Answer
$y=e^{-x^2/2+2}+1$
Work Step by Step
Find the initial condition:
$y(2)=2+\int_2^2[t-ty(t)]dt$
$y(2)=2+0$
$y(2)=2$
Construct a differential equation by differentiating the integral equation:
$\frac{d}{dx}(y(x))=\frac{d}{dx}(2+\int_2^x[t-ty(t)]dt$
$\frac{dy}{dx}=0+(x-xy(x))$
$\frac{dy}{dx}=x(1-y)$
Solve for $y$ by separating variables:
$\frac{1}{y-1}dy=-xdx$
$\int \frac{1}{y-1}dy=\int -xdx$
$\ln (y-1)=-\frac{x^2}{2}+C$
Substitute the initial condition:
$\ln(2-1)=-\frac{2^2}{2}+C$
$\ln 1=-2+C$
$0=-2+C$
$C=2$
We have now:
$\ln(y-1)=-\frac{x^2}{2}+2$
$y-1=e^{-x^2/2+2}$
$y=e^{-x^2/2+2}+1$
Thus, the solution is $y=e^{-x^2/2+2}+1$.
