(I looked for a post similar but couldn't find one. I apologize in advance if there's a post similar to this. Apologizes to the moderators for a weak post.)
I'm taking an ordinary differential equation class and as a challenge, my professor assigned this partial differential equation question. Since I have no idea how partial DE work, I need some help determining which steps I can take and which I cannot. The prompts have been written out but I will include (updated) photos of the original and my attempt.
Consider $x\frac{\partial u}{\partial y} - y\frac{\partial u}{\partial x}=0$
where$u(x,y)=f(x)g(y)$
(a) Use this assumption to convert the partial differential equation (3) into an equation that involves x, y, f, g, and only ordinary derivatives of f and g.
(b) Since f depends only on x and g only on y, the equation you obtained in part (a) should be now “separable.” Use some elementary algebra to separate the variables so that each side consists of functions that depend on only one variable.
(c) Here is a key observation: Your answer from part (b) is stating that a function of x is equal to a function of y. How can this be true? There is only one way this can be true! Can you see how? Based on your answer for this question, form two ordinary differential equations, one from each side, one for f in x and the other for g in y.
Just in case people cannot view the attempt photo:
(a)
$\frac{\partial u}{\partial y}(F(x)g(y))$ = $f(x)g'(y)$
$\frac{\partial u}{\partial x}(F(x)g(y))$ = $f'(x)g(y)$
Plugging into original equation$xf(x)g'(y)-yf'(x)g(y)=0$
(b)
Using algebra to "separate"$\frac{xf(x)}{f'(x)} = \frac{yg(y)}{g'(y)}$
Images:Part(a)Part (b)Part (c)My attempt
$\endgroup$1 Answer
$\begingroup$Replacing $u(x,y)=f(x)g(y)$ into the equation we get $$ xf(x)g'(y)-yf'(x)g(y)=0. $$By assuming that for all $x,y\in\mathbb{R}$ we have $f(x)\neq 0$ and $g(y)\neq0$, we can rewrite the previous equation as $$ \dfrac{f'(x)}{xf(x)}=\dfrac{g'(y)}{yg(y)}. $$Since the left-hand side depends only on $x$, while the right-hand side only on $y$, this implies that there exists a constant $\lambda\in\mathbb{R}$ such that for all $x\in\mathbb{R}$ and $y\in\mathbb{R}$ we have $$ f'(x)=\lambda xf(x) \quad \hbox{and}\quad g'(y)=\lambda yg(y). $$Then, solving these equations you immeadiately get $$ f(x)=C_1e^{-\frac{\lambda}{2}x^2} \quad \hbox{and}\quad g(y)=C_2e^{-\frac{\lambda}{2}y^2} \ \implies \ u(x,y)=C_3e^{-\frac{\lambda}{2}(x^2+y^2)}. $$
$\endgroup$ 2
