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

Question

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

$$ \ln(y(x))=x^2+C$$

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This I don't get, however: $$y(x) = C_1 \exp(x^2)$$

Where does this come from?

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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.

Answer 1

$$\ln(y(x))=x^2+C$$ $$e^{\ln(y(x))}=e^{x^2+C}$$ $$y(x)=e^{x^2}e^{C}$$

$e^{C}$ 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.

Answer 2

Strictly speaking, you get $\ln|y(x)|=x^2+C$, hence $y(x) = \pm e^{x^2}e^C$. Or $y\equiv 0$, which is a solution we lost when dividing by $y$. A neat way to put all these together is to write $y(x) = C_1e^{x^2}$ where $C_1$ is a new constant, $C_1=\pm e^C$ or $C_1=0$ (in other words, $C_1$ can be any real number). This is done all the time with ODE that lead to $\ln(y) = \dots$.

At the last stage we have multiple ways to write down the solution, because, e.g., $$3e^{x^2-1} = \frac{3}{e} e^{x^2}$$ Which form you choose does not matter; different people will arrive at different forms, depending on when how chose to use the initial condition.