Problem 3
Question
Show that the given functions are solutions of the system \(\mathbf{x}^{\prime}(t)=A(x) \mathbf{x}(t)\) for the given matrix \(A,\) and hence, find the general solution to the system (remember to check linear independence). If auxiliary conditions are given, find the particular solution that satisfies these conditions. $$\mathbf{x}_{1}(t)=\left[\begin{array}{l} e^{-t} \cos 2 t \\ e^{-t} \sin 2 t \end{array}\right], \quad \mathbf{x}_{2}(t)=\left[\begin{array}{c} -e^{-t} \sin 2 t \\ e^{-t} \cos 2 t \end{array}\right]$$, $$A=\left[\begin{array}{ll} 1 & -2 \\ 2 & 1 \end{array}\right], \quad \mathbf{x}(0)=\left[\begin{array}{l} 1 \\ 3 \end{array}\right]$$.
Step-by-Step Solution
Verified Answer
In short, we have shown that the given functions \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\) are solutions to the system of linear differential equations with the given matrix \(A\). We checked their linear independence using the Wronskian determinant and found that they are linearly independent. Thus, the general solution is given by \(\mathbf{x}(t) = C_1\mathbf{x}_1(t) + C_2\mathbf{x}_2(t)\). To find the particular solution, we used the initial condition \(\mathbf{x}(0) = \begin{bmatrix}1\\3\end{bmatrix}\) and determined the constants \(C_1 = 1\) and \(C_2 = 3\). Hence, the particular solution is given by:
\(\mathbf{x}(t) = \begin{bmatrix}e^{-t}\cos 2t - 3e^{-t}\sin 2t\\ e^{-t}\sin 2t + 3e^{-t}\cos 2t\end{bmatrix}\).
1Step 1: Find the derivatives of the given functions
To verify that \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\) are solutions to the given system, we need to compute their derivatives and see if they satisfy the given equation.
First, compute the derivative of \(\mathbf{x}_1(t)\):
\(\mathbf{x}_1'(t) = \left[\frac{d}{dt}(e^{-t}\cos 2t), \frac{d}{dt}(e^{-t}\sin 2t)\right]^T\)
Using the product rule, we get:
\(\mathbf{x}_1'(t) =\left[(-e^{-t}\cos 2t -2e^{-t}\sin 2t), (-e^{-t}\sin 2t +2e^{-t}\cos 2t)\right]^T\)
Similarly, compute the derivative of \(\mathbf{x}_2(t)\):
\(\mathbf{x}_2'(t) = \left[\frac{d}{dt}(-e^{-t}\sin 2t), \frac{d}{dt}(e^{-t}\cos 2t)\right]^T\)
Using the product rule, we get:
\(\mathbf{x}_2'(t) =\left[(-e^{-t}\cos 2t +2e^{-t}\sin 2t), (-e^{-t}\sin 2t -2e^{-t}\cos 2t)\right]^T\)
2Step 2: Check if the given functions satisfy the system
We have to check if \(\mathbf{x}_1'(t)\) and \(\mathbf{x}_2'(t)\) satisfy the differential equation system:
\(\mathbf{x}'(t) = A(x)\mathbf{x}(t)\)
First, for \(\mathbf{x}_1'(t)\), we have:
\begin{align*}
A(x)\mathbf{x}_1(t) &= \begin{bmatrix}1 & -2\\ 2 & 1\end{bmatrix} \begin{bmatrix}e^{-t}\cos 2t\\ e^{-t}\sin 2t\end{bmatrix}\\
&= \begin{bmatrix}e^{-t}\cos2t -2e^{-t}\sin2t\\2e^{-t}\cos2t + e^{-t}\sin2t\end{bmatrix}
\end{align*}
Comparing this with \(\mathbf{x}_1'(t)\), we see that they are equal. Thus, \(\mathbf{x}_1(t)\) is a solution of the system.
Similarly, for \(\mathbf{x}_2'(t)\), we have:
\begin{align*}
A(x)\mathbf{x}_2(t) &= \begin{bmatrix}1 & -2\\ 2 & 1\end{bmatrix} \begin{bmatrix}-e^{-t}\sin 2t\\ e^{-t}\cos2t\end{bmatrix}\\
&= \begin{bmatrix}-e^{-t}\sin2t -2e^{-t}\cos2t\\2e^{-t}\sin2t + e^{-t}\cos2t\end{bmatrix}
\end{align*}
Comparing this with \(\mathbf{x}_2'(t)\), we see that they are equal. Thus, \(\mathbf{x}_2(t)\) is a solution of the system.
