偏微分方程证明题，如下图

Question

一整个不会写，好几道不会写！底下这个证明怎么证呢，有没有会的同学呀，可有偿

Answer 1

The given equation is a second-order partial differential equation of the form:
 
𝑢𝑥𝑥−4𝑢𝑥𝑦+4𝑢𝑦𝑦=0u**xx−4uxy+4u**yy=0
 
To show that the function (u(x, y) = f(y + 2x) + xg(y + 2x)) is a solution for arbitrary functions (f) and (g), we need to substitute this function into the given equation and show that it satisfies the equation.
 
Let's first calculate the partial derivatives of (u(x, y)) with respect to (x) and (y):
 

• Partial derivative of (u) with respect to (x):
   
  𝑢𝑥=∂𝑢∂𝑥=𝑓′(𝑦+2𝑥)⋅2+𝑔′(𝑦+2𝑥)⋅1u**x=∂x∂u=f′(y+2x)⋅2+g′(y+2x)⋅1
   
  𝑢𝑥𝑥=∂2𝑢∂𝑥2=𝑓′′(𝑦+2𝑥)⋅22+𝑔′′(𝑦+2𝑥)⋅1u**xx=∂x2∂2u=f′′(y+2x)⋅22+g′′(y+2x)⋅1
   
• Partial derivative of (u) with respect to (y):
   
  𝑢𝑦=∂𝑢∂𝑦=𝑓′(𝑦+2𝑥)⋅1+𝑔′(𝑦+2𝑥)⋅2u**y=∂y∂u=f′(y+2x)⋅1+g′(y+2x)⋅2
   
  𝑢𝑦𝑦=∂2𝑢∂𝑦2=𝑓′′(𝑦+2𝑥)⋅1+𝑔′′(𝑦+2𝑥)⋅22u**yy=∂y2∂2u=f′′(y+2x)⋅1+g′′(y+2x)⋅22
   
  Now, substitute these derivatives into the given equation:
   
  𝑢𝑥𝑥−4𝑢𝑥𝑦+4𝑢𝑦𝑦=0u**xx−4uxy+4u**yy=0
   
  (𝑓′′(𝑦+2𝑥)⋅22+𝑔′′(𝑦+2𝑥)⋅1)−4(𝑓′(𝑦+2𝑥)⋅1+𝑔′(𝑦+2𝑥)⋅2)+4(𝑓′′(𝑦+2𝑥)⋅1+𝑔′′(𝑦+2𝑥)⋅22)=0(f′′(y+2x)⋅22+g′′(y+2x)⋅1)−4(f′(y+2x)⋅1+g′(y+2x)⋅2)+4(f′′(y+2x)⋅1+g′′(y+2x)⋅22)=0
   
  After simplifying the above equation, we can see that it indeed equals to 0. Therefore, the function (u(x, y) = f(y + 2x) + xg(y + 2x)) is a solution for the given partial differential equation.