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Chapter 2: Problem 33

Find a solution curve of the system $$ \begin{aligned} &\dot{x}_{1}=x_{1}+2 x_{2}+x_{3} \\ &\dot{x}_{2}=x_{2}+2 x_{3} \\ &\dot{x}_{3}=2 x_{3} \end{aligned} $$ which satisfies \(x_{1}(0)=x_{2}(0)=0\) and \(x_{3}(0)=1\).

Short Answer

Expert verified
The solution curve is given by \(x_1(t) = 5e^{2t} - 4te^{t} - 5e^{t}\), \(x_2(t) = 2(e^{2t} - e^t)\), and \(x_3(t) = e^{2t}\).

Step by step solution

01

Understanding the System

We are given a system of differential equations with three variables \(x_1\), \(x_2\), and \(x_3\). Each equation describes the rate of change of these variables.
02

Solving for \(x_3(t)\)

From the equation \(\dot{x}_3 = 2x_3\), we can separate variables and integrate:\[ \frac{1}{x_3} \frac{dx_3}{dt} = 2 \implies \int \frac{1}{x_3} dx_3 = \int 2 dt \]\[ \ln |x_3| = 2t + C_1 \implies x_3(t) = e^{2t+C_1} = Ce^{2t} \]. Given \(x_3(0) = 1\), we find \(C = 1\), so \(x_3(t) = e^{2t}\).
03

Solving for \(x_2(t)\)

Using \(x_3(t) = e^{2t}\) in the equation \(\dot{x}_2 = x_2 + 2x_3\):\[ \dot{x}_2 = x_2 + 2e^{2t} \]This is a linear differential equation of the form \(\dot{y} = y + g(t)\) which can be solved using an integrating factor. The integrating factor is \(e^{-t}\). Multiplying throughout gives: \[ e^{-t}\dot{x}_2 = e^{-t}x_2 + 2e^{t} \]\[ \frac{d}{dt}(e^{-t}x_2) = 2e^{t} \]Integrating both sides:\[ e^{-t}x_2 = \int 2e^{t} dt = 2e^{t} + C_2 \]Thus, \(x_2 = 2e^{2t} + C_2e^{t}\). Using \(x_2(0) = 0\), we find \(C_2 = -2\), so \(x_2(t) = 2(e^{2t} - e^t)\).
04

Solving for \(x_1(t)\)

Using \(x_2(t) = 2(e^{2t} - e^t)\) and \(x_3(t) = e^{2t}\) in the equation \(\dot{x}_1 = x_1 + 2x_2 + x_3\):\[ \dot{x}_1 = x_1 + 4(e^{2t} - e^{t}) + e^{2t} \]\[ \dot{x}_1 = x_1 + 5e^{2t} - 4e^{t} \]Applying an integrating factor \(e^{-t}\):\[ e^{-t}\dot{x}_1 = e^{-t}x_1 + 5e^{t} - 4 \]\[ \frac{d}{dt}(e^{-t}x_1) = 5e^{t} - 4 \]Integrating:\[ e^{-t}x_1 = \int (5e^{t} - 4) dt = 5e^{t} - 4t + C_3 \]Thus, \(x_1 = e^{t}(5e^{t} - 4t + C_3)\). Using \(x_1(0) = 0\), we find \(C_3 = -5\), so \(x_1(t) = 5e^{2t} - 4te^{t} - 5e^{t}\).
05

Solution Curve

Given the initial conditions, we found \(x_1(t) = 5e^{2t} - 4te^{t} - 5e^{t}\), \(x_2(t) = 2(e^{2t} - e^t)\), and \(x_3(t) = e^{2t}\). These functions describe the solution curve of the system of differential equations.

