Answer:

Step-by-step explanation:
We have that:
(Equation 1)
To resolve this integral equation, we need to use the second Fundamental Theorem of Calculus, which says:
![\frac{d}{dx} [\int\limits^x_a {f(t)} \, dt]=f(x)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5B%5Cint%5Climits%5Ex_a%20%7Bf%28t%29%7D%20%5C%2C%20dt%5D%3Df%28x%29)
So, we need to differentiate both sides of equation 1 with respect to x:
![\frac{dy}{dx} =\frac{d}{dx} [2+\int\limits^x_2 {[t-ty(t)]} \, dt]](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%5Cfrac%7Bd%7D%7Bdx%7D%20%5B2%2B%5Cint%5Climits%5Ex_2%20%7B%5Bt-ty%28t%29%5D%7D%20%5C%2C%20dt%5D)
![\frac{dy}{dx} =\frac{d(2)}{dx}+\frac{d}{dx} [\int\limits^x_2 {[t-ty(t)]} \, dt]](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%5Cfrac%7Bd%282%29%7D%7Bdx%7D%2B%5Cfrac%7Bd%7D%7Bdx%7D%20%20%5B%5Cint%5Climits%5Ex_2%20%7B%5Bt-ty%28t%29%5D%7D%20%5C%2C%20dt%5D)
We know that the derivate for a constant value is zero. And,
(Equation 2)
Using the second Fundamental Theorem of Calculus we know that:
![\frac{d}{dx} [\int\limits^x_2 {t} \, dt]=x\\ \frac{d}{dx} [\int\limits^x_2 {ty(t)} \, dt]=xy(x)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5B%5Cint%5Climits%5Ex_2%20%7Bt%7D%20%5C%2C%20dt%5D%3Dx%5C%5C%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5B%5Cint%5Climits%5Ex_2%20%7Bty%28t%29%7D%20%5C%2C%20dt%5D%3Dxy%28x%29)
So, we need to replace those equations in equation 2, and we obtain:
(Equation 3)
Now, we are going to resolve the equation 3 as a normal equation. So, we need to joint the same variables. I mean, variable y on one side and variable x on other side, as follows:

And, we integrate each side of the equation to obtain:
(Equation 4)
Now, we need to find the value of the constant C. And we know that we can find one point of the equation, replacing x=2 in equation 1, because the integral becomes zero, so:

And, we replace the value of y when x=2 in equation 4 and we obtain,

So, Equation 4 is:
(Equation 4')
Now, we need to clear the y variable from Equation 4', (we are going to asumme that y-1>0),
