@farzadaf No, +sin is correct. If you expand out the terms (1+i)(cos x + i*sin x), you get (cos x - sin x) + i*(cos x + sin x). The particular solution takes only the imaginary part, the term (cos x + sin x), discarding the real part (cos x - sin x).
This is because the original ODE (y) became the imaginary part of the "complexified" ODE (y-tilde).
also ich bin blond
SeritaCruzxq237 1 month ago
Equations make me dizzy. Hahahah
agapitoflores001 2 months ago
Which lesson should I watch to learn how to "complexfy" as the professor does at 17:58. Thank you very much.
RinaldiMeteoric 3 months ago
@RinaldiMeteoric It's the 6th. And by the way, you're the laziest folk i've ever seen !
FariFure 1 month ago
can someone tell me if that is first or second semster or whatever..?
h0wud0in2 4 months ago
woot he's wearing my favorite jacket again!!!
Liaomiao 8 months ago 3
the only reason i clicked on like on this video is when he mention it wants to be T but i make it X . lol funny. he is funny teacher.
nepsjg 10 months ago
@farzadaf No, +sin is correct. If you expand out the terms (1+i)(cos x + i*sin x), you get (cos x - sin x) + i*(cos x + sin x). The particular solution takes only the imaginary part, the term (cos x + sin x), discarding the real part (cos x - sin x).
This is because the original ODE (y) became the imaginary part of the "complexified" ODE (y-tilde).
kaugith 1 year ago
savior during finals
longshot1710 1 year ago
23:40 should be -sin i think
gorgolyt 1 year ago
@gorgolyt
yes I got -sin(x) as well
farzadaf 1 year ago
@gorgolyt No, he has it right. We want the imaginary part, aka the part with factor i.
The imaginary part of (1+i)(cos(x) + i sin(x)) is cos(x) + sin(x).
If we were looking for the real part, then indeed it would be cos(x) - sin(x).
mdfu1 3 weeks ago in playlist MIT 18.03 Differential Equations, Spring 2006
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kapilakaliu 1 year ago
he has a great sense of humor.
A huge thank you from Brazil.
pmshadow 1 year ago
Exelente!. Gracias, desde Perú.
Gurrataxxx 2 years ago
Buenísimo. Gracias, también desde Argentina
eordano 2 years ago
Why is it that we can assume 'a' to be a root of p(D) in general?
GirlyVoice 2 years ago
@GirlyVoice it isn't. p(a) is normally nonzero, so a isn't a root. he covers the special case of when it is a root in the second half of the video.
gorgolyt 1 year ago
Comment removed
GirlyVoice 2 years ago
)) Preps are all around vwry much alike. I had almost the same one of Physics ))
rambutanic 2 years ago
gracias de verdad uy este tipo es lo maximo jeje explica muy bien!!!
danielk320 2 years ago
Muy lindo ! Gracias ,desde argentina.
bellinivernon 3 years ago