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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=c1eax+c2ebx+c3ecx

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)].

Short Answer

Expert verified

For the case whena≠b≠cthe general solution isy=c1eax+c2ebx+c3ecx

For the case when a≠b=cthe general solution is

y=c1eax+c3x+c5ebx

Step by step solution

01

Given information from question

Given equation isD2+(1+2i)D+i-1=0

02

Definition of Differential equation 

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.

03

Solve the equation for the case, a≠b≠c

To solve this auxiliary equation for the case whena≠b≠cneed 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=0is a solution of the differential equation. Thus,

dydx=ayy=c1eax

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

dydx=byy=c2ebx

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

dydx=cyy=c3ecx

Therefore, the general solution is y=c1eax+c2ebx+c3ecx.

04

Solve the equation for the case

For the case whena≠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

dydx=ayy=c1eax

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

dydx=byy=c2ebx

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

(D-b)(D-b)y=(D-b)u=0dudx=buu=c3ebx

Substitutein the solution, therefore

dydx-by=c3ebx

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

I=∫(-b)dx=-bxel=e-bxyel=∫c3ebxe-bxdx=c3x+c4

Therefore, the general solution is y=c1eax+c3x+c5ebx.

05

Prove that the general solution will represent all the vector in the vector space. 

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

D-a1D-a2…..D-any=0

where to find the solution. First ifa1≠a2≠a3, andnrepresents the degree of the differential equation. The general solution in this case would be in the formy=c1ea1x+c2ea2x+……..+cneanx

In terms of vector space language eanxwill represent a basis vector of an n-dimensional vector space, and therefore, the general solution will represent all the vector in the vector space.

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