Exercise 3.(General Solution)Consider the system i = 4x – y, ÿ = 2x + ya) Write the system in matrix form, find the corresponding characteristic polynomial,and solve for the eigenvalues and eigenvectors.b) Find the general solution of the system of differential equations,c) Classify the fixed point at the origin.d) Solve the system with the initial condition xo = 3 and yo = 4.

Given system is dxdt=4x-ydydt=2x+y (i) In matrix form, system (i) can be written…

Answered: Exercise 3. (General Solution)Consider…

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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Exercise 3.
(General Solution)Consider the system i = 4x – y, ÿ = 2x + y
a) Write the system in matrix form, find the corresponding characteristic polynomial,
and solve for the eigenvalues and eigenvectors.
b) Find the general solution of the system of differential equations,
c) Classify the fixed point at the origin.
d) Solve the system with the initial condition xo = 3 and yo = 4.

Expert Solution

Step 1

Given system is

$\frac{d x}{d t} = 4 x - y \frac{d y}{d t} = 2 x + y$ (i)

In matrix form, system (i) can be written as

$x' = A x$ (ii)

Where

$x' = (\begin{matrix} \dot{x} \\ \dot{y} \end{matrix}), A = (\begin{matrix} 4 & - 1 \\ 2 & 1 \end{matrix}), x = (\begin{matrix} x \\ y \end{matrix})$

In order to find eigenvalues and eigenvectors, let us solve corresponding characteristic equation of matrix A

$\begin{array}{rcl} |A - k I| & = & 0 \\ |\begin{matrix} 4 - k & - 1 \\ 2 & 1 - k \end{matrix}| & = & 0 \\ k^{2} - 5 k + 6 & = & 0 \\ k & = & 2, 3 \end{array}$

For $k = 3$

$\begin{array}{rcl} (\begin{matrix} 4 - k & - 1 \\ 2 & 1 - k \end{matrix}) & = & (\begin{matrix} 1 & - 1 \\ 2 & 0 \end{matrix}) \\ ~ (\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix}) \end{array}$

On solving the matrix equation $\begin{array}{rcl} (\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix}) \end{array} (\begin{matrix} u \\ v \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ , taking $u = t$ gives $v = t$ . So, eigenvector corresponding to the eigenvalue $k = 3$ is $(\begin{matrix} 1 \\ 1 \end{matrix})$

For $k = 2$ , corresponding eigenvector is $(\begin{matrix} \frac{1}{2} \\ 1 \end{matrix})$

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