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Chapter 4: Problem 9

Consider the initial value problem \(\mathbf{y}^{\prime}=A \mathbf{y}+\mathbf{g}(t)\),\(\mathbf{y}(0)=\mathbf{y}_{0}\)(a) Form the complementary solution. (b) Construct a particular solution by assuming the form suggested and solving for the undetermined constant vectors \(\mathbf{a}, \mathbf{b}\), and \(\mathbf{c}\). (c) Form the general solution. (d) Impose the initial condition to obtain the solution of the initial value problem.\(\mathbf{y}^{\prime}=\left[\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 0 \\\ 0 & 0 & 1\end{array}\right] \mathbf{y}+\left[\begin{array}{l}1 \\ t \\\ 0\end{array}\right], \quad \mathbf{y}_{0}=\left[\begin{array}{l}1 \\ 0 \\\ 2\end{array}\right] . \quad \operatorname{Try} \mathbf{y}_{P}(t)=t \mathbf{a}+\mathbf{b}\)

Short Answer

Expert verified
Based on the given matrix differential equation problem: Determine the specific solution of the initial value problem:$$ \mathbf{y}(t) = \begin{bmatrix} \frac{3}{2} \\ \frac{1}{2} \\ 2 \end{bmatrix}e^{t} + t\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}+\begin{bmatrix} t-\frac{1}{2} \\ -\frac{1}{2} \\ 0 \end{bmatrix}. $$ given the initial condition $\mathbf{y}(0) = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}$.

Step by step solution

01

(a) Form the complementary solution

To find the complementary solution, we will first solve the homogeneous equation: $$ \mathbf{y}'=A\mathbf{y}. $$ The given matrix A is:$$ A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. $$Diagonalizing the matrix, we notice that it already has the diagonal form:$$ D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, $$and eigenvalues equal to \(1\). Therefore, the complementary solution is of the form:$$ \mathbf{y}_C(t) = \mathbf{C}e^{t}, $$where \(\mathbf{C}\) is a constant vector.
02

(b) Determine the matrix 饾憥 and 饾憦 of the particular solution

We are given that the particular solution has the form:$$ \mathbf{y}_P(t) = t\mathbf{a}+\mathbf{b}. $$Now, differentiate \(\mathbf{y}_P(t)\) with respect to \(t\) to get:$$ \mathbf{y}_P'(t) = \mathbf{a}. $$Substitute this into the non-homogeneous equation:$$ \mathbf{a} = A(t\mathbf{a}+\mathbf{b})+\mathbf{g}(t), $$where \(\mathbf{g}(t) = \begin{bmatrix} 1 \\ t \\ 0 \end{bmatrix}\). Expanding the equation, we get:$$ \mathbf{a} = At\mathbf{a}+A\mathbf{b}+\mathbf{g}(t), $$so$$ (1-tA)\mathbf{a} = A\mathbf{b}+\mathbf{g}(t), $$which yields$$ A\mathbf{b}+\mathbf{g}(t) = \begin{bmatrix} t \\ 0 \\ 0 \end{bmatrix}. $$Now, solve for 饾憦:$$ \begin{align*} A\mathbf{b} &= \begin{bmatrix} t-1 \\ -t \\ 0 \end{bmatrix} \\ \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\mathbf{b} &= \begin{bmatrix} t-1 \\ -t \\ 0 \end{bmatrix}. \end{align*}Solving the system, we find that$$ \mathbf{b} = \begin{bmatrix} t-\frac{1}{2} \\ -\frac{1}{2} \\ 0 \end{bmatrix}, $$and substituting this into the initial equation for 饾憥, we get$$ \mathbf{a} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}. $$So the particular solution is:$$ \mathbf{y}_P(t) = t\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}+\begin{bmatrix} t-\frac{1}{2} \\ -\frac{1}{2} \\ 0 \end{bmatrix}.
03

(c) Form the general solution

The general solution is the sum of the complementary and particular solutions:$$ \mathbf{y}(t) = \mathbf{y}_C(t) + \mathbf{y}_P(t) = \mathbf{C}e^{t} + t\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}+\begin{bmatrix} t-\frac{1}{2} \\ -\frac{1}{2} \\ 0 \end{bmatrix}.
04

(d) Impose the initial condition

We are given the initial condition \(\mathbf{y}(0) = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}\). We will substitute this into the general solution:$$ \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} = \mathbf{C}e^{0} + 0\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}+\begin{bmatrix} -\frac{1}{2} \\ -\frac{1}{2} \\ 0 \end{bmatrix}. $$Thus,$$ \mathbf{C} = \begin{bmatrix} \frac{3}{2} \\ \frac{1}{2} \\ 2 \end{bmatrix}. $$The solution of the initial value problem is$$ \mathbf{y}(t) = \begin{bmatrix} \frac{3}{2} \\ \frac{1}{2} \\ 2 \end{bmatrix}e^{t} + t\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}+\begin{bmatrix} t-\frac{1}{2} \\ -\frac{1}{2} \\ 0 \end{bmatrix}.

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Key Concepts

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

Complementary Solution
When tackling differential equations, one common approach is to separate the solution into two distinct parts: the complementary solution (\textbf{y}_C) and the particular solution (\textbf{y}_P). The complementary solution addresses the associated homogeneous equation, where the equation has zero on one side (i.e., without the non-homogeneous part).

