Answered: In the following problem apply the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

In the following problem apply the eigenvalue method to find a general solution of the given system. If initial values are given, find also the corresponding particular solution. use a computer system or graphingcalculator to construct a direction field and typical solutioncurves for the given system.

x'₁ = 9x₁ + 5x₂, x'₂ = -6x₁ - 2x₂; x₁(0) =1, x₂(0)=0

Expert Solution

Step 1

We have to find the particular solution of the given system of differential equations. First we have

to find the general solution.

The matrix form of the given system is $[\begin{matrix} x_{1}' \\ x_{2}' \end{matrix}] = [\begin{matrix} 9 & 5 \\ - 6 & - 2 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}]$ . Thus, the coefficient matrix is

$A = [\begin{matrix} 9 & 5 \\ - 6 & - 2 \end{matrix}]$ .

Step 2

Consider the matrix $A - λ I = [\begin{matrix} 9 - λ & 5 \\ - 6 & - 2 - λ \end{matrix}]$ . The determinant of this matrix is $(λ - 4) (λ - 3)$ . Solving

the equation $(λ - 4) (λ - 3) = 0$ , we get, $λ_{1} = 4$ and $λ_{2} = 3$ . These are the eigenvalues.

For $λ = 4$ , we have, $A - λ I = [\begin{matrix} 5 & 5 \\ - 6 & - 6 \end{matrix}]$ . The null space of this matrix is $\{[\begin{matrix} - 1 \\ 1 \end{matrix}]\}$ . It is an eigenvector

corresponding to the eigenvalue 4.

For $λ = 3$ , we have, $A - λ I = [\begin{matrix} 6 & 5 \\ - 6 & - 5 \end{matrix}]$ . The null space of this matrix is $\{[\begin{matrix} - \frac{5}{6} \\ 1 \end{matrix}]\}$ . It is an eigenvector

corresponding to the eigenvalue 3.

Let $C = [\begin{matrix} - 1 \\ 1 \end{matrix}]$ and $D = [\begin{matrix} - \frac{5}{6} \\ 1 \end{matrix}]$ .

Step 3

The general solution is given by $X = α C e^{λ_{1} t} + β D e^{λ_{2} t}$ where $X = [\begin{matrix} x_{1} (t) \\ x_{2} (t) \end{matrix}]$ . So we get,

$\begin{array}{rcl} X & = & α C e^{λ_{1} t} + β D e^{λ_{2} t} \\ [\begin{matrix} x_{1} (t) \\ x_{2} (t) \end{matrix}] & = & α [\begin{matrix} - 1 \\ 1 \end{matrix}] e^{4 t} + β [\begin{matrix} - \frac{5}{6} \\ 1 \end{matrix}] e^{3 t} \\ [\begin{matrix} x_{1} (t) \\ x_{2} (t) \end{matrix}] & = & [\begin{matrix} - α e^{4 t} - \frac{5 β}{6} e^{3 t} \\ α e^{4 t} + β e^{3 t} \end{matrix}] \end{array}$

It is given that $x_{1} (0) = 1, x_{2} (0) = 0$ . So we get,

$\begin{array}{rcl} [\begin{matrix} x_{1} (0) \\ x_{2} (0) \end{matrix}] & = & [\begin{matrix} - α e^{4 (0)} - \frac{5 β}{6} e^{3 (0)} \\ α e^{4 (0)} + β e^{3 (0)} \end{matrix}] \\ [\begin{matrix} 1 \\ 0 \end{matrix}] & = & [\begin{matrix} - α - \frac{5 β}{6} \\ α + β \end{matrix}] \end{array}$

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