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Key Concepts

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

Solution Curve
A solution curve in the context of differential equations is a collection of functions that satisfy the given system of equations along with the specified initial conditions. In geometric terms, it represents a path in the multi-dimensional space of the dependent variables that the system traverses over time.
For this exercise, we have determined the solution curve by separately solving for each variable: \(x_1(t), x_2(t),\) and \(x_3(t)\).
  • \(x_3(t) = e^{2t}\) is the result from solving a simplified equation, using the method of separation of variables.
  • \(x_2(t) = 2(e^{2t} - e^t)\) is obtained by integrating a linear first-order differential equation.
  • \(x_1(t) = 5e^{2t} - 4te^{t} - 5e^{t}\) involves solving a more complex form using an integrating factor.
Each of these functions contributes a dimension to the overall solution space, and when combined, they chart the complete trajectory of the system over time.
System of Differential Equations
A system of differential equations involves several equations that describe how multiple interdependent variables change simultaneously. In the given problem, we are dealing with a system containing three equations describing the rates of change of three variables: \(x_1\), \(x_2\), and \(x_3\).
This system can often be represented as:
  • \(\dot{x}_1 = x_1 + 2x_2 + x_3\)
  • \(\dot{x}_2 = x_2 + 2x_3\)
  • \(\dot{x}_3 = 2x_3\)
In a system like this, each differential equation is interconnected through the variables, meaning a change in one variable can directly or indirectly affect others. By solving the system, we derive a set of functions that describe the behavior and interaction of these variables over time. This solution is key to predicting future behavior based on initial circumstances.
Initial Conditions
Initial conditions are values that specify the starting point or state of a system of differential equations. They are crucial for determining the unique solution path the system will follow. In our given problem, the initial conditions are \(x_1(0) = 0\), \(x_2(0) = 0\), and \(x_3(0) = 1\).
  • \(x_1(0) = 0\) tells us where the solution curve for \(x_1\) starts on the time axis \(t=0\).
  • \(x_2(0) = 0\) provides a similar starting benchmark for \(x_2\).
  • \(x_3(0) = 1\) sets the initial value for \(x_3\).
These conditions are essential because they allow us to determine the specific constants when integrating each function. Without these pieces of information, we would have general solutions with arbitrary constants, which do not offer precise projections of the system's behavior.

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Most popular questions from this chapter

Find the form of the solution curves of the systems \(\dot{x}=A x\), \(x \in R^{4}\), where \(A\) equals: (a) \(\left[\begin{array}{llll}\lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda\end{array}\right]\); (b) \(\left[\begin{array}{llll}\alpha & -\beta & 0 & 0 \\ \beta & \alpha & 0 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda\end{array} \mid\right.\); (c) \(\left[\begin{array}{rrrr}\alpha & -\beta & 0 & 0 \\ \beta & \alpha & 0 & 0 \\ 0 & 0 & \gamma & -\delta \\ 0 & 0 & \delta & \gamma\end{array}\right]\).

Sketch phase portraits for the linear system $$ \left[\begin{array}{l} \dot{x}_{1} \\ \dot{x}_{2} \end{array}\right]=A\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right] $$ when \(A\) is given by: (a) \(\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]\) : (b) \(\left[\begin{array}{rr}-1 & 0 \\ 0 & 2\end{array}\right]\) (c) \(\left[\begin{array}{rr}3 & 1 \\ -1 & 3\end{array}\right]\) : (d) \(\left[\begin{array}{ll}2 & 1 \\ 0 & 2\end{array}\right]\) (e) \(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]:\) (f) \(\left[\begin{array}{rr}-3 & 0 \\ 0 & -3\end{array}\right]\) : (g) \(\left[\begin{array}{rr}0 & 2 \\ -2 & 0\end{array}\right]\); (h) \(\left[\begin{array}{rr}3 & 0 \\ 0 & -1\end{array}\right]\); (i) \(\left[\begin{array}{ll}3 & 0 \\ 0 & 0\end{array}\right]\); (j) \(\left[\begin{array}{rr}0 & -2 \\ 2 & 0\end{array}\right]\).

Find the solution curve \(\left(x_{1}(t), x_{2}(t)\right)\) which satisfies $$ \dot{x}_{1}=x_{1}+x_{2}+1, \quad \dot{x}_{2}=x_{1}+x_{2} $$ subject to the initial condition \(x_{1}(0)=a, x_{2}(0)=b\).

Let \(A\) be a \(3 \times 3\) real matrix with eigenvalues \(\lambda_{1}=\lambda_{2}=\lambda_{3}\). \(\left(=\lambda_{0}\right)\). Show that $$ \mathrm{e}^{A t}=\mathrm{e}^{i_{0}}\left(\boldsymbol{I}+t \boldsymbol{Q}+\frac{1}{2} t^{2} \boldsymbol{Q}^{2}\right) $$ where \(\boldsymbol{Q}=\boldsymbol{A}-\lambda_{0} \boldsymbol{I}\). Generalize this result to \(n \times n\) matrices.

Find the \(2 \times 2\) matrix \(A\) such that the system $$ \dot{x}=A x $$ has a solution curve $$ x(t)=\left[\begin{array}{c} e^{-t}(\cos t+2 \sin t) \\ e^{-t} \cos t \end{array}\right] . $$
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