Question

In Exercises \(11-20\) find a particular solution, given that \(Y\) is a fundamental matrix for the complementary system. $$ \mathbf{y}^{\prime}=\frac{1}{t}\left[\begin{array}{rr} 1 & t \\ -t & 1 \end{array}\right] \mathbf{y}+t\left[\begin{array}{c} \cos t \\ \sin t \end{array}\right] ; \quad Y=t\left[\begin{array}{rr} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right] $$

Short answer

Answer: The particular solution is: $$\mathbf{y_p}(t) = \left[\begin{array}{c}\cos t\left(-\frac{1}{t} + C_1\right) + \sin t C_2 \\ -\sin t\left(-\frac{1}{t} + C_1\right) + \cos t C_2\end{array}\right]$$

Step-by-step solution

Compute \(\mathbf{w}(t)\)

First, we need to find the inverse of the fundamental matrix \(Y\). The given fundamental matrix is: $$ Y = t\left[\begin{array}{rr} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right] $$ The determinant of \(Y\) is \(t^2\left(\cos^2t + \sin^2t\right) = t^2\). Therefore, the inverse of \(Y\) is: $$ Y^{-1} = \frac{1}{t^2}\left[\begin{array}{rr} \cos t & -\sin t \\ \sin t & \cos t \end{array}\right] $$ Now, we compute \(\mathbf{w}(t) = Y^{-1}\mathbf{g}(t)\): $$ \mathbf{w}(t) = \frac{1}{t^2}\left[\begin{array}{rr} \cos t & -\sin t \\ \sin t & \cos t \end{array}\right]\left[\begin{array}{c} \cos t \\ \sin t \end{array}\right] = \frac{1}{t^2}\left[\begin{array}{c} \cos^2t + \sin^2t \\ \cos t \sin t - \sin t \cos t \end{array}\right] = \left[\begin{array}{c} \frac{1}{t^2} \\ 0 \end{array}\right] $$

Integrate \(\mathbf{w}(t)\)

Next, we need to find the integral of \(\mathbf{w}(t)\) with respect to \(t\). This gives the vector function \(\mathbf{c}(t)\): $$ \mathbf{c}(t) = \int \mathbf{w}(t)dt = \int\left[\begin{array}{c} \frac{1}{t^2} \\ 0 \end{array}\right]dt = \left[\begin{array}{c} -\frac{1}{t} + C_1 \\ C_2 \end{array}\right] $$

Find the particular solution \(\mathbf{y_p}(t)\)

Finally, we can find the particular solution of the given differential equation by taking the product of the fundamental matrix \(Y\) and the vector \(\mathbf{c}(t)\): $$ \mathbf{y_p}(t) = Y\mathbf{c}(t) = t\left[\begin{array}{rr} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right]\left[\begin{array}{c} -\frac{1}{t} + C_1 \\ C_2 \end{array}\right] = \left[\begin{array}{c} \cos t\left(-\frac{1}{t} + C_1\right) + \sin t C_2 \\ -\sin t\left(-\frac{1}{t} + C_1\right) + \cos t C_2 \end{array}\right] $$ Thus, the particular solution of the given differential equation is: $$ \mathbf{y_p}(t) = \left[\begin{array}{c} \cos t\left(-\frac{1}{t} + C_1\right) + \sin t C_2 \\ -\sin t\left(-\frac{1}{t} + C_1\right) + \cos t C_2 \end{array}\right] $$

Key concepts

Fundamental Matrix

A fundamental matrix is a crucial concept in solving systems of linear differential equations. In the context of linear differential equations, the fundamental matrix provides a set of solutions that forms a basis for the solution space of the homogeneous system. This means that any other solution to the homogeneous system can be expressed as a linear combination of the solutions provided by this matrix.

