HSC Stuff > HSC Mathematics Extension 1
3U Maths Question Thread
fun_jirachi:
Hey there!
'I would initially think to take the whole power (x^2/2 +C) to the front and turn it into k, not just C' - note you can't actually do this (if I'm interpreting this correctly?)
The crux of what they're doing is using the fact that you can split \(e^{f(x)+g(x)}\) into \(e^{f(x)} \times e^{g(x)}\). Note that since C is a constant, we also have that \(e^C\) is also a constant - they just replace it with k to simplify things (ie. a combination of the two aforementioned ideas). It's a lot cleaner to look at and typically multiplication is easier to deal with in computation than addition (exceptions exist, of course). You can't simplify the polynomial (because no such constant exists) and it's often redundant to assign a variable that refers to it when you often need the polynomial for other things (differentiation, computing its value at a point, etc.)
Hope this helps :)
twelftholmes:
of course!! i just relaised that x^2/2 isn't a constant so you can't group that with the c.
your explanation is awesome thanks!
twelftholmes:
hey! back at it again with another differential equation question,
here is the link: https://imgur.com/44LZBRr
It's only a minor thing but in case it's important I'll ask just in case.
In the last step when they took the square root, why is it only negative instead of both plus/minus?
I know this is super simple but thanks for any clarification!
fun_jirachi:
Hey :)
They took the negative root because of the initial condition \(x = \frac{\pi}{2}, y = -1\). It's a little problematic to express \(f(y)\) as a function of x (in this case \(f(y) = y^2\)), so it's actually really important to simplify and make the distinction between the positive and negative root. Pay attention especially to the initial condition as it is (clearly, from this example) not only used to find constants of integration :)
twelftholmes:
wow I had no idea about that rule but it makes heaps of sense,
thanks for helping me again!
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