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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Find a general solution for the givenlinear system using the elimination method of Section 5.2.$\begin{array}{l}\frac{{d}^{3}x}{d{t}^{3}}-x+\frac{dy}{dt}+y=0\\ \frac{dx}{dt}-x+y=0\end{array}$

The general solution is $x\left(t\right)={A}_{1}+{A}_{2}{e}^{t}+{A}_{3}{e}^{-t}$

See the step by step solution

## Step 1: Elimination method

In the elimination method you either add or subtract the equations to get an equation in one variable. When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.

## Step 2: Solving linear system using the elimination method:

We will do the following question on the basis of solving linear system using the elimination method;

$\begin{array}{l}\frac{{d}^{3}x}{d{t}^{3}}-x+\frac{dy}{dt}+y=0\\ \frac{dx}{dt}-x+y=0\\ \\ {D}^{3}x-x+Dy+y=0\\ Dx-x+y=0\\ \left({D}^{3}-1\right)x+\left(D+1\right)y=0\\ \left(D-1\right)x+y=0\\ \\ \left(D-1\right)\left({D}^{3}-1\right)x+\left(D-1\right)\left(D+1\right)y-\left({D}^{3}-1\right)\left(D-1\right)x+\left({D}^{3}-1\right)y=0\\ \left({D}^{2}-{D}^{3}\right)y=0\\ \\ {m}^{2}-{m}^{3}=0\\ {m}^{3}\left(1-m\right)=0\\ ⇒m=0,0,1\\ \\ y\left(t\right)={C}_{1}{e}^{{m}_{1}t}+{C}_{2}{e}^{{m}_{2}t}+{C}_{3}{e}^{{m}_{3}t}\\ y\left(t\right)=\left({C}_{1}+t{C}_{2}\right)+{C}_{3}{e}^{t}\\ \\ \left({D}^{3}-1\right)x+\left(D+1\right)y-\left[\left(D-1\right)x+\left(D+1\right)y\right]=0\\ \left({D}^{3}-D\right)x=0\\ \\ {m}^{3}-m=0\\ m\left({m}^{2}-1\right)=0\\ ⇒m=0,±1\\ \\ x\left(t\right)={A}_{1}{e}^{{m}_{1}t}+{A}_{2}{e}^{{m}_{2}t}+{A}_{3}{e}^{{m}_{3}t}\\ ={A}_{1}+{A}_{2}{e}^{t}+{A}_{3}{e}^{-t}\end{array}$

$x\left(t\right)={A}_{1}+{A}_{2}{e}^{t}+{A}_{3}{e}^{-t}$