I am watching one of Ted Shifrin's multivariable calculus lectures and I have a question about some of the reasoning used in one of the examples. I will state the reasoning below.
Suppose $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable and $$x^2 \frac{\partial f}{\partial x} - y \frac{\partial f}{\partial y} = 0$$ everywhere.
We can use the gradient to determine what the graph of $f$ looks like. Note that this equation above may be rewritten as: $$\nabla{f} \cdot \begin{bmatrix}x^2 \\ -y\end{bmatrix}=0$$ Thus, we see that $\begin{bmatrix}x^2 \\ -y\end{bmatrix}$ is tangent to the level curve through $\begin{bmatrix}x \\ y\end{bmatrix}$.
On a level curve, the slope at $\langle{x,y}\rangle$ is $-\frac{y}{x^2}$.
Suppose that the level curve is given by $y = g(x)$. Then $$g'(x) = \frac{dy}{dx} = -\frac{y}{x^2}$$
We use separation of variables to compute $$ \begin{align*}\frac{dy}{dx} = -\frac{y}{x^2} &\iff \frac{1}{y} dy = -\frac{1}{x^2}dx \\ \log|y| &= \frac{1}{x} + C\end{align*} $$ Exponentiating, we see $$|y| = ce^{\frac{1}{x}}, \quad \text{ for } c \in \mathbb{R}$$
Consider an arbitrary differentiable function $\Phi(y), \enspace y\geq 0$. Setting $f(1,y) = \Phi(y)$ should uniquely determine $f$ for all $x,y \geq 0$.
Observe that $\Phi$ gives us information regarding the behavior of $f$ on the line $x=1$.
Given $\langle{X,Y}\rangle$, we can determine the constant associated with the level curve by considering the restriction $$Y=ce^{\frac{1}{X}} \implies c = e^{-\frac{1}{X}}Y$$ Now, along the line $x=1$, we know that it must intersect the level curve at $y = ce^{\frac{1}{1}}= ce$.
So, we see that $f(X,Y) = \Phi(e\cdot e^{-\frac{1}{X}} Y) $.
In summary: $$f(x,y) = \Phi(e^{1-\frac{1}{x}}y)$$
My questions are:
- How did we know to consider the form $y=g(x)$ for the level curve? In fact, we do not exactly get a function as our answer. Is the purpose of this just to specify some arbitrary function and then consider what possible functions satisfy the condition given by our knowledge of the slope at different points?
- How was $f(1,y) = \Phi(y)$ chosen? Could we have considered a line other than $x=1$ for our definition of $\Phi$ and determined functions that satisfy this differential equation? Why does considering the point $(1, ce)$ and $\Phi(ce)$ happen to give us the correct equation for $f(x,y)$?
