<s>[INST] is there a real solution to this differential equation? This is the equation:
$\frac {d^4y}{dx^4}=-y$
I know the solution to this equation:
$\frac {d^2y}{dx^2}=-y$
is this:
$y(x)=a\cdot \cos(x)+b\cdot \sin(x)$
and it involves $e^{ix}$, but I wondered if there was any solution to a higher-order version of this.
 [/INST] $$\frac {d^4y}{dx^4}=-y$$
The particular solutions on the form $y_p=e^{rx}$ are obtained according to the roots of $r^4=-1$, thus $r=\pm\frac{1\pm i}{\sqrt{2}}$.
$$y_p=\exp\left(\frac{\pm 1\pm i}{\sqrt{2}}x \right)=\exp\left(\pm\frac{x}{\sqrt{2}} \right)\left(\cos\left(\frac{x}{\sqrt{2}} \right)\pm i\sin\left(\frac{x}{\sqrt{2}} \right)\right)$$
So we have four independent particular solutions, wich leads to the general solution :
$y(x)=c_1\exp\left(\frac{x}{\sqrt{2}} \right)\cos\left(\frac{x}{\sqrt{2}} \right) +c_2\exp\left(-\frac{x}{\sqrt{2}} \right)\cos\left(\frac{x}{\sqrt{2}} \right) +c_3\exp\left(\frac{x}{\sqrt{2}} \right)\sin\left(\frac{x}{\sqrt{2}} \right) +c_4\exp\left(-\frac{x}{\sqrt{2}} \right)\sin\left(\frac{x}{\sqrt{2}} \right)$
Or on another equivalent form :
$y(x)=C_1\cosh\left(\frac{x}{\sqrt{2}} \right)\cos\left(\frac{x}{\sqrt{2}} \right)\:+C_2\sinh\left(\frac{x}{\sqrt{2}} \right)\cos\left(\frac{x}{\sqrt{2}} \right)\:+C_3\cosh\left(\frac{x}{\sqrt{2}} \right)\sin\left(\frac{x}{\sqrt{2}} \right)\:+C_4\sinh\left(\frac{x}{\sqrt{2}} \right)\sin\left(\frac{x}{\sqrt{2}} \right)$
All those solutions are real insofar the coefficients are real.
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