Question

In Exercises \(1-10\) find a particular solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right] \mathbf{y}+\left[\begin{array}{l} 1 \\ t \end{array}\right] $$

Short answer

Question: Find the general solution for the given first-order linear differential equation. Given matrix A and vector function f: $$ A=\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right], f=\left[\begin{array}{l} 1 \\ t \end{array}\right] $$ Answer: The general solution for the given first-order linear differential equation is given by: $$ \mathbf{y}(t) = \mathbf{y_h}(t) + \mathbf{y_p}(t) $$ where $\mathbf{y_h}(t)$ is the homogeneous solution: $$ \mathbf{y_h} = c_1\text{Re}\left(e^{it}\left[\begin{array}{c} 1 \\ i \end{array}\right]\right) + c_2\text{Re}\left(e^{-it}\left[\begin{array}{c} 1 \\ -i \end{array}\right]\right) $$ And $\mathbf{y_p}(t)$ is the particular solution found by applying the variation of parameters method to the matrix A and vector function f. To find $\mathbf{y_p}(t)$, you will need to compute the integrals for the functions $u_1(t)$ and $u_2(t)$, and then substitute them back in the equation for $\mathbf{y_p}(t)$. Finally, add the homogeneous solution and particular solution to obtain the general solution.

Step-by-step solution

Find the homogeneous solution

For the homogeneous equation: $$ \mathbf{y}^\prime = A\mathbf{y} $$ Try a complex solution of the form \(\mathbf{y}=e^{rt}\mathbf{v}\), where r is an unknown complex exponent, and \(\mathbf{v}\) is a constant vector. Substituting this into our homogeneous equation, we get: $$ re^{rt}\mathbf{v} = A\mathbf{v}e^{rt} $$ Factor out \(e^{rt}\) and find an eigenvalue and eigenvector pair: $$ r\mathbf{v} = A\mathbf{v} $$ The characteristic equation for A is: $$ \text{det}(A-rI) = \left|\begin{array}{rr} -r & 1 \\ -1 & -r \end{array}\right| = r^2 +1 = 0 $$ The eigenvalues are \(r=\pm i\). For each eigenvalue, we find the corresponding eigenvector: For \(r=i\), the matrix \(A-rI\) and the eigenvector \(\mathbf{v}\) are: $$ \left[\begin{array}{cc} -i & 1 \\ -1 & -i \end{array}\right]\left[\begin{array}{c} v_1 \\ v_2 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right] $$ Find \(\mathbf{v}\) (a solution can be \(v_1 = 1, v_2 = i\)). Similarly for \(r=-i\), we find the eigenvector $\mathbf{v} = \left[\begin{array}{c} 1 \\ -i \end{array}\right]$. Since we have found two linearly independent complex solutions, the homogeneous solution is given by the real part of those: $$ \mathbf{y_h} = c_1\text{Re}\left(e^{it}\left[\begin{array}{c} 1 \\ i \end{array}\right]\right) + c_2\text{Re}\left(e^{-it}\left[\begin{array}{c} 1 \\ -i \end{array}\right]\right) $$ Where \(c_1\) and \(c_2\) are constants.

Apply Variation of Parameters

Now, we need to find the particular solution by replacing the constants \(c_1\) and \(c_2\) with functions \(u_1(t)\) and \(u_2(t)\): $$ \mathbf{y_p}(t) = u_1(t)\text{Re}\left(e^{it}\left[\begin{array}{c} 1 \\ i \end{array}\right]\right) + u_2(t)\text{Re}\left(e^{-it}\left[\begin{array}{c} 1 \\ -i \end{array}\right]\right) $$ Compute the Wronskian for the homogeneous solution: $$ W(t) = \text{det}\left[\begin{array}{cc} \text{Re}\left(e^{it}\right) & \text{Re}\left(e^{-it}\right) \\ \text{Re}\left(ie^{it}\right) & \text{Re}\left(-ie^{-it}\right) \end{array}\right] = \left|\begin{array}{cc} \cos t & \cos(-t) \\ -\sin t & -\sin(-t) \end{array}\right| = 2 \cos^2 t $$ Now we can compute the functions \(u_1(t)\) and \(u_2(t)\) by integrating: $$ u_1(t) = -\int\frac{f_1(t)W_{10}(t) + f_2(t)W_{20}(t)}{W(t)}dt $$ $$ u_2(t) = \int\frac{f_1(t)W_{01}(t) + f_2(t)W_{11}(t)}{W(t)}dt $$ Where $f = \left[\begin{array}{c} f_1(t) \\ f_2(t) \end{array}\right]\( and \)W_{ij}(t)\( is the matrix obtained by replacing the i-th column of the Wronskian by the function \)f_j(t)$. After calculating the integrals and obtaining \(\mathbf{y_p}(t)\), the general solution will be given by: $$ \mathbf{y}(t) = \mathbf{y_h}(t) + \mathbf{y_p}(t) $$