In our case, we have the homogeneous differential equation \[ \mathbf{y}'=A\mathbf{y} \.\] The matrix A in the given problem is already diagonal, which simplifies the process significantly. We identify the eigenvalues, all of which are 1, and then construct the complementary solution using these eigenvalues. Since the eigenvalues are equal to 1, the complementary solution involves an exponential function with the eigenvalues as the exponent of e, the base of the natural logarithm. Mathematically, it is expressible as \[ \mathbf{y}_C(t) = \mathbf{C}e^{t}, \.\] where \(\mathbf{C}\) is a constant vector that must later be determined from initial conditions.
Particular Solution
The particular solution seeks to account for the non-homogeneous part of the differential equation, which is everything not considered in the complementary solution. It鈥檚 crafted to specifically fit the non-homogeneous component, in this case, \(\mathbf{g}(t)\), which is a function of t.

For our equation, a suggested form for the particular solution was \(\mathbf{y}_P(t)=t \mathbf{a}+\mathbf{b}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors that need to be found. After differentiating \(\mathbf{y}_P(t)\) to find \(\mathbf{y}_P'(t)\), and substituting it into the original differential equation, we compare coefficients to find the values of \(\mathbf{a}\) and \(\mathbf{b}\). Through a series of calculations, we are able to express these vectors in terms of t which leads us to a concrete particular solution that complements the complementary solution.
General Solution
In differential equations, the general solution (\textbf{y}(t)) is a combination of both the complementary and particular solutions. It captures the behavior of the equation across all possible scenarios and includes constants that are later specified by initial conditions.

Following the individual determination of \(\mathbf{y}_C(t)\) and \(\mathbf{y}_P(t)\), the general solution is simply their sum: \[ \mathbf{y}(t) = \mathbf{y}_C(t) + \mathbf{y}_P(t) \.\] Through this process, we incorporate the effects of both the homogeneous and non-homogeneous components of the differential equation. The next step is to apply the given initial conditions, which in our exercise is \(\mathbf{y}(0) = \mathbf{y}_0\), to determine the exact values of the constants present in the general solution. This results in a fully specified solution that satisfies both the differential equation and the initial conditions.

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Most popular questions from this chapter

Exercises 1-5: For the given matrix functions \(A(t), B(t)\), and \(\mathbf{c}(t)\), make the indicated calculations $$ A(t)=\left[\begin{array}{cc} t-1 & t^{2} \\ 2 & 2 t+1 \end{array}\right], \quad B(t)=\left[\begin{array}{cc} t & -1 \\ 0 & t+2 \end{array}\right], \quad \mathbf{c}(t)=\left[\begin{array}{c} t+1 \\ -1 \end{array}\right] $$ $$ A(t) \mathbf{c}(t) $$

In each exercise, (a) As in Example 3, rewrite the given scalar initial value problem as an equivalent initial value problem for a first order system. (b) Write the Euler's method algorithm, \(\mathbf{y}_{k+1}=\mathbf{y}_{k}+h\left[P\left(t_{k}\right) \mathbf{y}_{k}+\mathbf{g}\left(t_{k}\right)\right]\), in explicit form for the given problem. Specify the starting values \(t_{0}\) and \(\mathbf{y}_{0}\). (c) Using a calculator and a uniform step size of \(h=0.01\), carry out two steps of Euler's method, finding \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\). What are the corresponding numerical approximations to the solution \(y(t)\) at times \(t=0.01\) and \(t=0.02\) ?\(y^{\prime \prime}+y=t^{3 / 2}, \quad y(0)=1, \quad y^{\prime}(0)=0\)

We know that similar matrices have the same eigenvalues (in fact, they have the same characteristic polynomial). There are many examples that show the converse is not true; that is, there are examples of matrices \(A\) and \(B\) that have the same characteristic polynomial but are not similar. Show that the following matrices \(A\) and \(B\) cannot be similar: $$ A=\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $$

Each of the systems of linear differential equations can be expressed in the form \(\mathbf{y}^{\prime}=P(t) \mathbf{y}+\mathbf{g}(t) .\) Determine \(P(t)\) and \(\mathbf{g}(t)\) $$ A^{\prime \prime}(t)=\left[\begin{array}{cc} 2 t & \sin t \\ 0 & 0 \end{array}\right], \quad A(0)=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right], \quad A^{\prime}(0)=\left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right] $$

In each exercise, assume that a numerical solution is desired on the interval \(t_{0} \leq t \leq t_{0}+T\), using a uniform step size \(h\). (a) As in equation (8), write the Euler's method algorithm in explicit form for the given initial value problem. Specify the starting values \(t_{0}\) and \(\mathbf{y}_{0}\). (b) Give a formula for the \(k\) th \(t\)-value, \(t_{k}\). What is the range of the index \(k\) if we choose \(h=0.01\) ? (c) Use a calculator to carry out two steps of Euler's method, finding \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\). Use a step size of \(h=0.01\) for the given initial value problem. Hand calculations such as these are used to check the coding of a numerical algorithm.\(\mathbf{y}^{\prime}=\left[\begin{array}{cc}-t^{2} & t \\ 2-t & 0\end{array}\right] \mathbf{y}+\left[\begin{array}{l}1 \\\ t\end{array}\right], \quad \mathbf{y}(1)=\left[\begin{array}{l}2 \\\ 0\end{array}\right], \quad 1 \leq t \leq 4\)
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