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Chapter 4: Problem 11

Consider the initial value problem $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right] \mathbf{y}+\mathbf{g}(t), \quad \mathbf{y}\left(\frac{\pi}{2}\right)=\mathbf{y}_{0} $$ Suppose we know that $$ \mathbf{y}(t)=\left[\begin{array}{c} 1+\sin 2 t \\ e^{t}+\cos 2 t \end{array}\right] $$ is the unique solution. Determine \(\mathbf{g}(t)\) and \(\mathbf{y}_{0}\).

Short Answer

Expert verified
The values are \(\mathbf{y}_{0} = \left[\begin{array}{c} 1 \\ e^{\frac{\pi}{2}}-1 \end{array}\right]\) and \(\mathbf{g}(t) = \left[\begin{array}{c} -2e^{t}+4\cos 2t \\ e^{t}+4\sin 2t -2 \end{array}\right]\).

Step by step solution

01

Find \(\mathbf{y}_{0}\)

To find \(\mathbf{y}_{0}\), we will plug the initial condition \(t = \frac{\pi}{2}\) into the known solution for \(\mathbf{y}(t)\): $$\mathbf{y}\left( \frac{\pi}{2} \right)=\left[\begin{array}{c} 1+\sin(2(\frac{\pi}{2})) \\ e^{\frac{\pi}{2}}+\cos(2(\frac{\pi}{2})) \end{array}\right]=\left[\begin{array}{c} 1+\sin(\pi) \\ e^{\frac{\pi}{2}}+\cos(\pi) \end{array}\right]=\left[\begin{array}{c} 1 \\ e^{\frac{\pi}{2}}-1 \end{array}\right]$$ So, \(\mathbf{y}_{0} = \left[\begin{array}{c} 1 \\ e^{\frac{\pi}{2}}-1 \end{array}\right]\).
02

Find \(\mathbf{y'}(t)\)

To find the derivative of \(\mathbf{y}(t)\), we can apply the basic rules of differentiation for the matrix elements: $$\mathbf{y'}(t)=\left[\begin{array}{c} \frac{d}{dt}(1+\sin 2t) \\ \frac{d}{dt}(e^{t}+\cos 2t) \end{array}\right]=\left[\begin{array}{c} 2\cos 2t \\ e^{t}-2\sin 2t \end{array}\right]$$
03

Determine \(\mathbf{g}(t)\)

Now we can use the given equation for \(\mathbf{y'}(t)\) and substitute the known solution for \(\mathbf{y}(t)\) and the calculated value of \(\mathbf{y'}(t)\) to find \(\mathbf{g}(t)\): $$\mathbf{y'}(t) = \left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right] \mathbf{y}(t) + \mathbf{g}(t)$$ Substitute \(\mathbf{y}(t)\) and \(\mathbf{y'}(t)\) into the equation: $$\left[\begin{array}{c} 2\cos 2t \\ e^{t}-2\sin 2t \end{array}\right] = \left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right] \left[\begin{array}{c} 1+\sin 2t \\ e^{t}+\cos 2t \end{array}\right] + \mathbf{g}(t)$$ Compute the matrix multiplication: $$\left[\begin{array}{c} 2\cos 2t \\ e^{t}-2\sin 2t \end{array}\right] = \left[\begin{array}{c} 2(e^{t}+\cos 2t) \\ -2(1+\sin 2t) \end{array}\right] + \mathbf{g}(t)$$ Therefore, we can find \(\mathbf{g}(t)\) as: $$\mathbf{g}(t)=\left[\begin{array}{c} 2\cos 2t \\ e^{t}-2\sin 2t \end{array}\right] - \left[\begin{array}{c} 2(e^{t}+\cos 2t) \\ -2(1+\sin 2t) \end{array}\right] = \left[\begin{array}{c} -2e^{t}+4\cos 2t \\ e^{t}+4\sin 2t -2 \end{array}\right]$$ In conclusion, we have found that \(\mathbf{y}_{0} = \left[\begin{array}{c} 1 \\ e^{\frac{\pi}{2}}-1 \end{array}\right]\) and \(\mathbf{g}(t) = \left[\begin{array}{c} -2e^{t}+4\cos 2t \\ e^{t}+4\sin 2t -2 \end{array}\right]\).

