91Ó°ÊÓ

Given equation is.${(D+1)}^2(D^4−16)y=0$

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

$(D+1)(D+1)(D−2)(D+2)(D−i2)(D−i2)y=0.$

This is the sixth order differential equation which have five different roots, two of them are complex. To solve this equation, start by the simpler equation

$\begin{array}{c}(D+1)y=0\\ \frac{dy}{dx}=−y\\ y=C_1{e}^x\end{array}$

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

$\begin{array}{c}\frac{du}{dx}=−u\\ u=C_2{e}^{−x}\end{array}$

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

$\begin{array}{c}\frac{dy}{dx}+y=C_2{e}^{−x}\\ I=∫dx\\ =x\\ e^I=e^x\end{array}$

Solve further the equation

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

Take the equations, and, then

$\begin{array}{c}\frac{dy}{dx}=2y\\ y=C_4{e}^{2x}\\ \frac{dy}{dx}=−2y\\ y=C_5{e}^{−2x}\end{array}$

And finally, for$(D−i2)y=0$and,$(D+i2)y=0$then

$\begin{array}{c}\frac{dy}{dx}=i2y\\ y=C_6{e}^{2ix}\\ \frac{dy}{dx}=−i2y\\ y=C_7{e}^{−2ix}\end{array}$

Therefore, the general solution is$y=(C_{2}x+C_8)e^{−x}+C_4{e}^{2x}+C_5{e}^{−2x}+C_6{e}^{2ix}+C_7{e}^{−2ix}$