Question

For each of the exercises, (a) Rewrite the equations from the given exercise in vector form as \(\mathbf{y}^{\prime}(t)=A \mathbf{y}(t)\), identifying the constant matrix \(A\). (b) Rewrite the solution of the equations in part (a) in vector form as \(\mathbf{y}(t)=c_{1} \mathbf{y}_{1}(t)+\) $$ 2 y^{\prime \prime}+t y+e^{3 t}=y^{\prime \prime \prime}+(\cos t) y^{\prime} $$

Short answer

Short Answer: Since the given differential equation contains non-constant terms (e.g., \(ty_1\), \(e^{3t}\)), it cannot be written in a completely linear form using a constant matrix A. Hence, the system cannot be solved using the solution form of \(\mathbf{y}(t)=c_{1} \mathbf{y}_{1}(t)+\). An alternative method is required to solve this type of problem.

Step-by-step solution

Rewrite the given equation as a system of first-order differential equations

First, we need to rewrite the given equation as a system of linear first-order differential equations. Let \(y_1 = y\), \(y_2 = y_1'\), and \(y_3 = y_2'\), so we get the following system: \begin{align*} y_1' &= y_2 \\ y_2' &= y_3 \\ y_3' &= y_2'' \\ \end{align*} Now, rewrite the given equation in terms of \(y_1\), \(y_2\), and \(y_3\): $$ 2(y_3)+t(y_1)+e^{3 t}=(y_3')+(\cos t)(y_2') \Rightarrow y_3' - \cos t y_2' = 2y_3 + ty_1 + e^{3t} - \cos t y_2 $$ Now we have the following system of first-order differential equations: \begin{align*} y_1' &= y_2 \\ y_2' &= y_3 \\ y_3' &= 2y_3 + ty_1 + e^{3t} - \cos t y_2 \\ \end{align*}

Rewrite in vector form and identify constant matrix \(A\)

Now, we wish to rewrite the system in the form \(\mathbf{y}^{\prime}(t)=A \mathbf{y}(t)\). However, since we have non-constant terms in the third equation (\(ty_1\) and \(e^{3t}\)), we can't fully write it in matrix form with a constant matrix \(A\). In this case, we can only rewrite the system in a matrix-vector form, ignoring the additional non-linear terms: \(\begin{pmatrix}y_1' \\ y_2' \\ y_3'\end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ t & -\cos t & 2 \end{pmatrix} \begin{pmatrix}y_1 \\ y_2 \\ y_3\end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ e^{3t} \end{pmatrix}\) In this case, there isn't a constant matrix \(A\) such that the system can be fully written in the form \(\mathbf{y}^{\prime}(t)=A \mathbf{y}(t)\).

Rewrite the solution in vector form

Since there is no constant matrix \(A\) that could fully represent the system within the desired form, it is not appropriate to find the general solution in the form \(\mathbf{y}(t)=c_{1} \mathbf{y}_{1}(t)+\). An alternative solution approach is required for this problem which is beyond the scope of the given exercise.

Key concepts

Vector Form

When dealing with differential equations, rewriting them in vector form is a common practice. It allows us to handle systems of equations more systematically. In this context, converting a higher-order differential equation into a system of first-order equations is essential.

To illustrate, let's consider the original differential equation: \(2y'' + ty + e^{3t} = y''' + (\cos t) y''\). This is a third-order differential equation, which can be broken down into a system of first-order equations by introducing new variables:\

• Let \(y_1 = y\)
• \(y_2 = y_1'\)
• and \(y_3 = y_2'\)

Now, the system becomes:
\(y_1' = y_2\)
\(y_2' = y_3\)
The final equation results in:
\(y_3' = 2y_3 + ty_1 + e^{3t} - \cos ty_2\) which includes nonconstant terms. This makes us realize that the last part can't entirely map into the simplest vector form \(\mathbf{y}'(t) = A \mathbf{y}(t)\) with a constant \(A\), as we have noted.

First-order System

A first-order system is essentially a set of first-order equations that describe how several variables change with respect to another parameter, typically time. Converting higher-order differential equations into this form helps in simplifying the problem and using systematic methods to find solutions.

For our differential equation, we can rewrite it as a system of first-order equations using our chosen variables \(y_1, y_2,\) and \(y_3\). This involves introducing relations for each new variable's derivative, as we did earlier. By defining these first-order equations and variables, a complex differential equation becomes a simpler structure to work with.

Once transformed into a first-order system, it opens up various numerical methods for solutions and analysis, which are often more manageable compared to tackling the original form.

Matrix Equation

Matrix equations offer a structured way to write systems of equations. By framing our system in a matrix-vector form, it becomes clear and often easier to manage.

For instance, our system of equations can be written as \(\mathbf{y}'(t) = A\mathbf{y}(t) + \mathbf{g}(t)\). Here, \(A\) is a coefficient matrix and \(\mathbf{g}(t)\) represents any additional non-linear or non-constant terms. In our specific case, it becomes:
\[\begin{pmatrix} y_1' \ y_2' \ y_3' \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ t & -\cos t & 2 \end{pmatrix} \begin{pmatrix} y_1 \ y_2 \ y_3 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ e^{3t} \end{pmatrix} \]

This setup makes it evident where non-constant elements exist. It's also clearer to see where deviations from the simplest system form occur, such as \(\mathbf{g}(t) = \begin{pmatrix} 0 \ 0 \ e^{3t} \end{pmatrix}\). This complexity is due to the non-constant coefficients in the original equation.

Non-constant Coefficients

Non-constant coefficients in a differential equation are terms where the coefficients of the variables are functions of the independent variable, often the time \(t\). Unlike constant coefficients, these add a layer of complexity to the problem.

In our example, the equation \(2y'' + ty + e^{3t} = y''' + (\cos t) y''\) includes \(t\) and \(e^{3t}\), as well as \(\cos t\), indicating non-constant coefficients. These terms mean that the system's response changes as time progresses, making them trickier to handle with traditional methods.

Non-constant coefficients prevent the equation from being expressed purely in terms of a constant matrix \(A\) and force us to include additional terms, like \(\mathbf{g}(t)\). This section of complexity means that analytical solutions may be harder to find, and we might rely more on numerical methods to explore the system's behavior.