The given equation is a first-order partial differential equation (PDE) and can be solved using the method of characteristics. Let’s go step by step.

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Step 1: Write the PDE

The given equation is: $x u_x + (x + y) u_y = 1$ with the initial condition $u(1, y) = y$.

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Step 2: Characteristic equations

The method of characteristics converts the PDE into a system of ordinary differential equations (ODEs): $\frac{dx}{x} = \frac{dy}{x + y} = \frac{du}{1}.$

Let’s solve these equations one by one.

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Step 3: Solve $\frac{dx}{x} = \frac{dy}{x + y}$

This implies: $\frac{dy}{dx} = \frac{x + y}{x}.$ Rewriting: $\frac{dy}{dx} - \frac{y}{x} = 1.$ This is a linear first-order ODE for $y(x)$. The integrating factor is $\mu(x) = \frac{1}{x}$. Multiplying through by $\frac{1}{x}$: $\frac{d}{dx} \left( \frac{y}{x} \right) = \frac{1}{x}.$ Integrating both sides: $\frac{y}{x} = \ln|x| + C_1 \quad \implies \quad y = x \ln|x| + C_1 x,$ where $C_1$ is a constant.

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Step 4: Solve $\frac{dx}{x} = \frac{du}{1}$

This implies: $\frac{du}{1} = \frac{dx}{x}.$ Integrating: $u = \ln|x| + C_2,$ where $C_2$ is a constant.

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Step 5: Express the general solution

From Step 3, $C_1$ is constant along a characteristic curve. Let $C_1 = k$. Then: $y = x \ln|x| + kx.$ From Step 4, along the same characteristic curve: $u = \ln|x| + C_2.$ Expressing $C_2$ in terms of $k$, the solution is $u = u(x, y)$.

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Step 6: Apply initial condition $u(1, y) = y$

When $x = 1$, $u(1, y) = y$. Using the solution from Step 5, when $x = 1$, $y = 1 \ln|1| + k(1) = k$. Hence: $k = y.$ Substitute $k = y$ into the general solution: $u(x, y) = \ln|x| + y - x \ln|x|.$

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Final Solution:

The solution to the PDE is: $u(x, y) = \ln|x| + y - x \ln|x|.$

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Step 7: Is the solution defined everywhere?

• The solution is well-defined for $x > 0$ because of the $\ln|x|$ term, but it is undefined at $x \leq 0$.
• Therefore, the solution is not defined everywhere but is valid for $x > 0$.

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Would you like further clarifications? 😊
Here are 5 related questions to expand this topic:

1. What happens if the initial condition is changed to $u(x, 0) = x$?
2. How does the method of characteristics work for second-order PDEs?
3. Could you solve the equation in polar coordinates instead of Cartesian coordinates?
4. Why does the solution become undefined at $x \leq 0$?
5. How does the solution change for different forms of initial conditions, such as $u(x, y) = x^2 + y^2$?

Tip: When solving PDEs using characteristics, always check for regions where the solution might become singular or undefined.