Value of a function involving integration at a particular point Let $y = y(x) $ satisfy the equation 
$$y(x) + \int_1^x{y(t)}dt = x^2$$ then find the value of y(e).
My attempt: I tried to differentiate the given expression and reached to the following:
$$y'(x) + y(x) = 2x$$I don't have any clue how to find the value of $y(e)$ from here. Please help.
 A: You differentiated correctly. The expression you got after that, is called a "differential equation" and depending on its form, one can be solved in different ways.
A straight forward and common way to solve such a linear ordinary differential equation, is multiplying both sides by $e^x$, since one can see the following pattern :
$$y'(x) + y(x) = 2x \Leftrightarrow e^xy'(x)  + e^xy(x) = 2e^xx \Rightarrow [e^xy(x)]'=2e^xx \Rightarrow $$
$$\int[e^xy(x)]'dx=\int2e^xxdx \Rightarrow e^xy(x)= 2e^x(x-1) + c_1$$
To find the constant $c_1$, check your initial equation. Plugging $x=1$ will give you : 
$$y(1) + \int_1^1y(t)dt=1 \Rightarrow y(1)=1$$
This means that :
$$y(1)=c_1\cdot e^{-1}\Rightarrow c_1 = e$$
So, the final solution is : 
$$y(x) = 2(x-1) +\frac{e}{e^x}$$ 
and for $x=e$  :
$$y(e) = 2(e-1) + \frac{e}{e^e}=2(e-1) + e^{1-e}$$
A: Hint: This is a linear ODE. The solution is given as (homogeneouse+linear function ansatz for particular solution)
$$y(x)= c_1\exp(-x)+2x-2.$$
Plug this into the original equation and determine the value of $c_1$. Then finally use $x=e$ and determine $y(e)$.
