Question

Find y′′ for the curve ln(x) + y = ln(x^2) − y^2 at y = 0? Note that the domain for x is that x > 0. There will only be one point on the curve with y = 0

Answer 1

`(d^2y)/dx^2 = -(4y^2+4y+3)/(x^2(1+2y)^3)`

For `x=1` we have `y(1)=0` and `y''(1) = -3`

Explanation:

Given the implicit function:

`lnx +y = ln(x^2) -y^2`

defined for `x > 0`, note that based on the properties of logarithms:

`ln(x^2) = 2lnx`

so that the equation becomes:

`y^2+y = lnx`

and then for `x=1` we have `y=0`.

Differentiate both sides of the equation:

`2ydy/dx +dy/dx = 1/x`

`dy/dx (1+2y) = 1/x`

and again:

`(d^2y)/dx^2(1+2y) +2(dy/dx)^2 = -1/x^2`

substituting:

`dy/dx = 1/(x(1+2y) )`

we have:

`(d^2y)/dx^2(1+2y)= -2/(x(1+2y) )^2 -1/x^2`

`(d^2y)/dx^2 = -1/(x^2(1+2y)) (2/(1+2y) ^2 +1)`

`(d^2y)/dx^2 = -1/(x^2(1+2y)^3) (2+(1+2y) ^2 )`

`(d^2y)/dx^2 = -1/(x^2(1+2y)^3) (2+1+4y+4y^2 )`

`(d^2y)/dx^2 = -(4y^2+4y+3)/(x^2(1+2y)^3)`

For `x=1`:

`[(d^2y)/dx^2]_(x=1) = -3`

Answer 2

`(d^2y)/(dx^2)= pm 3`

Explanation:

`logx+y=logx^2-y^2rArrlog(xe^y)=log(x^2e^(-y^2))`

Assuming `x > 0` we have

`xe^y=x^2e^(-y^2)` or

`e^(y+y^2)=x` or

`y^2+y-log x=0` now solving for `y`

`y = 1/2(-1pmsqrt(1+4logx))` which is real for `1+4logx ge 0` or

`0 < x` and now

`(d^2y)/(dx^2) = pm(3 + 4 Logx)/(x^2 (1 + 4 Logx)^(3/2))`

Now for `e^(y+y^2)=x` we have `y=0 rArr x = 1` and then

`(d^2y)/(dx^2)= pm 3`