Chapter 11: Problem 19
In Problems 19-26, solve the given initial-value problem. \(\left[\begin{array}{c}\frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t}\end{array}\right]=\left[\begin{array}{rr}-3 & 0 \\ 4 & 2\end{array}\right]\left[\begin{array}{l}x_{1}(t) \\\ x_{2}(t)\end{array}\right]\) with \(x_{1}(0)=-5\) and \(x_{2}(0)=5\).
Short Answer
Step by step solution
Formulate the System of Equations
Solve the First Differential Equation
Solve the Second Differential Equation
Solve the Homogeneous Part of the Second Equation
Solve the Particular Solution of the Second Equation
Form the General Solution for \( x_2 \)
Apply Initial Conditions for \( x_2 \)
Write the Final Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
System of Differential Equations
The system we are dealing with is:
- \(\frac{dx_1}{dt} = -3x_1\)
- \(\frac{dx_2}{dt} = 4x_1 + 2x_2\)
Solving these systems typically involves finding each function's expression over time, usually using initial conditions given in the problem.
First-Order Linear Differential Equations
In this problem:
- \(\frac{dx_1}{dt} = -3x_1\) is treated as a straightforward linear differential equation.
- The second equation \(\frac{dx_2}{dt} = 4x_1 + 2x_2\) is considered a linear inhomogeneous equation because it includes a term that is not multiplied by \(x_2(t)\), namely \(4x_1(t)\).
Homogeneous Solution
For the second differential equation \(\frac{dx_2}{dt} = 4x_1 + 2x_2\), we first consider the homogeneous part:
- \(\frac{dx_2}{dt} = 2x_2\)
- \(x_2^h(t) = C_2 e^{2t}\)
Particular Solution
- \(\frac{dx_2}{dt} = -20e^{-3t} + 2x_2\)
- Solving gives \(A = 4\).