Diff eq with constants... Eulers identity...

[Rijad Hadzic]

Homework Statement

Find the general solution of the second order DE.

$y'' + 9y = 0$

Homework Equations

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

$m^2 + 9 = 0$
$m = 3i$ and $m = -3i$

so the general solution would be:

$c_1e^{3ix} + c_2e^{-3ix} = y$

my book gives me

$e^{i\theta} = cos(\theta) + isin(\theta)$

from there I get

$e^{iβx} = cos(βx) + isin(βx)$
$e^{i-βx} = cos(βx) - isin(βx)$

I have

$e^{i3x} = cos(3x) + isin(3x)$
$e^{i-3x} = cos(3x) - isin(3x)$

so I get $y = c_1cos(3x) + c_1isin(3x) + c_2cos(3x) -c_2isin(3x)$

but my book gives me

$y = c_1cos(3x) + c_2sin(3x)$

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]

Your $c_1$ and $c_2$ are just not the same constants as those of the book.

 

[Rijad Hadzic]

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]

Homework Statement

Find the general solution of the second order DE.

$y'' + 9y = 0$

Homework Equations

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

$m^2 + 9 = 0$
$m = 3i$ and $m = -3i$

so the general solution would be:

$c_1e^{3ix} + c_2e^{-3ix} = y$

my book gives me

$e^{i\theta} = cos(\theta) + isin(\theta)$

from there I get

$e^{iβx} = cos(βx) + isin(βx)$
$e^{i-βx} = cos(βx) - isin(βx)$

I have

$e^{i3x} = cos(3x) + isin(3x)$
$e^{i-3x} = cos(3x) - isin(3x)$

so I get $y = c_1cos(3x) + c_1isin(3x) + c_2cos(3x) -c_2isin(3x)$

but my book gives me

$y = c_1cos(3x) + c_2sin(3x)$

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.

 

[Rijad Hadzic]

Gotcha. Ty everyone!