91Ó°ÊÓ

Given equation is${(D^4−1)}^{2}y=0$

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.

Rewrite the auxiliary equation

$(D+1)(D+1)(D−1)(D−1)(D+i)(D+i)(D−i)(D−i)y=0.$

Which is the eighth order differential equation with four different roots.

To solve this equation, start by taking the simpler equation,

$\begin{array}{c}\frac{dy}{dx}=−y\text{ }\\ \text{ }y=b_1{e}^x\end{array}$

Then, to find the second solution form${(D+1)}^{2}y=0$, assume$(D+1)y=u$, therefore it will become. $(D+1)u=0$Thus,

$\begin{array}{c}\frac{du}{dx}=−u\\ u=b_2{e}^{−x}\end{array}$

Substitute$(D+1)y=u$into the solution, so get a first order linear differential equation

$\begin{array}{c}\frac{dy}{dx}+y=b_2{e}^{−x}\\ I=∫dx=x\\ e^l=e^x\end{array}$

Solve further

$\begin{array}{c}y{e}^l=∫\left(b_2{e}^{−x}\right)e^{x}dx\\ =b_{2}x+C_2\\ y=(C_{1}x+C_3)e^{−x}\end{array}$

Therefore, the solution is

$\begin{array}{c}(D+1)(D+1)y=0\text{ }\\ y=(C_{1}x+C_3)e^{−x}\end{array}$

Use the same method for the three roots

$\begin{array}{c}(D−1)(D−1)y=0\\ y=(C_{3}x+C_4)e^x\\ (D+i)(D+i)y=0\\ y=(C_{5}x+C_6)e^{−ix}\end{array}$

Solve further the equation

$\begin{array}{c}(D−i)(D−i)y=0\text{ }\\ y=(C_{7}x+C_8)e^{ix}\end{array}$

The general solution is a linear combination of all of these independent solutions

$y=(C_{1}x+C_3)e^{−x}+(C_{3}x+C_4)e^x+(C_{5}x+C_6)e^{−ix}+(C_{7}x+C_8)e^{ix}$