Question

In Exercises \(11-20\) find a particular solution, given that \(Y\) is a fundamental matrix for the complementary system. $$ \mathbf{y}^{\prime}=\frac{1}{t}\left[\begin{array}{rrr} 1 & 1 & 0 \\ 0 & 2 & 1 \\ -2 & 2 & 2 \end{array}\right] \mathbf{y}+\left[\begin{array}{r} 1 \\ 2 \\ 1 \end{array}\right] \quad Y=\left[\begin{array}{rrr} t^{2} & t^{3} & 1 \\ t^{2} & 2 t^{3} & -1 \\ 0 & 2 t^{3} & 2 \end{array}\right] $$

Short answer

Answer: \(\mathbf{y_p}=\left[\begin{array}{r} -t^2\\2t^2-3\\2-2t\end{array}\right]\)

Step-by-step solution

Calculate the inverse of Y

To find the inverse of the fundamental matrix Y, compute \(Y^{-1}\), given by: $$ \mathbf{Y}^{-1} = \frac{1}{\operatorname{det}(\mathbf{Y})}\text{adj}(\mathbf{Y}), $$ where \(\operatorname{det}(\mathbf{Y})\) is the determinant of \(\mathbf{Y}\) and \(\text{adj}(\mathbf{Y})\) is the adjugate of \(\mathbf{Y}\). Compute the determinant of \(\mathbf{Y}\): $$ \operatorname{det}(\mathbf{Y})=-t^{3}(0-t^{2}2)+t^{3}2t^{3} $$ $$ \operatorname{det}(\mathbf{Y})=-2t^{5}+2t^{6}. $$ Now compute adjugate of \(\mathbf{Y}\): $$ \text{adj}(\mathbf{Y})=\left[\begin{array}{rrr} 8 t^{3} & -4 t^{3} & 2 t^{5} \\ -4 t^{3} & 4 t^{3} & 2 t^{6} - t^{5} \\ 4 t^{3} & -2 t^{3} & -2 t^{6} + 2 t^{5} \end{array}\right] $$ Now, compute the inverse: $$ \mathbf{Y}^{-1} = \frac{1}{-2t^{5}+2t^{6}}\left[\begin{array}{rrr} 8 t^{3} & -4 t^{3} & 2 t^{5} \\ -4 t^{3} & 4 t^{3} & 2 t^{6} - t^{5} \\ 4 t^{3} & -2 t^{3} & -2 t^{6} + 2 t^{5} \end{array}\right]. $$

Find the inhomogeneous term g(t)

We are given the inhomogeneous term as a column vector: $$ \mathbf{g}(t) = \left[\begin{array}{r}1\\2\\1\end{array}\right]. $$

Compute the product of Y inverse and g(t)

Now, compute the product of \(\mathbf{Y}^{-1}\) and \(\mathbf{g}(t)\): $$ \mathbf{Y}^{-1}\mathbf{g}(t)=\frac{1}{-2t^{5}+2t^{6}}\left[\begin{array}{rrr} 8 t^{3} & -4 t^{3} & 2 t^{5} \\ -4 t^{3} & 4 t^{3} & 2 t^{6} - t^{5} \\ 4 t^{3} & -2 t^{3} & -2 t^{6} + 2 t^{5} \end{array}\right]\left[\begin{array}{r}1\\2\\1\end{array}\right]. $$ After performing the matrix-vector multiplication, we get: $$ \mathbf{Y}^{-1}\mathbf{g}(t)=\frac{1}{-2t^{5}+2t^{6}}\left[\begin{array}{r}6t^{3}+2t^{5}\\-4t^{5}+8t^{3}+2t^{6}-t^{5}\\2t^{5}-2t^{6}-2t^{3}\end{array}\right]. $$

Integrate the result

Integrate the resulting vector to obtain the particular solution \(\mathbf{y_p}\): $$ \mathbf{y_p} = \int \mathbf{Y}^{-1}\mathbf{g}(t)\, dt = \int \frac{1}{-2t^{5}+2t^{6}}\left[\begin{array}{r}6t^{3}+2t^{5}\\-4t^{5}+8t^{3}+2t^{6}-t^{5}\\2t^{5}-2t^{6}-2t^{3}\end{array}\right] dt $$ The result is: $$ \mathbf{y_p}=\left[\begin{array}{r} -t^2\\2t^2-3\\2-2t\end{array}\right]. $$ So, we found the particular solution \(\mathbf{y_p}\) for the given inhomogeneous system of first-order linear differential equations.

Key concepts

Particular Solution

Finding a particular solution for a differential equation involves identifying a specific solution that satisfies the non-homogeneous part of the equation. In our example, we are dealing with a system of first-order linear differential equations. Each differential equation in this system describes a particular type of change. We already found the general solution for the associated homogeneous system using the fundamental matrix. Now, the focus is on finding a specific solution that also accounts for the presence of any external functions or forcing terms.

To do this, we utilized the inverse of the fundamental matrix, which allowed us to incorporate the inhomogeneous term given as a column vector. The resulting product gave us a vector function which we then integrated to obtain the specific particular solution. This solution is crucial because it details how the system behaves under the influence of external forces. In practice, the particular solution may represent actual quantities of interest in physical applications or engineering scenarios.

Fundamental Matrix

The fundamental matrix is a key concept in understanding systems of first-order linear differential equations. It serves as a building block for solutions to such systems. Essentially, a fundamental matrix is a matrix whose columns are linearly independent solutions to the homogeneous system of differential equations.

In the provided exercise, the fundamental matrix \(Y\) was given, and it played a pivotal role in finding the particular solution. This matrix is known as fundamental because it encapsulates the intrinsic properties of the system without influences from any external forces or inhomogeneous terms. By calculating the inverse of this matrix, we were able to resolve the contributions from the inhomogeneous component and find the unique particular solution.

Remember, the fundamental matrix allows us to express the general solution for the differential system. This is made possible by multiplying the fundamental matrix by a constant vector, which accounts for initial conditions or specific settings of the problem.

First-order Linear Differential Equations

First-order linear differential equations are equations where the highest derivative is the first derivative. These equations form the simplest type of linear differential equations you can encounter, yet they are foundational in understanding more complex systems.

In the context of the exercise, we dealt with a system of such equations. This means we had multiple first-order linear equations involving our function's derivatives, organized as a vector equation. Solving these systems requires understanding how these equations interact through their coefficients and the matrix representations.

One distinguishing feature of these systems is their linearity, which allows solutions to be added together to form new solutions. This property is vital because it underlies the solution process of using matrices and inverses, like the fundamental matrix we discussed earlier. Linear systems also tend to have well-established methods for finding solutions, making them practical and widely applicable for real-world scenarios.