1. **Problem statement:** Solve the first order partial differential equation (PDE) given by: $$3U_x - 2U_y + U = 2x + 1$$ where $U_x = \frac{\partial U}{\partial x}$ and $U_y = \frac{\partial U}{\partial y}$. 2. **Formula and method:** This is a linear first order PDE. We use the method of characteristics. The characteristic equations are: $$\frac{dx}{3} = \frac{dy}{-2} = \frac{dU}{2x + 1 - U}$$ 3. **Solve the characteristic curves:** From the first equality: $$\frac{dx}{3} = \frac{dy}{-2} \implies 3 dy = -2 dx \implies 3y + 2x = C_1$$ where $C_1$ is a constant along characteristics. 4. **Solve the ODE along characteristics:** Using the second equality: $$\frac{dU}{ds} = 2x + 1 - U$$ where $s$ parameterizes the characteristic curve. But since $3y + 2x = C_1$, we can express $y$ in terms of $x$ and $C_1$ or vice versa. Rewrite the PDE along the characteristic: $$\frac{dU}{ds} + U = 2x + 1$$ This is a linear ODE in $U$ with integrating factor $e^{s}$. 5. **Express $x$ in terms of $s$:** From $\frac{dx}{ds} = 3$, integrate: $$x = 3s + C_2$$ Similarly, from $\frac{dy}{ds} = -2$: $$y = -2s + C_3$$ Using the relation $3y + 2x = C_1$, substitute $x$ and $y$: $$3(-2s + C_3) + 2(3s + C_2) = C_1 \implies -6s + 3C_3 + 6s + 2C_2 = C_1 \implies 3C_3 + 2C_2 = C_1$$ Constants $C_2, C_3$ relate to initial conditions; for simplicity, treat $s$ as parameter along characteristic. 6. **Solve the ODE for $U$:** $$\frac{dU}{ds} + U = 2(3s + C_2) + 1 = 6s + 2C_2 + 1$$ Multiply both sides by $e^{s}$: $$\frac{d}{ds}(U e^{s}) = e^{s}(6s + 2C_2 + 1)$$ Integrate both sides: $$U e^{s} = \int e^{s}(6s + 2C_2 + 1) ds + C$$ 7. **Integrate the right side:** Use integration by parts for $\int 6s e^{s} ds$: $$\int 6s e^{s} ds = 6 (s e^{s} - e^{s}) + C = 6 e^{s}(s - 1) + C$$ Also, $$\int e^{s} ds = e^{s} + C$$ So, $$\int e^{s}(6s + 2C_2 + 1) ds = 6 e^{s}(s - 1) + (2C_2 + 1) e^{s} + C$$ 8. **Solve for $U$:** $$U e^{s} = 6 e^{s}(s - 1) + (2C_2 + 1) e^{s} + C$$ Divide both sides by $e^{s}$: $$U = 6(s - 1) + 2C_2 + 1 + C e^{-s}$$ 9. **Rewrite in terms of $x$ and $y$:** Recall $x = 3s + C_2 \implies s = \frac{x - C_2}{3}$. Substitute $s$: $$U = 6\left(\frac{x - C_2}{3} - 1\right) + 2C_2 + 1 + C e^{-\frac{x - C_2}{3}}$$ Simplify: $$U = 2(x - C_2 - 3) + 2C_2 + 1 + C e^{-\frac{x - C_2}{3}} = 2x - 2C_2 - 6 + 2C_2 + 1 + C e^{-\frac{x - C_2}{3}}$$ $$U = 2x - 5 + C e^{-\frac{x - C_2}{3}}$$ 10. **Express $C$ and $C_2$ in terms of $3y + 2x$:** Since $3y + 2x = C_1$, and $C_2$ is related to initial conditions, the general solution is: $$U = 2x - 5 + F(3y + 2x) e^{-\frac{x}{3}}$$ where $F$ is an arbitrary function determined by initial/boundary conditions. **Final answer:** $$\boxed{U(x,y) = 2x - 5 + e^{-\frac{x}{3}} F(3y + 2x)}$$ where $F$ is an arbitrary function of the characteristic variable $3y + 2x$.