Finding the solution of a partial derivative equation

Question

Let $f : \mathbb R^2 → \mathbb R$ be a function that has partial derivatives that are continuous on $\mathbb R^2$

Answer the following questions:

1. Find all the $g$ s such that $\frac{∂g(u, v)}{∂u} = 0$ [ Hint: Turn the problem into an ODE by fixing $v$ and by considering $h(u) := g(u, v)$ ]

2. Find all the solutions of $3. \frac{∂f(x, y)}{∂x} − 2. \frac{∂f(x,y)}{∂y} = 0$ [ Hint: Let $(u, v) := (x + y, 2x + 3y)$ and let $g(u, v) := f(x, y)$]

My attempts:

1. If $\frac{∂g(u, v)}{∂u} = 0$ then $\frac{∂h(u)}{∂u} = 0$ (because $h(u) = g(u,v)$), then $h(u)$ is a constant function and since $v$ is fixed, we can take $h(u) = v$, and hence $g(u,v) = v$. Is this correct? Any other answers?

2.I don't know how to start sovling this using the chain rule. Any help please?

Answer 1

1. Consider $v^2$, $v^3$, or $v^n$. Any function of v will work here, as will any sum of functions of $v$, just as long as there is no $u$. So we can say that $g(u,v) = h(v)$, where $h(v)$ is any arbitrary function of $v$ (with no $u$ involved).
2. Start with the equality, and then take the $(x,y)$ partials of both sides. $$g(u,v) = f(x,y)$$ $$\frac{\partial g(u,v)}{\partial x} = \frac{\partial f(x,y)}{\partial x}, \frac{\partial g(u,v)}{\partial y} = \frac{\partial f(x,y)}{\partial y} $$ $$\frac{\partial g(u,v)}{\partial x} = \frac{\partial g(u,v)}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial g(u,v)}{\partial v}\frac{\partial v}{\partial x} = 1\frac{\partial g(u,v)}{\partial u}+2\frac{\partial g(u,v)}{\partial v}$$ $$\frac{\partial g(u,v)}{\partial y} = \frac{\partial g(u,v)}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial g(u,v)}{\partial v}\frac{\partial v}{\partial y} = 1\frac{\partial g(u,v)}{\partial u}+3\frac{\partial g(u,v)}{\partial v}$$ So: $$3\frac{\partial f(x,y)}{\partial x}-2\frac{\partial f(x,y)}{\partial y} = 3(1\frac{\partial g(u,v)}{\partial u}+2\frac{\partial g(u,v)}{\partial v})-2(1\frac{\partial g(u,v)}{\partial u}+3\frac{\partial g(u,v)}{\partial v})=\frac{\partial g(u,v)}{\partial u}=0.$$ It's probably no coincidence that this is exactly the same as part 1.

So now we have that $g(u,v) = h(v)$, where $h(v)$ is any arbitrary function of $v$ (with no $u$ involved). So $$f(x,y) = h(2x+3y),$$ and you can check that this does satisfy the partial differential equation.