Which of the following functions is a solution to the differential equation ?
Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 39m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
13. Intro to Differential Equations
Basics of Differential Equations
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Given the linear system of differential equations find the most general real-valued solution.
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Step 1: Start by writing the system of differential equations in matrix form. Represent the system as \( \frac{d}{dt} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \). This allows us to analyze the system using linear algebra techniques.
Step 2: Find the eigenvalues of the coefficient matrix \( \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \). To do this, solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( A \) is the matrix and \( \lambda \) represents the eigenvalues. The determinant is \( \det \begin{bmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{bmatrix} = (2 - \lambda)^2 - 1 \).
Step 3: Simplify the characteristic equation \( (2 - \lambda)^2 - 1 = 0 \) to find the eigenvalues. Expand and solve for \( \lambda \): \( \lambda^2 - 4\lambda + 3 = 0 \). Factorize the quadratic equation to obtain \( \lambda = 3 \) and \( \lambda = 1 \). These are the eigenvalues of the matrix.
Step 4: For each eigenvalue, find the corresponding eigenvector by solving \( (A - \lambda I) \vec{v} = 0 \). For \( \lambda = 3 \), solve \( \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 \). For \( \lambda = 1 \), solve \( \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 \). Normalize the eigenvectors if necessary.
Step 5: Construct the general solution using the eigenvalues and eigenvectors. The solution takes the form \( \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = C_1 e^{\lambda_1 t} \vec{v}_1 + C_2 e^{\lambda_2 t} \vec{v}_2 \), where \( \lambda_1 \) and \( \lambda_2 \) are the eigenvalues, and \( \vec{v}_1 \) and \( \vec{v}_2 \) are the corresponding eigenvectors. Substitute the values of \( \lambda \) and \( \vec{v} \) to express \( x(t) \) and \( y(t) \) explicitly.
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