Question

Repeat Exercise 7 with $$ A=\left[\begin{array}{rr} -4 & -10 \\ 3 & 7 \end{array}\right], \quad \mathbf{y}_{1}=\left[\begin{array}{r} -5 e^{2 t} \\ 3 e^{2 t} \end{array}\right], \quad \mathbf{y}_{2}=\left[\begin{array}{c} 2 e^{t} \\ -e^{t} \end{array}\right], \quad \mathbf{k}=\left[\begin{array}{r} -19 \\ 11 \end{array}\right] $$

Short answer

Based on the given matrix \(A\), two solution vectors \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\), and a constant vector \(\mathbf{k}\), we verified that \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are solutions to the homogeneous system \(\mathbf{\dot y} = A\mathbf{y}\). We also found a particular solution \(\mathbf{y}_{p} = \begin{bmatrix} 3 \\ -2 \end{bmatrix}\) to the nonhomogeneous system \(\mathbf{\dot y} = A\mathbf{y} + \mathbf{k}\). The general solution of the nonhomogeneous system is given by: $$ \mathbf{y} = c_{1}\left[\begin{array}{r} -5 e^{2 t} \\ 3 e^{2 t} \end{array}\right] + c_{2}\left[\begin{array}{r} 2 e^{t} \\ -e^{t} \end{array}\right] + \left[\begin{array}{r} 3 \\ -2 \end{array}\right] $$ where \(c_{1}\) and \(c_{2}\) are constants that depend on the initial conditions.

Step-by-step solution

Check if \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are solutions to the homogeneous system

To check if \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are solutions to the homogeneous system, we need to compute the derivatives of these vectors and check if these derivatives are equal to the product \(A\mathbf{y}\). For \(\mathbf{y}_{1}\), the derivative is: $$ \mathbf{\dot y}_{1} = \frac{d}{dt} \left[\begin{array}{r} -5 e^{2 t} \\ 3 e^{2 t} \end{array}\right] = \left[\begin{array}{r} -10 e^{2 t} \\ 6 e^{2 t} \end{array}\right] $$ And \(A\mathbf{y}_{1}\) is: $$ A\mathbf{y}_{1} = \left[\begin{array}{rr} -4 & -10 \\ 3 & 7 \end{array}\right]\left[\begin{array}{r} -5 e^{2 t} \\ 3 e^{2 t} \end{array}\right] = \left[\begin{array}{r} -10 e^{2 t} \\ 6 e^{2 t} \end{array}\right] $$ Since \(\mathbf{\dot y}_{1} = A\mathbf{y}_{1}\), \(\mathbf{y}_{1}\) is a solution to the homogeneous system. For \(\mathbf{y}_{2}\), the derivative is: $$ \mathbf{\dot y}_{2} = \frac{d}{dt} \left[\begin{array}{r} 2 e^{t} \\ -e^{t} \end{array}\right] = \left[\begin{array}{r} 2 e^{t} \\ -e^{t} \end{array}\right] $$ And \(A\mathbf{y}_{2}\) is: $$ A\mathbf{y}_{2} = \left[\begin{array}{rr} -4 & -10 \\ 3 & 7 \end{array}\right]\left[\begin{array}{r} 2 e^{t} \\ -e^{t} \end{array}\right] = \left[\begin{array}{r} 2 e^{t} \\ -e^{t} \end{array}\right] $$ Since \(\mathbf{\dot y}_{2} = A\mathbf{y}_{2}\), \(\mathbf{y}_{2}\) is a solution to the homogeneous system.

Find a particular solution of the nonhomogeneous system

We will guess that a particular solution \(\mathbf{y}_{p}\) to the nonhomogeneous system is of the form \(A\mathbf{y}\). Now we have: $$ \mathbf{\dot y}_{p} = A\mathbf{y}_{p}+\mathbf{k} $$ Substituting our guess into the equation, we get: $$ A\mathbf{y}_{p} = A\mathbf{y}_{p}+\mathbf{k} $$ Subtracting \(A\mathbf{y}_{p}\) from both sides, we are left with: $$ \mathbf{k} = 0 $$ However, the given vector \(\mathbf{k}\) is non-zero, so our guess of a particular solution was incorrect. We need to try another method to find a particular solution.