In this exercise, the fundamental matrix \(Y\) is given as:

• \(Y = t\begin{bmatrix} \cos t & \sin t \ -\sin t & \cos t \end{bmatrix}\)

The matrix is multiplied by \(t\), which indicates it scales with the independent variable \(t\). Here, \(\cos t\) and \(\sin t\) represent the trigonometric functions that effectively capture the oscillatory nature of solutions typical in harmonic motion or wave-like systems.

Fundamental matrices are essential for understanding the behavior of the system as they also set the stage for finding a particular solution, which involves factoring in the inhomogeneous part of the differential equation.

Particular Solution

The particular solution of a linear differential equation is one approach to solving the non-homogeneous equation. The objective of finding a particular solution is to address the non-homogeneous part separately from the homogeneous solutions supplied by the fundamental matrix.

In this case, the process begins once we have determined the inverse of the fundamental matrix \(Y^{-1}\) and computed \(\mathbf{w}(t)\) using \(Y^{-1}\) and the inhomogeneous term \(\mathbf{g}(t) = \begin{bmatrix} \cos t \ \sin t \end{bmatrix}\). This was calculated to be \(\mathbf{w}(t) = \begin{bmatrix} \frac{1}{t^2} \ 0 \end{bmatrix}\).

After obtaining \(\mathbf{w}(t)\), integrate it with respect to \(t\) to get \(\mathbf{c}(t)\), giving:

• \(\mathbf{c}(t) = \int \mathbf{w}(t)dt = \begin{bmatrix} -\frac{1}{t} + C_1 \ C_2 \end{bmatrix}\)

Finally, the product \(Y\mathbf{c}(t)\) produces the particular solution \(\mathbf{y_p}(t)\):

• \(\mathbf{y_p}(t) = \begin{bmatrix} \cos t(-\frac{1}{t} + C_1) + \sin t C_2 \ -\sin t(-\frac{1}{t} + C_1) + \cos t C_2 \end{bmatrix}\)

Hence, solving both the homogeneous and non-homogeneous parts gives a comprehensive solution to the system.

Inverse Matrix

Finding the inverse of the fundamental matrix is a pivotal step in solving a system with non-homogeneous linear differential equations. The inverse of a matrix \(Y\), denoted \(Y^{-1}\), exists only if \(Y\) is a square matrix and its determinant is non-zero.

For our fundamental matrix:

• The determinant is \(t^2(\cos^2 t + \sin^2 t) = t^2\), which is indeed non-zero for \(t eq 0\).
• The inverse is calculated as follows:\[Y^{-1} = \frac{1}{t^2}\begin{bmatrix} \cos t & -\sin t \ \sin t & \cos t \end{bmatrix}\]

In linear algebra, the inverse matrix plays a crucial role in transforming and manipulating systems of equations. In differential equations, the inverse matrix helps relate the fundamental matrix and the vector function \(\mathbf{g}(t)\) to determine \(\mathbf{w}(t)\).

This ability to "reverse" the effects of a matrix operation makes inverses extremely valuable in finding particular solutions to differential equations.

Vector Integration

Vector integration involves integrating each component of a vector function independently. This is essential in solving differential equations where the integral of a vector function must be evaluated to obtain the particular solution.

In the given context:

• We compute the integral of \(\mathbf{w}(t) = \begin{bmatrix} \frac{1}{t^2} \ 0 \end{bmatrix}\) with respect to \(t\).

The integration yields:

• \(\mathbf{c}(t) = \int \mathbf{w}(t)\,dt = \begin{bmatrix} -\frac{1}{t} + C_1 \ C_2 \end{bmatrix}\)

Each component is integrated separately, emphasizing the nature of vector operations as component-wise calculations.

Understanding vector integration is not only crucial for solving differential equations but also enhances one's grasp of multivariable calculus, where such integrations become commonplace in applications across physics and engineering. By integrating \(\mathbf{w}(t)\) appropriately, we capture how the influence of the inhomogeneous part varies across time \(t\), allowing us to conclude with a correct representation of the system's particular solution.