complex nuuuuumbers (1 Viewer)

[Trebla]

On second read, you're right. I thought I had seen it there, though I couldn't find it in the complex numbers section, so I thought it wasn't there.

That does seem slightly confusing though, rather than teaching it in the context of complex numbers.

That's mainly because you'll need to use it quite often when solving second order ODEs but can sort of get away without using it in complex numbers.

A simple example:
d²y/dt² + 4y = 0

Upon testing y = Pe^kt for some general constant P, the equation reduces to
k² + 4 = 0
which has solutions k = - 2i, 2i
Thus the general solution by principle of superposition (don't worry about why it looks like this if you haven't covered it before) is:
y = Ae^2it + Be^-2it
= A (cos 2t + isin 2t) + B (cos 2t - isin 2t)
= (A + B) cos 2t + i(A - B) sin 2t
= C cos 2t + D sin 2t
(since C and D are just arbitary constants)
Note that it is possible to to write C cos 2t + D sin 2t in the form R cos (nt + α) which means that the solutions are oscillatory which makes sense because you should recognise that the original equation
d²y/dt² + 4y = 0
(i.e. d²y/dt² = - 4y)
describes simple harmonic motion if t represents time and y represents displacement.

 

[jet]

That's mainly because you'll need to use it quite often when solving second order ODEs but can sort of get away without using it in complex numbers.

A simple example:
d²y/dt² + 4y = 0

Upon testing y = Pe^kt for some general constant P, the equation reduces to
k² + 4 = 0
which has solutions k = - 2i, 2i
Thus the general solution by principle of superposition (don't worry about why it looks like this if you haven't covered it before) is:
y = Ae^2it + Be^-2it
= A (cos 2t + isin 2t) + B (cos 2t - isin 2t)
= (A + B) cos 2t + i(A - B) sin 2t
= C cos 2t + D sin 2t
(since C and D are just arbitary constants)
Note that it is possible to to write C cos 2t + D sin 2t in the form R cos (nt + α) which means that the solutions are oscillatory which makes sense because you should recognise that the original equation
d²y/dt² + 4y = 0 (i.e. d²y/dt² = - 4y) describes simple harmonic motion if t represents time and y represents displacement.

I've covered it by myself - not doing maths for a year gets boring but thanks for the heads up.

 

[micuzzo]

That's mainly because you'll need to use it quite often when solving second order ODEs but can sort of get away without using it in complex numbers.

A simple example:
d²y/dt² + 4y = 0

Upon testing y = Pe^kt for some general constant P, the equation reduces to
k² + 4 = 0
which has solutions k = - 2i, 2i
Thus the general solution by principle of superposition (don't worry about why it looks like this if you haven't covered it before) is:
y = Ae^2it + Be^-2it
= A (cos 2t + isin 2t) + B (cos 2t - isin 2t)
= (A + B) cos 2t + i(A - B) sin 2t
= C cos 2t + D sin 2t
(since C and D are just arbitary constants)
Note that it is possible to to write C cos 2t + D sin 2t in the form R cos (nt + α) which means that the solutions are oscillatory which makes sense because you should recognise that the original equation
d²y/dt² + 4y = 0
(i.e. d²y/dt² = - 4y)
describes simple harmonic motion if t represents time and y represents displacement.

what do you study at uni, just out of curiosity?

 

[kaz1]

This new syllabus looks awesome. No conics (most tedious topic) and harder 3unit (this shit gave me nightmares).

Lucky cunts.

 

[shaon0]

what do you study at uni, just out of curiosity?

Maths, probably.

 

[untouchablecuz]

its a shame harder 3 unit was culled, always allowed for the best questions

tho, DEs are pretty good

 

[untouchablecuz]

Maths, probably.

Adv Maths at usyd i think it was

 

[Aquawhite]

I don't know what conics will be like but I am certainly glad to see the harder 4U topic leaving!

I've seen Euler's in older questions I have. I had no idea what was going on and my teacher didn't know either I think - she was in a rush though.