   SEARCH HOME Math Central Quandaries & Queries  Hei. I don’t speak lot of english but here is my question,hope u understand: f(x) + f ''''(x)=0. so, my question. what is f(x), where f ''''(x) is f(x) derivative by four time ? i tried to find the answer and i knew f(x) is something like that f(x)=e^x*sinx but miss something. i don`t know. Hello Andrea,
To solve this problem we need to have an understanding of solving High Order Homogeneous Linear Differential Equations, and specifically when we have complex roots.
If we re-write the questions $f(x) + f ’’’’(x) = 0$ into the form $y^{(4)} +y =0,$ we can see that this is a Fourth Order Linear Differential Equation. Therefore, we can assume that the solutions will be of the form $y(x) = e^{rx}.$ To find the value(s) of r we must set up the characteristic equation for this differential equation, which is  $r^4 + 1=0,$ with $r = (-1)^{1/4.}$

To finish solving the problem we need to be able to evaluate the four roots of -1.
METHOD 1

To do this we must recall how to solve characteristic equations of the form $r^n – a = 0.$ To do this we use the fact that a in polar form is $a = R(e^{i \alpha}),$ where $R = |a|,$ and where $\alpha = 0$ when $a > 0,$ and $\alpha = \pi$ when $a < 0.$
From Euler’s formula $e^{i \theta} = \cos \theta + i \sin \theta,$ we can conclude that $e^{i2k \pi}= 1$, with $k = 0, 1, 2, 3, …$
From that we can conclude that $a = R(e^{i(\alpha+2k \pi)})$ and  $a^{1/n} = R^{1/n}(e^{i(\alpha+2k \pi)/n}),$ with
$k = 0, 1, 2 ,3,…, n – 1$
Applying this formula to our situation we have $r^4 – (– 1) = 0,$ so $a = – 1, R = 1,$ and $\alpha = \pi$  since $a < 0.$ Using the formula    $a^{1/n} = R(e^{i(\alpha+2k \pi)/n}),$ with $n = 4,$ we use this and Euler’s formula to find the four roots. $(k = 0, 1, 2, 3).$

METHOD 2

If we solve the quartic equation $r^4 + 1= 0,$ we end up getting 4 different imaginary roots. These roots are $\pm \sqrt i$  and  $\pm i \sqrt i.$ In order to simplify this and use our complex roots general solution for differential equations, we need to be able to evaluate $\sqrt i$. To do this we can use Euler’s formula $e^{i \theta} = \cos \theta + i \sin \theta$ Plugging in $\theta = \frac{\pi}{2}$. we get that $e^{i \frac{\pi}{2}} = i,$ that means that
$\sqrt i = i^{1/2} = \left( e^{i \frac{\pi}{2}} \right)^{\frac12} = e^{i \frac{\pi}{4}}$. Evaluating this back into Euler’s formula we get that $\sqrt i = e^{i \frac{\pi}{4}} = \cos\left( \frac{\pi}{4}\right) + i \sin\left( \frac{\pi}{4}\right),$ which gives us the result that$\sqrt i = \frac{1+i}{\sqrt 2}.$ Using this fact we can determine the complex value of the other three roots.

Once we have our four roots we can write our general solution to the Homogeneous problem. Since there will be four roots, there will be 4 different terms in our solution.
Hope this helps.
Brennan Yaremko     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.