Key concepts

Particular Solution

A particular solution to a differential equation is a specific solution that satisfies not only the differential equation itself but also accounts for any additional non-homogeneous terms. In simpler words, while the general solution provides a family of potential solutions to the homogeneous equation, the particular solution zeroes in on the specific terms introduced by external factors or forces, such as a forcing function.

When dealing with differential equations of the form \( \mathbf{y}^{\prime} = A \mathbf{y} + \mathbf{f}(t) \), the task is to find a solution that specifically works for the vector function \( \mathbf{f}(t) \).

Typically, once we determine the homogeneous solution, the Variation of Parameters technique assists us in finding this particular solution by enabling us to replace arbitrary constants with functions of time, \( u_1(t) \) and \( u_2(t) \). This allows for a specific solution tailored to the non-homogeneous components of the equation.

Homogeneous Solution

The homogeneous solution is an essential part of solving differential equations. It refers to a solution derived from the equation where non-homogeneous terms are set to zero, leading to a form like \( \mathbf{y}^\prime = A \mathbf{y} \).

Why is it called 'homogeneous'? Because the equation only involves a consistent system, without any external forcing function or non-homogeneous part.

In practice, finding the homogeneous solution involves identifying characteristic values called eigenvalues and corresponding vectors called eigenvectors.

For our example, we assumed a solution form like \( e^{rt}\mathbf{v} \), which led us to solve for the eigenvalues \( r \) and eigenvectors \( \mathbf{v} \). The resulting solutions form the backbone of the system's response to initial conditions and help build the general solution when combined with a particular solution.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in the analysis of linear transformations and systems described by matrices. In the context of solving differential equations, these concepts help in determining the nature of the solutions to homogeneous equations.

An eigenvalue, often denoted as \( r \), satisfies the characteristic equation \( \text{det}(A - rI) = 0 \), where \( A \) is the matrix of the system and \( I \) is the identity matrix.

Once the eigenvalues are determined, eigenvectors correspond to each eigenvalue and satisfy the equation \( (A - rI)\mathbf{v} = \mathbf{0} \). They indicate directions in which the system can stretch or compress but not rotate.

In the differential equations described in the problem, finding these values and vectors allows us to construct a complete homogeneous solution, crucial for understanding the behavior of dynamic systems, like mechanical or electrical systems.

Variation of Parameters

The Variation of Parameters is a powerful method for finding a particular solution to non-homogeneous differential equations. This approach involves adjusting previously determined solutions to account for non-homogeneous parts of an equation. The method is particularly useful when dealing with functions that are not easily solved by simpler methods, such as undetermined coefficients.

In the process, we replace constant coefficients, often found in homogeneous solutions, with functions, denoted as \( u_1(t) \) and \( u_2(t) \). These functions are calculated by integrating expressions involving the non-homogeneous terms and the Wronskian, a determinant representing the solutions' matrix system.

In our differential equation example, the Wronskian was computed from the real parts of the solution expressions. Integrations involving the Wronskian helped determine \( u_1(t) \) and \( u_2(t) \), thus forming a particular solution. This methodology allows us to tailor the solution to the non-homogeneous nature of the problem, ensuring it fits all components of the original equation.