Try a different method to find a particular solution

Let's assume a particular solution of the form \(\mathbf{y}_{p} = \left[\begin{array}{r}a \\ b\end{array}\right]\), where \(a\) and \(b\) are constants that we need to determine. From the nonhomogeneous system, we have \(\mathbf{\dot y}_{p} = A\mathbf{y}_{p}+\mathbf{k}\). Since \(\mathbf{y}_{p}\) is constant, its derivative is zero. We can rewrite the system as: $$ 0 = A\mathbf{y}_{p}+\mathbf{k} $$ Substituting \(\mathbf{y}_{p}\) into the equation, we get: $$ 0 = \left[\begin{array}{rr} -4 & -10 \\ 3 & 7 \end{array}\right]\left[\begin{array}{r} a \\ b \end{array}\right]+\left[\begin{array}{r} -19 \\ 11 \end{array}\right] $$ To solve for \(a\) and \(b\), we can set up a system of linear equations: $$ \begin{cases} -4a - 10b = 19 \\ 3a + 7b = -11 \end{cases} $$ After solving this system, we find that \(a=3\) and \(b=-2\). So, the particular solution is: $$ \mathbf{y}_{p} = \left[\begin{array}{r} 3 \\ -2 \end{array}\right] $$

Write the general solution of the nonhomogeneous system

Now that we have found the particular solution \(\mathbf{y}_{p}\) and verified that \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are solutions to the homogeneous system, we can write the general solution of the nonhomogeneous system as a sum of the homogeneous and particular solutions: $$ \mathbf{y} = c_{1}\mathbf{y}_{1} + c_{2}\mathbf{y}_{2} + \mathbf{y}_{p} $$ Here, \(c_{1}\) and \(c_{2}\) are constants that depend on the initial conditions. So, the general solution is: $$ \mathbf{y} = c_{1}\left[\begin{array}{r} -5 e^{2 t} \\ 3 e^{2 t} \end{array}\right] + c_{2}\left[\begin{array}{r} 2 e^{t} \\ -e^{t} \end{array}\right] + \left[\begin{array}{r} 3 \\ -2 \end{array}\right] $$

Key concepts

Homogeneous Systems

Homogeneous systems in the context of linear differential equations involve a set of equations in which each is equal to zero. Think of it as a scenario where there are no outside influences affecting the system—only the innate rules dictate the behavior. In simpler terms, in a homogeneous system, you're looking at a world with balance and symmetry.

To check if a solution fits a homogeneous system, we compute the derivative of each component of the vector solution and verify that it's equivalent to multiplying the matrix with the vector solution itself. In our example, both \( \mathbf{y}_1 \) and \( \mathbf{y}_2 \) turned out to be solutions when tested. This means they naturally satisfy the system's inherent rules without any external forces."

• The derivatives of solutions need to fit perfectly within the framework defined by the system matrix.
• This characteristic is what makes a system homogeneous—consistent behavior driven by intrinsic factors alone.

Nonhomogeneous Systems

Unlike homogeneous systems, nonhomogeneous systems integrate an additional vector that acts much like an external force shifting the natural dynamics. The system is described not only by its internal structure, represented by the matrix, but also by the influence of an extraneous term—the vector \( \mathbf{k} \).

When dealing with nonhomogeneous systems, the challenge lies in finding solutions that account for these added forces. Solutions to these systems won't simply be multiples of the natural solutions, as seen in homogeneous equations. Instead, they require a specific adjustment which is typically manifested in what's called a particular solution.

• Nonhomogeneous systems reflect real-world scenarios more often since additional "forces" or influences frequently act on systems, demanding solutions that account for such complexities.
• We need both a particular solution and the natural behavior embedded through homogeneous solutions to describe the system wholely.

Particular Solutions

A particular solution in the context of nonhomogeneous systems refers to a specific solution that accounts for the external terms affecting the system, which means satisfying not just the system matrix \( A \) but also the vector \( \mathbf{k} \).

Finding a particular solution often involves guessing a solution form and testing to see if it satisfies the system conditions. If a guess doesn't work, another method must be tried. In our given problem, by setting a constant solution and solving the resulting system of equations, we found \( \mathbf{y}_p= \begin{bmatrix} 3 \ -2 \end{bmatrix} \). This particular solution balances out the effects of the external vector, \( \mathbf{k} \).

• Guessing a form for the particular solution can sometimes be straightforward but often requires adjustments and alternative approaches if the initial guesses do not work.
• A correct particular solution will compensate for the effects introduced by \( \mathbf{k} \).

General Solutions

The general solution of a nonhomogeneous system incorporates both the natural and particular behaviors the system exhibits. It forms a complete picture by combining solutions to the homogeneous equation with the particular solution to account for any external influences.

In essence, the general solution is expressed as \( \mathbf{y} = c_1 \mathbf{y}_1 + c_2 \mathbf{y}_2 + \mathbf{y}_p \), where \( c_1 \) and \( c_2 \) are constants determined by initial conditions, \( \mathbf{y}_1 \) and \( \mathbf{y}_2 \) are homogeneous solutions, and \( \mathbf{y}_p \) is the particular solution. This expression means we are layering intrinsic system dynamics with additional solved influences to grasp what happens when both internal and external factors are in play.

• The general solution is necessary to encapsulate all potential scenarios and influences affecting the system.
• Understanding and solving for constants \( c_1 \) and \( c_2 \) requires specific starting conditions, enabling the solution to be tailored to the system's initial state.