Let $$y = y(x)$$ be the solution of the differential equation $$x^3 dy + (xy - 1)dx = 0$$, $$x \gt 0$$, $$y\left(\frac{1}{2}\right) = 3 - e$$. Then $$y(1)$$ is equal to

We need to solve the differential equation $$x^3 dy + (xy - 1)\,dx = 0$$ with the initial condition $$y\bigl(\tfrac12\bigr) = 3 - e$$.

We start by rewriting the equation as $$x^3 \frac{dy}{dx} + xy - 1 = 0$$ which leads to $$\frac{dy}{dx} + \frac{y}{x^2} = \frac{1}{x^3}\,.$$

This is a linear ODE of the form $$\frac{dy}{dx} + P(x)y = Q(x)$$ with $$P(x)=\frac{1}{x^2}$$ and $$Q(x)=\frac{1}{x^3}\,.$$

Next, we compute the integrating factor $$\mathrm{IF} = e^{\int \!P(x)\,dx} = e^{\int \frac{1}{x^2}\,dx} = e^{-1/x}\,.$$

Multiplying both sides of the differential equation by this integrating factor gives

$$y\,e^{-1/x} = \int \frac{1}{x^3}\,e^{-1/x}\,dx\,.$$

Substituting $$t = \frac{1}{x}$$, $$dt = -\frac{1}{x^2}\,dx$$, $$dx = -\frac{dt}{t^2}$$ transforms the integral into $$\int \frac{1}{x^3}e^{-1/x}\,dx = \int t^3 e^{-t}\cdot\Bigl(-\frac{dt}{t^2}\Bigr) = -\int t\,e^{-t}\,dt\,.$$

Using integration by parts, we have

$$\int t\,e^{-t}\,dt = -t\,e^{-t} + \int e^{-t}\,dt = -t\,e^{-t} - e^{-t} + C\,.$$ Thus $$-\int t\,e^{-t}\,dt = t\,e^{-t} + e^{-t} + C = \frac{e^{-1/x}}{x} + e^{-1/x} + C\,.$$ This yields $$y\,e^{-1/x} = \frac{e^{-1/x}}{x} + e^{-1/x} + C \quad\Longrightarrow\quad y = \frac{1}{x} + 1 + C\,e^{1/x}\,.$$ Next, we apply the initial condition $$y\bigl(\tfrac{1}{2}\bigr)=3-e$$: $$3 - e = \frac{1}{\tfrac{1}{2}} + 1 + C\,e^{2} = 2 + 1 + C\,e^{2} = 3 + C\,e^{2}\,,\quad\text{so}\quad C\,e^{2} = -e\,,\quad C = -e^{-1}\,.$$ Finally, the value at $$x=1$$ is $$y(1) = 1 + 1 + \bigl(-e^{-1}\bigr)e^{1} = 2 - 1 = 1\,.$$ Therefore, $$y(1) = 1\,. $$