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If they give you the graph, you can look at it to see if an endpoint is included or not.

Depends on whether the domain includes the endpoint or not.

I think if you get below 30 ATAR, they don't tell you your exact ATAR, they just tell you you got a * ATAR, which means you got less than 30.

Re: Several Variable Calculus Yeah, pretty much. Two of those integrals would be equal by symmetry, of course.

Re: Several Variable Calculus $\noindent That would tell us that when $\phi = 0$ for example, $\rho$ is going to go to $\infty$, but our region is bounded, so we should not have $\rho$ going to infinity for this integral.$

$\noindent For interval notation, an open bracket means you exclude that endpoint and a closed bracket means you include that endpoint. For example, if $x$ is in $(3, 4]$, this means that $3< x \leq 4$.$

Well if you have something you think is the answer, you can check it by subbing it into the given ODE and also checking if it satisfies the initial conditions.

We don't use the initial conditions now, they're not what determine the A and B. The ODE is what determines them. We need to substitute our particular solution into the ODE and use that to solve for A and B. Then we will be able to get our general solution with two arbitrary constants in it...

$\noindent Remember, the particular solution in this case has a different form. Of course we get $0$ if trying a particular solution of the form $A \sin \left(\sqrt{20}t\right)$ -- this is because that is a solution to the homogenous equation! The particular solution in this case will have the...

Yes, that's what you solve for part (d).

$\noindent We found the general solution to be$ $$q(t) = c_{1} \cos (\omega_{0} t) + c_{2}\sin (\omega_{0} t) + \frac{100}{20 - \omega^{2}}\sin (\omega t),$$ $\noindent where $\omega_{0} = \sqrt{20}$ and $\omega^{2} \neq 20$. We are given the initial conditions $q(0) = 2$ and $q'(0) = 0$...

Current is rate of flow of charge, so initially 0 current means q'(0) = 0. Also, initially 2 coulombs of charge means q(0) = 2.

$\noindent In general we'd try a particular solution of the form $A \cos \omega t + B\sin \omega t$ for the given right-hand side of the ODE, but since the LHS of the ODE is what it is, it suffices to just try $A \sin \omega t$ (if you add a cosine term in, it'll still work, you'll just find the...

$\noindent For a particular solution for that, you can try something of the form $a\sin \omega t$. Note that this is not a solution to the homogenous equation if $\omega \neq \pm \sqrt{20}$.$

$\noindent The ODE to solve is $q'' + 20 q = 100\sin \omega t$, where $\omega \neq \pm \sqrt{20}$.$

$\noindent For c), you'd do it by solving it with a general $\omega$ where $\omega \neq \omega_{0}$, where $\omega_{0}$ is the natural frequency, which is the \textbf{square root} of $20$. I.e. solve $q'' + 20 q = 100 \sin \omega t$, where $\omega \neq \sqrt{20}$ (with the given initial values)...

Yeah there's only one positive value, but you could also use -20 and that'd still give resonance. Usually omega is thought of as positive I guess. Using a negative omega here just makes the RHS of your ODE become negative sine, since sine is odd. This means your ''particular solution'' becomes...

$\noindent Here's an example (different numbers to your Q.). Suppose we are given the differential equation$ $$2x'' + 0.5x = \cos \left(\omega t\right).$$ $\noindent Here $m = 2$ and $k = 0.5$, so the ``natural frequency'' of the system is $\omega_{0} = \sqrt{\frac{0.5}{2}} =...

$\noindent Remember, there will be resonance iff the value of $\omega$ is equal to the ``natural frequency'' $\omega_{0}$ of the system. The ODE $mx'' + kx = f(t)$ (where $f(t)$ is the external force, or input) has natural frequency $\omega_{0} = \sqrt{\frac{k}{m}}$, if $k,m$ are positive...

For Q1) you basically find a particular solution and add it on to the solution to the homogenous equation to get the overall solution. I don't want to do the calculations now but I assume you know the steps. And for Q2), you should probably review your resonance theory (and did you manage to...