Differential equation with limit 
Find a solution y=f(x) to the differential equation 
  $$\frac{1}{x^2}y'-\frac{2}{x^3}y+\frac{1}{x^2+x}=0 ,\qquad x>0$$ satisfying
  $\lim_{x\to\infty} \frac{f(x)}{x^2}=1$.

I have $y=x^2(\ln(x+1)-\ln(x)+C)$ so far but can someone please explain the process from beginning to end? Thanks!
 A: That limits lets you find $C$. Just evaluate $$\lim_{x\to\infty}\frac{f(x)}{x}=1\iff \ln (x+1)-\ln x +c=1 \iff \lim_{x\to\infty} \ln \left (\frac{x+1}{x}\right )+c=1 \stackrel{\ln x \text{ cont.}}{\iff} \ln \left (\lim_{x\to\infty}\frac{x+1}{x}\right )+c_1=1\iff 0+c=1 \iff c=1$$
So $$y=x^2(\ln (x+1)-\ln x+1)$$
A: $$\lim_{x \to \infty}\frac{x^2(\ln(x+1)-\ln(x)+C)}{x^2}=\lim_{x \to \infty}\ln(x+1)-\ln(x)+C=C+\lim_{x \to \infty}\ln(\frac{x+1}{x})$$
But $\frac{x+1}{x} \to 1$, so $\ln(\frac{x+1}{x}) \to 0$, so:
$$C+\lim_{x \to \infty}\ln(\frac{x+1}{x})=C$$
If you put $C=1$ you have $\lim_{x \to \infty}\frac{y(x)}{x^2}=1$
A: You've got a solution to the differential equation:
$$
f(x)=x^2(\ln(x+1)-\ln(x)+C)
$$
but it involves an arbitrary constant $C$.  The limit condition is there so that the constant $C$ is defined to have a particular value - so there's a unique solution rather than a family of solutions differing by a constant.  
We have
$$
f(x)/x^2=(\ln(x+1)-\ln(x)+C)
$$
There's only one possible value of $C$ such that the limit of this expression as $x\to\infty$ is $1$.  What is it?  Hint: Remember the rules for manipulating logarithms!
