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!

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 $$y=x^2(\ln (x+1)-\ln x+1)$$
