By the method used in solving (5.4) to get. (5.9), show that the solution of the third-order equation

$(D-a)(D-b)(D-c)y=0$

is

$y=c_1{e}^{ax}+c_2{e}^{bx}+c_3{e}^{cx}$

if a, b, c are all different, and find the solutions if two or three of the roots of the auxiliary equation are equal. Generalize the result to higher-order equations. State your results in vector space language [see comment following equation (5.9)].

Given equation is$D^2+(1+2i)D+i-1=0$

Differential equation:

A differential equation is a formula that includes one or more functions and their derivatives. The rate of change of a function at a point is defined by its derivatives.

To solve this auxiliary equation for the case when$a\ne b\ne c$need first to solve the similar equation. If$(D-a)y=0$, then$(D-b)(D-c)·0=0$, so any solution of$(D-a)y=0$is a solution of the differential equation. Thus,

$$\begin{gathered} \frac{dy}{dx}=ay \\ y=c_1{e}^{ax} \end{gathered}$$

Similarly, if$(D-b)y=0$, then$(D-a)(D-c)·0=0$, so any solution of$(D-b)y=0$is a solution of the differential equation. Thus,

$$\begin{gathered} \frac{dy}{dx}=by \\ y=c_2{e}^{bx} \end{gathered}$$

Also, if$(D-c)y=0$, then$(D-a)(D-b)·0=0$, so any solution of$(D-c)y=0$is a solution of the differential equation. Thus,

$$\begin{gathered} \frac{dy}{dx}=cy \\ y=c_3{e}^{cx} \end{gathered}$$

Therefore, the general solution is $y=c_1{e}^{ax}+c_2{e}^{bx}+c_3{e}^{cx}$.

For the case when$a\ne b=c$(so the auxiliary equation becomes$(D-b)(D-b)·y=0)$, now to find the solution. First if, for this case the solution is

$$\begin{gathered} \frac{dy}{dx}=ay \\ y=c_1{e}^{ax} \end{gathered}$$

Then if$(D-b{)}^2=(D-b)(D-b)y=0$, then find the two solutions. Thus, the second solution is

$$\begin{gathered} \frac{dy}{dx}=by \\ y=c_2{e}^{bx} \end{gathered}$$

And it can find the third solution if let$(D-b)y=u$, therefore,

$$\begin{gathered} (D-b)(D-b)y=(D-b)u=0 \\ \frac{du}{dx}=bu \\ u=c_3{e}^{bx} \end{gathered}$$

Substitutein the solution, therefore

$\frac{dy}{dx}-by=c_3{e}^{bx}$

Which has become a first order differential equation and can solve it,

$$\begin{gathered} I=\int (-b)dx=-bx \\ {e}^l=e^{-bx} \\ y{e}^l=\int \left(c_3{e}^{bx}\right)e^{-bx}dx=\left(c_{3}x+c_4\right) \end{gathered}$$

Therefore, the general solution is $y=c_1{e}^{ax}+\left(c_{3}x+c_5\right)e^{bx}$.

If generalized the solution for higher order equation which would be in the form

$\left(D-a_1\right)\left(D-a_2\right)\dots ..\left(D-a_n\right)y=0$

where to find the solution. First if$a_1\ne a_2\ne a_3$, and$n$represents the degree of the differential equation. The general solution in this case would be in the form$y=c_1{e}^{a_{1}x}+c_2{e}^{a_{2}x}+\dots \dots ..+c_n{e}^{a_{n}x}$

In terms of vector space language $e^{a_{n}x}$will represent a basis vector of an n-dimensional vector space, and therefore, the general solution will represent all the vector in the vector space.