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I want to get a particular solution to the differential equation $$ y''+2y'+2y=2e^x cos(x) $$ and therefore I would like to 'complexify' the right hand side. This means that I want to write the right hand side as $q(x)e^{\alpha x}$ with $q(x)$ a polynomial. How is this possible?

The solution should be $(1/4)e^x(\sin(x)+\cos(x))$ but I cannot see that.

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Ana Lytics, you appear to have flagged your own question for moderator attention as "too localized". Was that intentional? –  Zev Chonoles Jun 28 '11 at 0:38
    
Yes, I had the impression that it may not help anybody else since it is very localized. –  Anna Lytics Jun 28 '11 at 0:44
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I think it's a fine question. If it is a homework question, we should add the "homework" tag though. –  Zev Chonoles Jun 28 '11 at 0:48
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3 Answers

The point is that (for real $x$) $2 e^x \cos(x)$ is the real part of $2 e^x e^{ix} = 2 e^{(1+i)x}$. Find a particular solution of $y'' + 2 y' + 2 y = 2 e^{(1+i)x}$, and its real part is a solution of $y'' + 2 y' + 2 y = 2 e^x \cos(x)$.

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This is right. You don't need to "complexify"... just need to realize the solution you're looking for is the real component of $2e^{(1+i)x}$, this follows from the linearity of the D.E... keep in mind you still have to find the kernel solution though :-) Don't forget to add that! –  ae0709 Jun 28 '11 at 0:56
    
The passage from $2 e^x \cos(x)$ to $2 e^{(1+i)x}$ could be considered as "complexifying". –  Robert Israel Jun 29 '11 at 5:32
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As $\cos x=\frac{e^{ix}+e^{-ix}}{2}$, $2e^x \cos x = e^{x+ix}+e^{x-ix}$, but that is not of the requested form. Is it close enough?

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Unfortunately this does not help me much. Or is there a way to get to a solution with this form? Thanks. –  Anna Lytics Jun 28 '11 at 0:15
    
Find the particular solution with RHS $e^{x+ix}$, then find the particular solution with RHS $e^{x-ix}$. Your final particular solution is then a linear combination of these. –  GEdgar Jun 28 '11 at 0:37
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I know that this might be awfully late, but I just started learning about complexification this term and thought I would put up my solution--please excuse any incorrect language I might use as it's the idea I am trying to get across.

First we start off by defining a complex analogue to your function:

eq {1}: $$ z''+2z'+2z=2e^xe^{ix} $$

where $$z=RE(y)+i*IM(y)$$

Basically, we can recover the original diffEq by extracting the real part of our complex diffEq. The next step is to use the method of undetermined coefficients to find a guess for what our particular complex solution might be. Guess:

eq {2}: $$ z=Ae^{x}e^{ix} $$ so that:

$$ z'=Ae^{x}e^{ix}+iAe^{x}e^{ix} $$ and $$ z''=i2Ae^{x}e^{ix} $$

Plugging this into {1}:

$$ i2Ae^{x}e^{ix} + 2(Ae^{x}e^{ix}+iAe^{x}e^{ix}) + 2(Ae^{x}e^{ix})= 2e^xe^{ix} $$

We can simplify by removing the common factor of $ e^{x}e^{ix} $:

$$ A(4+4i)=2 => A= \frac{1}{2 + 2i}$$

Convert A to complex polar form:

$$ A=\frac{\sqrt[]{2}}{4}e^{-i\frac{\pi}{4}} $$

Plugging this into {2}:

$$ z=\frac{\sqrt[]{2}}{4}e^{-i\frac{\pi}{4}}e^{x}e^{ix} $$

This can be simplified to eq {3}:

$$ z=\frac{\sqrt[]{2}}{4}e^{x}e^{i(x-\frac{\pi}{4})} $$

Since our particular solution should be of the form $cos(x)$, we take the real part of {3} and call that our particular x-solution:

$$ x = RE(z) = \frac{\sqrt[]{2}}{4}e^{x}cos(x-\frac{\pi}{4}) $$

Finally using our difference of cosine identity:

$$ \frac{\sqrt[]{2}}{4}e^{x}(cos(x)\frac{\sqrt[]{2}}{2} + sin(x)\frac{\sqrt[]{2}}{2}) $$ $$ \frac{\sqrt[]{2}}{4}\frac{\sqrt[]{2}}{2}e^{x}(cos(x) + sin(x)) $$ $$ \frac{2}{8}e^{x}(cos(x) + sin(x)) $$ $$ \frac{1}{4}e^{x}(cos(x) + sin(x)) $$

$QED$

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