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Diff eq with constants.. Eulers identity..

1. The problem statement, all variables and given/known data
Find the general solution of the second order DE.

[itex] y'' + 9y = 0 [/itex]
2. Relevant equations


3. The attempt at a solution

Problem is straight forward I just don't get why my answer is different than the books.

So you get

[itex] m^2 + 9 = 0 [/itex]
[itex] m = 3i [/itex] and [itex] m = -3i [/itex]

so the general solution would be:

[itex] c_1e^{3ix} + c_2e^{-3ix} = y [/itex]

my book gives me

[itex] e^{i\theta} = cos(\theta) + isin(\theta) [/itex]

from there I get

[itex] e^{iβx} = cos(βx) + isin(βx) [/itex]
[itex] e^{i-βx} = cos(βx) - isin(βx) [/itex]

I have

[itex] e^{i3x} = cos(3x) + isin(3x) [/itex]
[itex] e^{i-3x} = cos(3x) - isin(3x) [/itex]

so I get [itex] y = c_1cos(3x) + c_1isin(3x) + c_2cos(3x) -c_2isin(3x) [/itex]

but my book gives me

[itex] y = c_1cos(3x) + c_2sin(3x) [/itex]

I feel like my answer is still valid for some reason.. I just don't know how they got their answer from my answer. I used the correct identity..
 

Orodruin

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Your ##c_1## and ##c_2## are just not the same constants as those of the book.
 
Your ##c_1## and ##c_2## are just not the same constants as those of the book.
So I guess that means my answer is still valid then... I just don't get what they used for c1 and c2... did they use i or something? because I notice there is no i term in their answer..
 

Ray Vickson

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1. The problem statement, all variables and given/known data
Find the general solution of the second order DE.

[itex] y'' + 9y = 0 [/itex]
2. Relevant equations


3. The attempt at a solution

Problem is straight forward I just don't get why my answer is different than the books.

So you get

[itex] m^2 + 9 = 0 [/itex]
[itex] m = 3i [/itex] and [itex] m = -3i [/itex]

so the general solution would be:

[itex] c_1e^{3ix} + c_2e^{-3ix} = y [/itex]

my book gives me

[itex] e^{i\theta} = cos(\theta) + isin(\theta) [/itex]

from there I get

[itex] e^{iβx} = cos(βx) + isin(βx) [/itex]
[itex] e^{i-βx} = cos(βx) - isin(βx) [/itex]

I have

[itex] e^{i3x} = cos(3x) + isin(3x) [/itex]
[itex] e^{i-3x} = cos(3x) - isin(3x) [/itex]

so I get [itex] y = c_1cos(3x) + c_1isin(3x) + c_2cos(3x) -c_2isin(3x) [/itex]

but my book gives me

[itex] y = c_1cos(3x) + c_2sin(3x) [/itex]

I feel like my answer is still valid for some reason.. I just don't know how they got their answer from my answer. I used the correct identity..
Just write ##a = c_1 + c_2## and ##b = i c_1 - i c_2##. Your solution becomes ##a \cos 3x + b \sin 3x## for two constants ##a## and ##b##.

BTW: when using LaTeX, do NOT write ##cos(\theta)##, etc.; write, instead, ##\cos( \theta)## or ##\cos \theta##, which you get by typing "\cos" instead of "cos", The notation ##cos \theta## looks ugly and is hard to read, but ##\cos \theta## looks good and is clear.
 
Gotcha. Ty everyone!
 

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