Find the error in the following integration problem Given that $y(x)$ is differentiable  And
$$ x\int_{1}^x y(t) dt = (x+1)\int_1^x ty(t)dt$$
 A: Since you don't write clearly all the steps of your calculus, one cannot find where is the mistake. 
Obviously, $y(x)=0$ and $x=1$ are two distinct solutions of the equation. The question is : Is there other(s) solution(s) ?
$$ x\int_{1}^x y(t) dt = (x+1)\int_1^x ty(t)dt  \tag 1$$
This is an integral equation which can be solved for $y(x)$.
The differentiation leads to : $\quad \int_{1}^x y(t) dt +xy(x)=\int_1^x ty(t)dt+(x+1)xy(x)$
$\quad \int_{1}^x y(t) dt =\int_1^x ty(t)dt+x^2y(x)$
Differentiating again leads to : $\quad y(x)=xy(x)+2xy(x)+x^2y'(x)$
$$(3x-1)y(x)+x^2y'(x)=0$$
The general solution of this first order linear ODE is : 
$$y(x)=c\:x^{-3}e^{-1/x} $$
Any constant $c$. All these functions are not solution of Eq.$(1)$, but the solutions are among them. 
$c$ has to be determined in order to satisfy Eq.$(1)$.
$ x\int_{1}^x c\:t^{-3}e^{-1/t} dt = (x+1)\int_1^x tc\:x^{-3}e^{-1/t}dt $
$c(x+1)e^{-1/x}-2cxe^{-1}=c(x+1)e^{-1/x}-c(x+1)e^{-1} \quad\to\quad c(x-1)e^{-1}=0$
They are only two possibilities : 


*

*$c=0\quad\implies\quad y(x)=0$ is a solution. 

*$x=1$, with any integrable function $y(x)$, is another solution.

A: The error happened when you divided top and bottom of y(x)/(xy(x)) by y(x). How did you know that y(x) was not zero in a punctured neighborhood of x=1? 
