( xln(x) ) ] / ( x ln(x) )²