<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] $$y''''+y=0$$
Characteristic equation is
$$(r^4+1)=0$$
$$(r^2-i)(r^2+i)=0$$
$$(r^2-i)=(r-(\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}))(r+(\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}))$$
$$(r^2+i)=(r-i(\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}))(r+i(\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}))$$
So the solution is
$$y_1(x)=c_1e^{\frac {\sqrt 2}{2}x}\cos(\frac {\sqrt 2}{2}x)+c_2e^{\frac {\sqrt 2}{2}x}\sin(\frac {\sqrt 2}{2}x)$$
And
$$y_2(x)=c_3e^{-\frac {\sqrt 2}{2}x}\cos(\frac {\sqrt 2}{2}x)+c_4e^{-\frac {\sqrt 2}{2}x}\sin(\frac {\sqrt 2}{2}x)$$
$$y(x)=y_1(x)+y_2(x)$$

You can also use exponetials:
$$r^4=-1 \implies (\rho e^{i\beta})^4=e^{i\pi}$$
We have that $\rho=1$ and:
$$4\beta=\pi +2k\pi ,k=0,1,2.....$$
$$\beta =\frac {\pi(1+2k)}4$$
We find four values for $\beta$ :
$$\beta=\frac {\pi}4,\frac {3\pi}4,\frac {5\pi}4,\frac {7\pi}4$$
The solution for example for $\beta=\frac {\pi}4$
$$r=e^{i\pi/4}=\cos \frac {\pi}4+ i \sin \frac {\pi}4=\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}$$
$$y_1(x)=c_1e^{\frac {\sqrt 2}{2}x}\left (\cos(\frac {\sqrt 2}{2}x)+i\sin(\frac {\sqrt 2}{2}x) \right)$$
Do this for all values of $\beta$ to find the complete answer.
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