0
$\begingroup$

The ODE I'm trying to solve is: $y''+2y'+2y = 3$. I've never tried to solve an ODE with complex roots until this problem so it's challenge for me. These are my steps for getting r: $$a_2r^2 +a_1r+a_0=0$$ $$r^2+2r+2=0$$ $$(r+1)^2 = -1$$ $$r = \pm i-1$$ $Q_2(x) = \frac32$ and these are my last few steps plugging everything into the final equation: $$y = c_1e^{r_1x}+c_2e^{r_2x}+Q_k(x)$$ $$y = c_1e^{(i-1)x} + c_2e^{(-i-1)x} + \frac32$$ $$y = c_1 e^{-x} (\cos{x}+i\sin{x})+ c_2 e^{-x} (\cos{x}-i\sin{x})+\frac32$$

I think somehow the imaginary answers should cancel each other out, but they don't. Wolfram alpha gives

$$y = c_1 e^{-x}\sin{x} + c_2e^{-x}\cos{x}+\frac32$$

Where did I go wrong?

$\endgroup$
  • $\begingroup$ With complex roots, the coefficients can be complex, so that ultimately the solution will be a real-valued function. $\endgroup$ – Bernard Mar 23 '17 at 23:59
0
$\begingroup$

I don't think that you went wrong anywhere but the solution for complex roots $r = \alpha \pm i \beta$ can be expressed as $C_1 e^{\alpha x}\cos(\beta x) + C_2 e^{\alpha x}\sin(\beta x)$. Your solution $r = \pm i -1= -1 \pm i$ gives $\alpha = -1$ and $\beta = 1$, which leads to $C_1 e^{- x}\cos (x) + C_2 e^{- x}\sin( x)$.

$\endgroup$
0
$\begingroup$

Ff you rearrange the solution you found, you'll get

$y = (c_1 + c_2) e^{-x} \cos x + i(c_1 - c_2) e^{-x} \sin x + \frac{3}{2}$

Then, use the substitution $a_1 = c_1 + c_2$, $a_2 = i(c_1 - c_2)$ to get the solution Wolfram Alpha found.

But, you may ask, doesn't that mean that some of those constants might be complex? Well, yes. But that's fine - nothing in there prevented any of that happening, and it can happen no matter how you rearrange the formula (there's nothing stopping $a_1$ and $a_2$ from being complex). However, this is just the general form of the solution. Particular solutions could wind up having completely real coefficients, depending on how they're defined.

$\endgroup$
0
$\begingroup$

$y = $$C_1e^{-x}(\cos x + i \sin x) + C_2e^{-x}(\cos x - i \sin x)+\frac32\\ (C_1 + C_2) e^{-x}(\cos x + i \sin x) + (iC_1 - iC_2) e^{-x} \sin x + \frac 32$

$(C_1 + C_2)$ is just an arbitrary constant. As is $(iC_1 - iC_2). $ There nothing that says that arbitrary constants cannot be complex.

$\endgroup$

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.