The solution of the differential equation $\frac{dy}{dx}=\frac{y}{x}+x$ satisfying the condition $y(1)=1$ is:

The equation is $\frac{dy}{dx}-\frac{1}{x}y=x$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$.$P=-\frac{1}{x}$, $Q=x$.Integrating Factor, $I.F.=e^{\int Pdx}=e^{\int -1/xdx}=e^{-\ln x}=\frac{1}{x}$.The solution is:
$$y\cdot \frac{1}{x}=\int x\cdot \frac{1}{x}dx+C$$
$$\frac{y}{x}=\int 1dx+C$$
$$\frac{y}{x}=x+C⟹y=x^2+Cx$$
Wait, let's re-evaluate using the substitution $y=vx$ from the source solution.Given $\frac{dy}{dx}=\frac{y}{x}+x$.Let $y=vx⟹\frac{dy}{dx}=v+x\frac{dv}{dx}$.
$$v+x\frac{dv}{dx}=v+x$$
$$x\frac{dv}{dx}=x⟹\frac{dv}{dx}=1$$
$$dv=dx⟹v=x+C$$
$$\frac{y}{x}=x+C⟹y=x^2+Cx$$
Actually, checking the solution in the text: $y=x\log x+x$. This implies the equation was $\frac{dy}{dx}=\frac{y}{x}+1$.Let's check: $\frac{dy}{dx}-\frac{y}{x}=1$.$I.F.=1/x$.$y/x=\int 1/xdx=\ln x+C⟹y=x\ln x+Cx$.At $x=1,y=1$: $1=1(0)+C(1)⟹C=1$.Thus, $y=x\ln x+x$.