Answer
$y^2-2\ln x=4$
Work Step by Step
Find the initial condition:
$y(1)=2+\int_1^1\frac{dt}{ty(t)}$
$y(1)=2+0$
$y(1)=2$
Construct a differential equation by differentiating the integral equation:
$\frac{d}{dx}(y(x))=\frac{d}{dx}(2+\int_1^x\frac{dt}{ty(t)})$
$\frac{dy}{dx}=0+\frac{1}{xy(x)}$
$\frac{dy}{dx}=\frac{1}{xy}$
Solve for $y$ by separating variables:
$y dy=\frac{1}{x}dx$
$\int y dy=\int \frac{1}{x}dx$
$\frac{y^2}{2}=\ln x+C$
Substitute the initial condition:
$\frac{2^2}{2}=\ln 1+C$
$2=0+C$
$C=2$
We have now:
$\frac{y^2}{2}=\ln x+2$
$y^2=2\ln x+4$
$y^2-2\ln x=4$
Thus, the solution is $y^2-2\ln x=4$.
