关于求解一阶线性偏微分方程$2u_x − u_y = 10u + 5e^{x−3y}$的合适方法咨询

Hey there! Let's work through this problem together—first off, this is a nonhomogeneous first-order linear partial differential equation, and the Lagrange auxiliary equations method does apply here, so you were on the right track! Chances are you just hit a snag in handling the nonhomogeneous term. Let's break down the correct step-by-step approach:

Step 1: Solve the homogeneous PDE first #

Start with the corresponding homogeneous equation:
$$2u_x - u_y = 10u$$
The Lagrange auxiliary equations for this are:
$$\frac{dx}{2} = \frac{dy}{-1} = \frac{du}{10u}$$

• First, find the characteristic curves by solving $\frac{dx}{2} = \frac{dy}{-1}$:
  Integrate both sides: $-dx = 2dy$ → $x + 2y = C_1$ (where $C_1$ is a constant).
• Next, relate $dx$ and $du$ using $\frac{dx}{2} = \frac{du}{10u}$:
  Rearrange to $\frac{du}{u} = 5dx$, integrate to get $\ln|u| = 5x + C_2$ → $u e^{-5x} = C_2$.

The homogeneous solution is formed by making $C_2$ a function of $C_1$, so:
$$u_h = \phi(x + 2y) e^{5x}$$
where $\phi$ is any differentiable function of its argument.

Step 2: Find a particular solution for the nonhomogeneous term #

The nonhomogeneous part is $5e^{x-3y}$. Since this is an exponential function, we can guess a particular solution of the form $u_p = A e^{x-3y}$ (where $A$ is a constant to be determined).

Calculate the partial derivatives:

• $u_{p_x} = A e^{x-3y}$
• $u_{p_y} = -3A e^{x-3y}$

Substitute into the original PDE:
$$2(A e^{x-3y}) - (-3A e^{x-3y}) = 10(A e^{x-3y}) + 5e^{x-3y}$$

Simplify both sides:
$$5A e^{x-3y} = (10A + 5) e^{x-3y}$$

Cancel out the exponential term (since it's never zero) and solve for $A$:
$$5A = 10A + 5 → -5A = 5 → A = -1$$

So the particular solution is:
$$u_p = -e^{x-3y}$$

Step 3: Combine homogeneous and particular solutions #

The general solution to the original PDE is the sum of the homogeneous and particular solutions:
$$u(x,y) = \phi(x + 2y) e^{5x} - e^{x-3y}$$

Why your earlier attempts might have failed #

• If the coordinate method didn't work, maybe you didn't properly transform variables to align with the characteristic curves (using $\xi = x + 2y$, $\eta = y$ for example would simplify the PDE significantly).
• For Lagrange's method, it's easy to get stuck when handling nonhomogeneous terms—remember that after finding the homogeneous solution, you need to either use variation of parameters or guess a form for the particular solution that matches the nonhomogeneous term's structure.

备注：内容来源于stack exchange，提问作者starry41