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Chapter 6: Problem 7

\(x^{\prime}=x-3 y+e^{t}, y^{\prime}=-x+6 y, x(0)=0\), \(y(0)=1, t=1\)

Short Answer

Expert verified
For \(t = 1\), \( x = -2\) and \( y = 7\).

Step by step solution

01

- Determine the initial conditions

Given initial conditions are: \(x(0) = 0\) and \(y(0) = 1\)
02

- Write down the differential equations

The system of differential equations is: \( x' = x - 3y + e^t \) \( y' = -x + 6y \)
03

- Substitute initial conditions into the equations

Substitute \(t = 0\), \(x(0) = 0\), and \(y(0) = 1\): \[ x' = 0 - 3(1) + e^0 \] \[ x' = -3 + 1 \] \[ x' = -2 \] Now, \[ y' = -(0) + 6(1) \] \[ y' = 6 \]
04

- Solve the system at \(t = 1\)

For \(t = 1\), we integrate: \( x(t) = \text{Integral of (-2)} \times t \to x(t) = -2t \) \( y(t) = \text{Integral of (6)} \times t \to y(t) = 6t + 1 \text{ (considering initial condition)} \)
05

- Calculate values at \(t = 1\)

For \(t = 1\): \( x(1) = -2(1) = -2 \) \( y(1) = 6(1) + 1 = 7 \)

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Key Concepts

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

initial conditions
When solving a system of differential equations, initial conditions are crucial. They define the starting point of the functions. Here, the initial conditions are given as:
鈥� \(x(0)=0\)
鈥� \(y(0)=1\).
These initial values are plugged into our differential equations to find specific constants or to simplify the equations.

For instance, knowing \(x(0)=0\) helps ground our solution specifically to this initial condition, ensuring uniqueness. Similarly, \(y(0)=1\) allows us to solve for \(y\) accurately.
integration
Integration is a fundamental tool when dealing with systems of differential equations. When you reach a step that requires integration, you essentially find the antiderivative of a given function.

In our solution, the differential equation for \(x\) given by \(x' = -2\) integrates to:
鈥� \( \text{Integral of } -2 dt = -2t\).
For \(y\), we integrate the differential:
鈥� \( \text{Integral of } 6 dt = 6t\).
However, we must also consider the initial condition (\(y(0) = 1\)), so the final solution becomes:
鈥� \(y(t) = 6t + 1\).

Thus, integration allows us to revert the differential equations back to their original forms.
first-order differential equations
In this exercise, we deal with first-order differential equations, which involve the first derivative, denoted as \(x'\) or \(y'\). These equations help describe how \(x(t)\) and \(y(t)\) change over time.

The system given includes:
鈥� \(x' = x - 3y + e^t\)
鈥� \(y' = -x + 6y\)
These equations help determine the behavior of \(x\) and \(y\) as functions of time \(t\). By solving these differential equations, we get insight into how the dynamic system progresses.
First-order differential equations often appear in various real-world scenarios like population growth, circuit analysis, and others.

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Most popular questions from this chapter

. \(\mathbf{X}^{\prime}=\left(\begin{array}{cc}3 & 2 \\ -5 & 1\end{array}\right) \mathbf{X}+\left(\begin{array}{c}0 \\\ 10\end{array}\right), \mathbf{X}(0)=\left(\begin{array}{c}1 \\\ -2\end{array}\right)\)

Suppose that substance \(X\) decays into substance \(Y\) at rate \(k_{1}>0\) which in turn decays into another substance at rate \(k_{2}>\) 0 . If \(x(t)\) and \(y(t)\) represent the amount of \(X\) and \(Y\), respectively, then the system \(\left\\{\begin{array}{l}x^{\prime}=-k_{1} x \\ y^{\prime}=k_{1} x-k_{2} y\end{array} \quad\right.\) is solved to determine \(x(t)\) and \(y(t)\). Show that \((0,0)\) is the equilibrium solution of this system. Find the eigenvalues of the system and classify the equilibrium solution. Also, solve the system. Find \(\lim _{t \rightarrow \infty} x(t)\) and \(\lim _{t \rightarrow \infty} y(t)\). Do these limits correspond to the physical situation?

\(\mathbf{X}^{\prime}=\left(\begin{array}{ll}3 & -1 \\ 4 & -1\end{array}\right) \mathbf{X}+\left(\begin{array}{c}\cos t \\ \sin t\end{array}\right), \mathbf{X}(0)=\left(\begin{array}{l}0 \\\ 0\end{array}\right)\)

\(x^{\prime}=-x-2 y, y^{\prime}=x+y\)

\(\Phi(t)=\left(\begin{array}{cc}t^{-1} & 1 \\ 1 & t^{2}\end{array}\right)\), $$ \mathbf{A}=\left(\begin{array}{cc} -(t-1)^{-1} & t^{-2}(t-1)^{-1} \\ -2 t(t-1)^{-1} & -2(t-1)^{-1} \end{array}\right) \text {, } $$ $$ \left\\{\begin{array}{l} (t-1) t^{2} x^{\prime}=-t^{2} x+y \\ (t-1) y^{\prime}=-2 t x+2 y \\ x(2)=1, y(2)=0 \end{array}\right. $$
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