<s>[INST] Functional with 4th grade characteristic equation I get a functional example, which looks like this:
$\mathcal{F}(u) = \int_0^{\frac{\pi}{4}} (x^2 + 16u^2 - u''^2) dx $
with conditions: $u(0) = 1, u'(0) = 0, u(\frac{\pi}{4}) = 0, u(\frac{\pi}{4}) = -2 $
Using Euler-Lagrange equation :
$\frac{\partial F}{\partial u} - \frac{d^2}{dx^2} \cdot \frac{\partial F}{\partial u''} = 0 $
got equation:
$2u^{(4)} + 16u = 0$
$u^{(4)} +8u = 0$
I tried to solve it with characteristic equation
$\lambda^4 + 8 = 0$, but probably I failed.
These are my next steps. Could someone tell me whether they correct or not? And how to solve it?
$r = \lambda^2$
$ r^2 + 8 = 0 $
$\Delta = -4 \cdot 8 = -32 $
$\sqrt{\Delta} = \sqrt{-1 \cdot 32} = \pm t \cdot 4\sqrt{2}$
$r_1 = 2\sqrt{2} \cdot t$
$r_2 = - 2\sqrt{2} \cdot t$
$\lambda^2 = 2\sqrt{2} \cdot t $
$\lambda^2 = - 2\sqrt{2} \cdot t $
$\lambda_1 = \sqrt{2\sqrt{2} \cdot t}$
$\lambda_2 = - \sqrt{2\sqrt{2} \cdot t}$
$\lambda_3 = \sqrt{ - 2\sqrt{2} \cdot t}$
$\lambda_4 = - \sqrt{ - 2\sqrt{2} \cdot t}$
 [/INST] Your method isn't very clear. You introduce a delta, then it goes away with no explanation. You seem to be using t where i is standard.  Solving the equation $\lambda^4 + 8 = 0$ is pretty simple; just move the 8 to the other side, and then take the fourth root. 
$\lambda^4 = 8^{(1/4)}\omega $ 
where $\omega^4 = -1$ 
Note that there are four solutions, corresponding to the four roots of -1.
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