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Chapter 6: Problem 5

\(\left\\{\begin{array}{l}x_{1}^{\prime}=-3 x_{1}, x_{1}(0)=-1 \\\ x_{2}^{\prime}=1, x_{2}(0)=1\end{array}\right.\)

Short Answer

Expert verified
The solutions are \(x_1(t) = -e^{-3t}\) and \(x_2(t) = t + 1\).

Step by step solution

01

Solve the differential equation for x1

The differential equation for \(x_1\) is given by \(x_1' = -3x_1\). This is a first-order linear differential equation. The general solution for such an equation is of the form \(x_1(t) = Ce^{-3t}\).
02

Apply Initial Condition for x1

Using the initial condition \(x_1(0) = -1\), substitute \(t = 0\) and solve for \(C\): \(x_1(0) = Ce^{-3 \times 0} = C = -1\). Thus, the solution for \(x_1(t)\) is \(x_1(t) = -e^{-3t}\).
03

Solve the differential equation for x2

The differential equation for \(x_2\) is \(x_2' = 1\). This can be integrated directly to \(x_2(t) = t + C\).
04

Apply Initial Condition for x2

Using the initial condition \(x_2(0) = 1\), substitute \(t = 0\) and solve for \(C\): \(x_2(0) = 0 + C = 1 \), so \(C = 1\). Thus, the solution for \(x_2(t)\) is \(x_2(t) = t + 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-Order Linear Differential Equation
A first-order linear differential equation is an equation involving a function and its first derivative. It can generally be expressed as:
\(\frac{dy}{dt} + P(t)y = Q(t)\)
where P(t) and Q(t) are functions of t. These equations often describe exponential growth or decay processes.
In our exercise, we have x_1' = -3x_1, which fits this form. Using known methods, we solve it to find x_1(t). The solution involves an exponential function due to the linearity and constant coefficient of the equation.
Initial Conditions
Initial conditions specify the value of the function at a particular point, usually at the beginning. They are crucial for finding the particular solution to a differential equation.
For instance, we have x_1(0) = -1 and x_2(0) = 1. These values allow us to solve for the constant in our general solutions.
By applying the initial condition to our solved equation, we're able to determine the precise behavior of the solution from the very start.
General Solution
The general solution of a differential equation encompasses all possible solutions before applying initial conditions. It's represented with arbitrary constants that are determined by these conditions.
For x_1' = -3x_1, the general solution is x_1(t) = Ce^{-3t}. Applying x_1(0) = -1 gives us C = -1, so our specific solution is x_1(t) = -e^{-3t}. Similarly, solving x_2' = 1 gives x_2(t) = t + C. The initial condition x_2(0) = 1 helps find that C = 1, thus our specific solution is x_2(t) = t + 1.

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Most popular questions from this chapter

\(\mathbf{X}^{\prime}=\left(\begin{array}{cc}-4 & 0 \\ 2 & 4\end{array}\right) \mathbf{X}, \mathbf{X}(0)=\left(\begin{array}{l}8 \\ 0\end{array}\right)\)

\(\left\\{\begin{array}{l}x^{\prime}=-2 x+4 y+2 z-2 w \\ y^{\prime}=6 x-10 y-7 z+4 w \\ z^{\prime}=-6 x+10 y+7 z-4 w \\ w^{\prime}=9 x-16 y-10 z+7 w\end{array}\right.\)

If \(\mathbf{A}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) and \(|\mathbf{A}|=a d-b c \neq 0\), then show that $$ \mathbf{A}^{-1}=\frac{1}{a d-b c}\left(\begin{array}{cc} d & -b \\ -c & a \end{array}\right) $$

Show that in the \(3 \times 3\) system \(\mathbf{X}^{\prime}=\mathbf{A X}\) with the eigenvalue \(\lambda\) of multiplicity 3 with one corresponding (linearly independent) eigenvector \(\mathbf{v}_{1}\), three linearly independent solutions of the system are \(\mathbf{X}_{1}=\mathbf{v}_{1} e^{\lambda t}, \mathbf{X}_{2}=\) \(\left(\mathbf{v}_{1} t+\mathbf{w}_{2}\right) e^{\lambda t}\), and \(\mathbf{X}_{3}=\left(\frac{1}{2} \mathbf{v}_{1} t^{2}+\mathbf{w}_{2} t+\right.\) \(\left.\mathbf{u}_{3}\right) e^{\lambda t}\), where \(\mathbf{u}_{3}\) satisfies \((\mathbf{A}-\lambda \mathbf{I}) \mathbf{u}_{3}=\mathbf{w}_{2}\) and \(\mathbf{w}_{2}\) satisfies \((\mathbf{A}-\lambda \mathbf{I}) \mathbf{w}_{2}=\mathbf{v}_{1}\).

\(x^{\prime}=2 y+x^{-1} y^{-2}, y^{\prime}=-2 x-x^{-2} y^{-1} ; R=\) \([0.01,5] \times[0.01,5]\)
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