Toronto Math Forum
APM346--2019 => APM346--Lectures & Home Assignments => Home Assignment 1 => Topic started by: Brittany Palandra on January 14, 2019, 10:04:54 PM
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Problem 4 was to find the general solution to $uu_{xy} = u_xu_y$. I used the hint and got $\frac{\partial}{\partial y} \ln u_x = \frac{\partial}{\partial y} \ln u$, but I'm not sure where to go from here.
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I have a guess but also not quite sure if I'm in the right way. Integrate the equation you got
$$\int_ {}{}\frac{(uu_x)_y}{uu_x} dy = \int {}{}\frac{u_y}{u}dy \qquad\qquad\qquad\qquad \color{red}{\text{Error in the left part; the rest incorrect. V.I.}}$$
$$ \ln uu_x = \ln u + f(x)$$
where $f$ is any functions that is only of $x$,
$$uu_x = u \cdot g(x)$$
where $g$ is a function of $f$ mentioned above. Divide $u$ from both sides,
$$ u_x = g(x)$$
then integrate both sides,
$$ u = G(x) + h(y)$$
where $G'(x) = g(x)$, and $h$ is any function that is only of y.
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This is my new try: Follow the hint and I got,
$$\frac{u_{xy}}{u_x} = \frac{u_y}{u}$$
integrate both sides,
$$\ln (u_x) = \ln (u) + f(x)$$
where $f(x)$ is any function only of x. Then, we have,
$$u_x = g(x) \cdot u$$
where $g(x)$ is a function of $f(x)$ (i.e. $g(x) = e^{f(x)}$). Rewrite this equation,
$$\frac{\partial u}{\partial x} = g(x) \cdot u$$
Above is ok, but below is not: you should wright $dx$ and $du$ (because standalone $\partial u$ and $\partial x$ do not make sense), and you need to integrate rather than differentiate
$$\frac{\partial u}{u} = g(x) \partial x$$
$$-\frac{1}{u^2} = g'(x) + h(y)$$
where $h$ is any function that only of y. Therefore,
$$u = \frac{1}{\sqrt{-g'(x) - h(y)}}$$
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Not sure if this will work
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Junjing, you are right but I am not sure if anyone but me would be able to read your solution
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Junjing, you are right but I am not sure if anyone but me would be able to read your solution
Okay. Next time I will make it 'skinnier'. Sorry.