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Chapter 9: Problem 16

Use hand calculations to find a fundamental set of solutions for the system \(\mathbf{y}^{\prime}=A \mathbf{y}\), where \(A\) is the matrix given. \(A=\left(\begin{array}{rr}2 & 0 \\ -4 & -2\end{array}\right)\)

Short Answer

Expert verified
The fundamental set of solutions is \( (e^{2t}, 0) \) and \( (0, e^{-2t}) \).

Step by step solution

01

Write the System of Differential Equations

The system of differential equations is given by \(\mathbf{y}' = A \mathbf{y}\). Here, \(A\) is the matrix \(\begin{pmatrix}2 & 0 \ -4 & -2\end{pmatrix}\). This results in the system: \(\begin{pmatrix} y_1' \ y_2' \end{pmatrix} = \begin{pmatrix} 2 & 0 \ -4 & -2 \end{pmatrix} \begin{pmatrix} y_1 \ y_2 \end{pmatrix}\). Expanding, we get: \(y_1' = 2y_1\) and \(y_2' = -4y_1 - 2y_2\).
02

Solve the First Equation

The first equation is \(y_1' = 2y_1\). This is a first-order linear differential equation, whose solution is of the form \(y_1 = C_1 e^{2t}\), where \(C_1\) is a constant.
03

Solve the Second Equation

Substitute \(y_1 = C_1 e^{2t}\) into the second equation \(y_2' = -4y_1 - 2y_2\), which becomes \(y_2' = -4C_1 e^{2t} - 2y_2\). This can be rewritten as \(y_2' + 2y_2 = -4C_1 e^{2t}\).
04

Solve the Non-Homogeneous Equation for \(y_2\)

The equation \(y_2' + 2y_2 = -4C_1 e^{2t}\) is a linear first-order non-homogeneous differential equation. Using an integrating factor \(e^{\int 2 dt} = e^{2t}\), transform the equation to \(d/dt (e^{2t} y_2) = -4C_1 e^{4t}\). Integrate both sides to find \(e^{2t} y_2 = -C_1 e^{4t} + C_2\). Thus, \(y_2 = -C_1 e^{2t} + C_2 e^{-2t}\).
05

Combine Solutions to Form the General Solution

The fundamental set of solutions is comprised of solutions \((y_1, y_2)\). For this system, a fundamental set is given by \((e^{2t}, 0)\) and \( (0, e^{-2t})\). So the general solution is \(y_1 = C_1 e^{2t}\) and \(y_2 = C_2 e^{-2t}\), where \(C_1\) and \(C_2\) are constants.

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Key Concepts

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

Linear Differential Equations
Linear differential equations are a fundamental class of differential equations in mathematics. These are equations involving derivatives of a function and are linear in terms of the unknown function and its derivatives. An example of a linear differential equation is:
\[ y' = ay + b \]where \(a\) and \(b\) are constants or functions of the independent variable, and \(y\) is the unknown function.
  • The characteristic feature is that the function \(y\) and its derivatives appear to the first power and are not multiplied together.
  • These equations can be solved using integrating factors, transformations, or characteristic equations when the equation is homogeneous.
Solving these equations often involves finding an exponential function as a solution by employing methods like separation of variables for simpler types, or integrating factors for non-homogeneous equations.
Non-Homogeneous Differential Equations
Non-homogeneous differential equations are similar to their homogeneous counterparts but include an additional function on the right-hand side of the equation. The general form of a non-homogeneous linear differential equation is:
\[ y' + p(t)y = g(t) \]where \(p(t)\) and \(g(t)\) are given functions, and \(y\) is the unknown function.
  • The function \(g(t)\) represents the non-homogeneous part, distinguishing it from homogeneous equations where \(g(t) = 0\).
  • Toolbox methods like undetermined coefficients or variation of parameters often help solve these equations.
  • For instance, in the exercise, the non-homogeneous differential equation for \(y_2\) was tackled using an integrating factor, which is \(e^{\int 2 dt} = e^{2t}\).
This method allowed transforming the equation into a form that can be directly integrated, paving the way to find a solution that includes terms combining constants and exponentials.
Fundamental Set of Solutions
In the context of systems of differential equations, a fundamental set of solutions is a collection of solutions that form a basis for the solution space of the differential equation system. This means any solution to the system can be expressed as a linear combination of the fundamental solutions.
  • In the exercise, we found the fundamental set of solutions for the given system by solving the differential equations \(\mathbf{y}' = A \mathbf{y}\).
  • The solutions \((e^{2t}, 0)\) and \((0, e^{-2t})\) were identified as the fundamental set.
  • These represent independent solution vectors, meaning they are linearly independent over the interval of interest.
Understanding and identifying such sets are crucial since they help describe all possible behaviors of solutions in a system of linear differential equations.

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Most popular questions from this chapter

For the matrices, use a computer to help find a fundamental set of solutions to the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). \(A=\left(\begin{array}{rrr}-7 & 2 & -4 \\ 42 & -11 & 18 \\ 38 & -10 & 18\end{array}\right)\)

Each of the matrices has only one eigenvalue \(\lambda\). In each exercise, determine the smallest \(k\) such that \((A-\lambda I)^{k}=0\). The use the fact that $$ e^{t A}=e^{\lambda t}\left[I+t(A-\lambda I)+\frac{t^{2}}{2 !}(A-\lambda I)^{2}+\cdots\right]$$ to compute \(e^{t A}\). \(A=\left(\begin{array}{rrrr}-5 & 0 & -1 & 4 \\ -4 & 0 & 1 & 5 \\ 4 & -4 & -5 & -4 \\ 0 & -1 & -1 & -2\end{array}\right)\)

Each equation has a characteristic equation possessing distinct real roots. Find the general solution of each equation. \(y^{\prime \prime \prime}-2 y^{\prime \prime}-y^{\prime}+2 y=0\)

Classify the equilibrium point of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) based on the position of \((T, D)\) in the trace-determinant plane. Sketch the phase portrait by hand. Verify your result by creating a phase portrait with your numerical solver. \(A=\left(\begin{array}{ll}2 & -4 \\ 8 & -6\end{array}\right)\)

For the matrices, use a computer to help find a fundamental set of solutions to the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). \(A=\left(\begin{array}{rrr}-1 & -2 & 4 \\ -1 & 0 & -4 \\ -1 & 2 & -6\end{array}\right)\)
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