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.


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 $y'' + 2 y' + 2 y = 2 e^x \cos(x)$.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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! $\endgroup$ – mathmath8128 Jun 28 '11 at 0:56
  • 1
    $\begingroup$ The passage from $2 e^x \cos(x)$ to $2 e^{(1+i)x}$ could be considered as "complexifying". $\endgroup$ – Robert Israel Jun 29 '11 at 5:32

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?

| cite | improve this answer | |
  • $\begingroup$ Unfortunately this does not help me much. Or is there a way to get to a solution with this form? Thanks. $\endgroup$ – Anna Lytics Jun 28 '11 at 0:15
  • $\begingroup$ 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. $\endgroup$ – GEdgar Jun 28 '11 at 0:37

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:

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


| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.