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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

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

Step by step solution

01

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: $$ \mathbf{Y}^{-1} = \frac{1}{-2t^{5}+2t^{6}}\left[\begin{array}{rrr} 8 t^{3} & -4 t^{3} & 2 t^{5} \\ -4 t^{3} & 4 t^{3} & 2 t^{6} - t^{5} \\ 4 t^{3} & -2 t^{3} & -2 t^{6} + 2 t^{5} \end{array}\right]. $$
02

Find the inhomogeneous term g(t)

We are given the inhomogeneous term as a column vector: $$ \mathbf{g}(t) = \left[\begin{array}{r}1\\2\\1\end{array}\right]. $$
03

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: $$ \mathbf{Y}^{-1}\mathbf{g}(t)=\frac{1}{-2t^{5}+2t^{6}}\left[\begin{array}{r}6t^{3}+2t^{5}\\-4t^{5}+8t^{3}+2t^{6}-t^{5}\\2t^{5}-2t^{6}-2t^{3}\end{array}\right]. $$
04

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Find the general solution. \(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}-2 & 2 & -6 \\ 2 & 6 & 2 \\ -2 & -2 & 2\end{array}\right] \mathbf{y}\)

Describe and graph trajectories of the given system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 9 & -3 \\ -1 & 11 \end{array}\right] \mathbf{y} $$

Suppose the \(n \times n\) matrix function \(A\) and the \(n-\) vector function \(\mathbf{f}\) are continuous on \((a, b)\). Let \(t_{0}\) be in \((a, b),\) let \(\mathbf{k}\) be an arbitrary constant vector, and let \(Y\) be a fundamental matrix for the homogeneous system \(\mathbf{y}^{\prime}=A(t) \mathbf{y}\). Use variation of parameters to show that the solution of the initial value problem $$ \mathbf{y}^{\prime}=A(t) \mathbf{y}+\mathbf{f}(t), \quad \mathbf{y}\left(t_{0}\right)=\mathbf{k} $$ is $$ \mathbf{y}(t)=Y(t)\left(Y^{-1}\left(t_{0}\right) \mathbf{k}+\int_{t_{0}}^{t} Y^{-1}(s) \mathbf{f}(s) d s\right) . $$

Suppose the matrix $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ has a repeated eigenvalue \(\lambda_{1}\) and the associated eigenspace is one- dimensional. Let \(\mathbf{x}\) be a \(\lambda_{1}-\) eigenvector of \(A .\) Show that if \(\left(A-\lambda_{1} I\right) \mathbf{u}_{1}=\mathbf{x}\) and \(\left(A-\lambda_{1} I\right) \mathbf{u}_{2}=\mathbf{x},\) then \(\mathbf{u}_{2}-\mathbf{u}_{1}\) is parallel to \(\mathbf{x}\). Conclude from this that all vectors \(\mathbf{u}\) such that \(\left(A-\lambda_{1} I\right) \mathbf{u}=\mathbf{x}\) define the same positive and negative half-planes with respect to the line \(L\) through the origin parallel to \(\mathbf{X}\).

In Exercises \(33-40\) find vectors \(\mathbf{U}\) and \(\mathbf{V}\) parallel to the axes of symmetry of the shadow trajectories, and plot a typical trajectory. $$ \text { C/G } \mathbf{y}^{\prime}=\left[\begin{array}{rr} -1 & -5 \\ 20 & -1 \end{array}\right] \mathbf{y} $$

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