91Ó°ÊÓ

Given equation is.$(D^4+4)y=0$

A differential equation is a formula that connects the derivatives of one or more unknown functions. Functions are used to represent physical quantities, derivatives are used to characterise their rates of change, and differential equations are used to define a relationship between them in applications.

Rewriting the auxiliary equation

$\begin{array}{c}[{\left(D^2\right)}^2−{\left(i2\right)}^2]y=(D^2−i2)(D^2+i2)y\\ =0\end{array}$

To find the general solution of this forth order differential equation. It needs to treat $(D^2+\sqrt{i2})y=0$and$(D^2+\sqrt{i2})y=0$separately. For$(D^2+i2)y=0$it has the roots$(D^2−\sqrt{i2})(D^2+\sqrt{i2})y=0$

Any complex number's root could be written as:

$\begin{array}{c}{z}^{1/2}={\left(r{e}^{i\mathrm{θ}}\right)}^{1/2}\\ =\sqrt{r}(\cos \frac{\mathrm{θ}}{n}+i\sin \frac{\mathrm{θ}}{n})\end{array}$

The polar coordinates for $2i$are$r=2$ and $\mathrm{θ}=\mathrm{π}/2$therefore

$\begin{array}{c}2e^{\mathrm{π}/2}=\sqrt{2}{e}^{\mathrm{π}/4}\\ =\sqrt{2}(\cos \frac{\mathrm{θ}}{4}+i\sin \frac{\mathrm{θ}}{4})\\ =1+i\end{array}$

Therefore, roots of the equation become, .$(D−1−i)(D+1+i)y=0$

Therefore, the solution is

$y=(c_1{e}^{ix}+c_2{e}^{−ix})e^x$

Where for the part of ,$D^{2}y=0$ the roots would be

$\begin{array}{c}(D−i\sqrt{i2})(D+i\sqrt{i2})y=0\\ (D−i)(1+i)(D+i(1+i\left)\right)y=0\\ (D−i+1)(D+i−1)y=0\end{array}$

Therefore, the solution is,

$y(c_3{e}^{ix}+c_4{e}^{−ix})e^{ix}$

Thus, the general solution is

$y=(c_1{e}^{ix}+c_2{e}^{−ix})e^x+(c_3{e}^{ix}+c_4{e}^{−ix})e^{−x}$