Find the general solution of the given system. dx = 4x + y dt dy = -2x + 2y dt (x(t), y(t)) =
Find the general solution of the given system. dx = 4x + y dt dy = -2x + 2y dt (x(t), y(t)) =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![To solve the given system of differential equations, follow these steps:
1. **System of Differential Equations:**
\[
\begin{align}
\frac{dx}{dt} &= 4x + y \\
\frac{dy}{dt} &= -2x + 2y
\end{align}
\]
2. **General Solution:**
We aim to find \((x(t), y(t))\), which represents the solution as functions of \(t\).
3. **Matrix Representation:**
The system can be written in matrix form:
\[
\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ -2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}
\]
Let \(\mathbf{X} = \begin{pmatrix} x \\ y \end{pmatrix}\), then:
\[
\frac{d\mathbf{X}}{dt} = A\mathbf{X} \quad \text{where} \quad A = \begin{pmatrix} 4 & 1 \\ -2 & 2 \end{pmatrix}
\]
4. **Eigenvectors and Eigenvalues:**
To find the general solution, we find the eigenvalues \(\lambda\) and eigenvectors \(\mathbf{v}\) of matrix \(A\).
\[
\det(A - \lambda I) = 0
\]
\[
\det \begin{pmatrix} 4 - \lambda & 1 \\ -2 & 2 - \lambda \end{pmatrix} = (4 - \lambda)(2 - \lambda) - (-2)(1) = 0
\]
\[
(4 - \lambda)(2 - \lambda) + 2 = \lambda^2 - 6\lambda + 10 = 0
\]
Solve the quadratic equation for \(\lambda\).
5. **Using the Eigenvalues and Eigenvectors:**
Assume eigenvalues are \(\lambda_1\) and \(\lambda_2\) with corresponding eigenvectors \(\mathbf{v}_1\) and \(\mathbf{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77d2f4c8-4183-4052-877b-b5f84c5df55f%2F9573a52c-616f-4b86-a032-b94b89e4e180%2Fx7y3sko_processed.jpeg&w=3840&q=75)
Transcribed Image Text:To solve the given system of differential equations, follow these steps:
1. **System of Differential Equations:**
\[
\begin{align}
\frac{dx}{dt} &= 4x + y \\
\frac{dy}{dt} &= -2x + 2y
\end{align}
\]
2. **General Solution:**
We aim to find \((x(t), y(t))\), which represents the solution as functions of \(t\).
3. **Matrix Representation:**
The system can be written in matrix form:
\[
\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ -2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}
\]
Let \(\mathbf{X} = \begin{pmatrix} x \\ y \end{pmatrix}\), then:
\[
\frac{d\mathbf{X}}{dt} = A\mathbf{X} \quad \text{where} \quad A = \begin{pmatrix} 4 & 1 \\ -2 & 2 \end{pmatrix}
\]
4. **Eigenvectors and Eigenvalues:**
To find the general solution, we find the eigenvalues \(\lambda\) and eigenvectors \(\mathbf{v}\) of matrix \(A\).
\[
\det(A - \lambda I) = 0
\]
\[
\det \begin{pmatrix} 4 - \lambda & 1 \\ -2 & 2 - \lambda \end{pmatrix} = (4 - \lambda)(2 - \lambda) - (-2)(1) = 0
\]
\[
(4 - \lambda)(2 - \lambda) + 2 = \lambda^2 - 6\lambda + 10 = 0
\]
Solve the quadratic equation for \(\lambda\).
5. **Using the Eigenvalues and Eigenvectors:**
Assume eigenvalues are \(\lambda_1\) and \(\lambda_2\) with corresponding eigenvectors \(\mathbf{v}_1\) and \(\mathbf{
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