3Step 3: Check the linear independence and find the general solution
To find the general solution of the given system, we have to check whether \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\) are linearly independent. We can use the Wronskian determinant, which is defined as follows for two functions:
\(W[\mathbf{x}_1(t), \mathbf{x}_2(t)] = \begin{vmatrix}e^{-t}\cos 2t& -e^{-t}\sin 2t\\ e^{-t}\sin 2t&e^{-t}\cos 2t\end{vmatrix}\)
Calculating the Wronskian determinant, we see that it is nonzero:
\(W[\mathbf{x}_1(t), \mathbf{x}_2(t)] = (e^{-t}\cos 2t)(e^{-t}\cos 2t) - (-e^{-t}\sin 2t)(e^{-t}\sin 2t) = e^{-2t}\)
Since the Wronskian determinant is nonzero, \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\) are linearly independent. Thus, the general solution for the system is given by:
\(\mathbf{x}(t) = C_1\mathbf{x}_1(t) + C_2\mathbf{x}_2(t)\)
4Step 4: Find the particular solution
We are given the initial condition \(\mathbf{x}(0) = \begin{bmatrix}1\\3\end{bmatrix}\). To find the particular solution, we need to determine the constants \(C_1\) and \(C_2\) that satisfy this condition. Plug in \(t = 0\) into the general solution and set it equal to the initial condition:
\begin{bmatrix}1\\3\end{bmatrix} = C_1\begin{bmatrix}1\\0\end{bmatrix} + C_2\begin{bmatrix}0\\1\end{bmatrix}\)
From this, we see that \(C_1 = 1\) and \(C_2 = 3\). Therefore, the particular solution is given by:
\(\mathbf{x}(t) = \mathbf{x}_1(t) + 3\mathbf{x}_2(t) = \begin{bmatrix}e^{-t}\cos 2t - 3e^{-t}\sin 2t\\ e^{-t}\sin 2t + 3e^{-t}\cos 2t\end{bmatrix}\)
Key Concepts
Linear IndependenceGeneral SolutionParticular SolutionWronskian Determinant
Linear Independence
When we talk about linear independence in the context of solutions to a system of differential equations, we are checking if the solutions are not just simple multiples of each other.
In terms of functions, two functions are linearly independent if no scalar multiple of one function can produce the other function.
This concept is crucial in determining the general solution for differential equations.
To determine linear independence, we can use the Wronskian determinant, which is a determinant of a matrix whose rows are composed of the functions and their derivatives.
In terms of functions, two functions are linearly independent if no scalar multiple of one function can produce the other function.
This concept is crucial in determining the general solution for differential equations.
To determine linear independence, we can use the Wronskian determinant, which is a determinant of a matrix whose rows are composed of the functions and their derivatives.
- If the Wronskian determinant is non-zero at some point in the interval, the functions are linearly independent.
- If it's zero everywhere, they are dependent.
General Solution
The general solution to a system of differential equations provides a comprehensive description of all possible solutions.
For linear systems, the general solution is typically a linear combination of linearly independent solutions.
In our exercise, \[\mathbf{x}(t) = C_1 \mathbf{x}_1(t) + C_2 \mathbf{x}_2(t)\]is the general solution where \(C_1\) and \(C_2\) are constants.
This format means that any particular solution to the system can be constructed by choosing appropriate values for these constants.
For linear systems, the general solution is typically a linear combination of linearly independent solutions.
In our exercise, \[\mathbf{x}(t) = C_1 \mathbf{x}_1(t) + C_2 \mathbf{x}_2(t)\]is the general solution where \(C_1\) and \(C_2\) are constants.
This format means that any particular solution to the system can be constructed by choosing appropriate values for these constants.
- The coefficients \(C_1\) and \(C_2\) offer flexibility to fit auxiliary or initial conditions, tailoring the solution to specific requirements.
- Finding the general solution is a fundamental step before solving for particular solutions.
Particular Solution
A particular solution of a differential equation is a specific solution that satisfies not only the equation itself but also any given initial or boundary conditions.
To find it, we substitute specific values into the general solution to determine the unknown constants.
In our exercise, we used the initial condition \(\mathbf{x}(0) = \begin{bmatrix}1\3\end{bmatrix}\) to find \(C_1\) and \(C_2\):
To find it, we substitute specific values into the general solution to determine the unknown constants.
In our exercise, we used the initial condition \(\mathbf{x}(0) = \begin{bmatrix}1\3\end{bmatrix}\) to find \(C_1\) and \(C_2\):
- Plugging \(t = 0\) into the general solution and comparing it against the initial condition helps solve the constants.
- By solving \(\begin{bmatrix}1\3\end{bmatrix} = C_1 \mathbf{x}_1(0) + C_2 \mathbf{x}_2(0)\), you find the specific values for \(C_1\) and \(C_2\) that fit the condition.
Wronskian Determinant
The Wronskian determinant provides a powerful tool for checking linear independence of solutions in a differential equation system.
It is composed of the solutions of the differential system arranged in a matrix with their derivatives, and the determinant is then computed.
For example, for two functions \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\), the Wronskian \(W(\mathbf{x}_1(t), \mathbf{x}_2(t))\) is calculated as follows:\[W(\mathbf{x}_1, \mathbf{x}_2) = \begin{vmatrix} e^{-t} \cos 2t & -e^{-t} \sin 2t \ e^{-t} \sin 2t & e^{-t} \cos 2t \end{vmatrix}\]
It is composed of the solutions of the differential system arranged in a matrix with their derivatives, and the determinant is then computed.
For example, for two functions \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\), the Wronskian \(W(\mathbf{x}_1(t), \mathbf{x}_2(t))\) is calculated as follows:\[W(\mathbf{x}_1, \mathbf{x}_2) = \begin{vmatrix} e^{-t} \cos 2t & -e^{-t} \sin 2t \ e^{-t} \sin 2t & e^{-t} \cos 2t \end{vmatrix}\]
- In our exercise, the Wronskian is \(e^{-2t}\), which is nonzero, verifying the linear independence of the solutions.
- A non-zero Wronskian indicates a set of functions that can form a basis for the space of solutions to the differential equation.
Other exercises in this chapter
Problem 2
Solve the given system of differential equations. $$x_{1}^{\prime}=2 x_{1}-3 x_{2}, \quad x_{2}^{\prime}=x_{1}-2 x_{2}$$
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Show that the given vector functions are linearly independent on \((-\infty, \infty)\). $$\begin{array}{l} \mathbf{x}_{1}(t)=\left[\begin{array}{c} t+1 \\ t-1 \
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Determine the general solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\) $$\left[\begin{array}{rr} -3 & -2 \\ 2 & 1 \end{arr
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