First order differential equation involving natural logs where y(x)=0

Question

I am assigned to a question where I need to solve a 1st-order differential equation where $y(x)=0$

I am a little confused, since in order to solve for $C_1$, I need to take the natural log of a negative number. My calculator cannot compute this, so I'm wondering if it's even possible to solve for this equation.

The question:

$y'(4x-4-x^2)=9y^2-1, y(4)=0$

My solution:

$\int\frac{dy}{9y^2-1}=\int\frac{dx}{4x-4-x^2}$

$\int\frac{dy}{(3y+1)(3y-1)}=-\int\frac{dx}{(x-2)^2}$

$\frac{1}{6} ln \frac{3y-1}{3y+1}=\frac{1}{x-2}+c_1$

$y(4)=0$

$\frac{1}{6} ln \frac{3*0-1}{3*0+1}=\frac{1}{4-2}+c_1$

$\frac{1}{6} ln \frac{-1}{1}=\frac{1}{2}+c_1$

$c_1=-\frac{1}{2}$

Can this question be solved like this?

Thanks.

Answer 1

Typically in calculus, when the result of integration is the natural log of an expression, the absolute value of the expression is used instead. Here, you would have $\ln |\frac{3y - 1}{3y + 1}|$ instead of just $\ln \frac{3y - 1}{3y + 1}$. And with the absolute value sign, you would be dealing with only positive numbers.

Answer 2

The issue is that the integral of $1/x$ is not $\ln x$, but $\ln |x|$