Q:

Solving the integral equation $y(x) = 3 + 2\int_1^x t y(t) dt $ by reducing it to a differential equation

Solve the integral equation
  $$y(x) = 3 + 2\int_1^x t \ y(t) \ dt  $$

First I solved for the integral equation. Then  I'm told to differentiate and I get 
$${dy \over dx} = 2 x  y(x) $$
Then I see that they're separable and I use that so I take both of the integrals after arranging the functions. What I get is
$$ \ln(y(x))=x^2+C$$

This I don't get, however:
$$y(x) = C_1  \exp(x^2)$$
Where does this come from?

Now I choose an initial value for the first equation and I choose $ x=1$
$$ y(1) = 3 + 0 = 3$$
Now here's the problem
After this it's supposed to be $C_1= 3/e$
and the solution is
$$ y(x) = 3 \exp(x^2-1)$$
I understand everything up til choosing the initial value but i'm guessing it might be arbitrary however what happens with the $C_1$? How is this calculated? Everything gets really confusing here.

A:

$$\ln(y(x))=x^2+C$$
$$e^{\ln(y(x))}=e^{x^2+C}$$
$$y(x)=e^{x^2}e^{C}$$
$e^{C}$ is just a number, so let $C_{1}=e^{C}$:
$$y(x)=C_1e^{x^2}$$
$C_1$ is just another way of expressing our constant of integration.