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Key Concepts

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

Initial Value Problems
Initial value problems (IVPs) are an essential concept in differential equations, describing a scenario where you determine a function that satisfies a differential equation and an initial condition. This means you're not only looking for a function that meets the equation, but also a specific solution that begins at a particular point in time.For the given initial value problem: - We have a differential equation of the form \( \mathbf{y}^{\prime}= \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix} \mathbf{y} + \mathbf{g}(t) \), where \( \mathbf{y}(\frac{\pi}{2}) = \mathbf{y}_{0} \).- The task involves finding the solution to the matrix differential equation which satisfies both this equation and the initial condition. By solving the IVP, using the provided expression for \( \mathbf{y}(t) \), we determine that at the initial condition \( t = \frac{\pi}{2} \), \( \mathbf{y}(t) \) takes the form that satisfies the given condition. This allows us to find our initial condition vector \( \mathbf{y}_{0} \). It is a crucial step in understanding how initial values influence the solution to differential equations.
Matrix Operations
Matrix operations are an integral part of solving differential equations in systems with multiple interdependent variables. In the problem, we use a constant coefficient matrix to represent the system's dynamic behavior. Here, the operations you need to understand include matrix multiplication and addition.- Matrix multiplication, for instance, is needed to describe the interaction of different components of the system. It involves performing row-by-column operations across matrices. For example: \[ \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix} \begin{bmatrix} 1 + \sin(2t) \ e^{t} + \cos(2t) \end{bmatrix} = \begin{bmatrix} 2(e^{t} + \cos 2t) \ -2(1 + \sin 2t) \end{bmatrix} \].- Understanding these operations is crucial, and they are conducted to handle the interactions in a multi-variable differential system.Grasping these matrix operations allows for a correct computation of results, forming the foundation to proceed with solving for \( \mathbf{g}(t) \). Knowing matrix operations simplifies the step where we substitute into the original equation.
Solution Verification
Solution verification is the step where you ensure that your calculated solution is both correct and satisfies all parts of the problem. In other words, it's about seeking consistency between what you propose as a solution and what the problem initially presents. - In our example, we verify by plugging \( \mathbf{y}(t) \) into the differential equation and checking if it fulfills both the derivative calculated and the dynamic equation condition. - To verify, we must ensure that substituting back into the original differential equation, considering both the left hand (the derivative \( \mathbf{y}'(t) \)) and right hand side (the result of matrix operation plus \( \mathbf{g}(t) \)), aligns. Verification confirms that our derived \( \mathbf{y}_{0} \) and \( \mathbf{g}(t) \) are correct. This step eliminates errors by ticking off all conditions, showing the solution reliable.
Differentiation Rules
Differentiation rules help in calculating the rate of change in any given function. When dealing with matrices and systems as in this problem, applying these rules accurately becomes crucial.- For scalar functions like \(1 + \sin 2t\) and \(e^{t} + \cos 2t\), differentiation rules are applied element-wise: - The derivative of \( \sin 2t \) is \( 2 \cos 2t \). - The derivative of \( e^{t} \) is simply \( e^{t} \). - The derivative of \( \cos 2t \) becomes \(-2 \sin 2t \).- These principles transform understanding into the matrix counterpart as we built \( \mathbf{y}'(t) \): \[ \mathbf{y'}(t) = \begin{bmatrix} 2 \cos 2t \ e^{t} - 2 \sin 2t \end{bmatrix} \].Applying differentiation rules to matrix-valued functions accurately helps tackle complex dynamic systems and is necessary whenever you tackle problems involving differential equations. This accuracy is what allows us to align our solutions with mathematical expectations.

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Most popular questions from this chapter

The given matrix \(A\) is diagonalizable. (a) Find \(T\) and \(D\) such that \(T^{-1} A T=D\). (b) Using (12c), determine the exponential matrix \(e^{A t}\).\(A=\left[\begin{array}{rr}1 & 2 \\ -2 & 1\end{array}\right]\)

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 \(\mathbf{g}(t)\) $$ \begin{aligned} &y_{1}^{\prime}=t^{-1} y_{1}+\left(t^{2}+1\right) y_{2}+t \\ &y_{2}^{\prime}=4 y_{1}+t^{-1} y_{2}+8 t \ln t \end{aligned} $$

The exact solution of the initial value problem \(\mathbf{y}^{\prime}=\left[\begin{array}{cc}0.5 & 1 \\ 1 & 0.5\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}1 \\\ 0\end{array}\right] \quad\) is given by \(\quad \mathbf{y}(t)=\frac{1}{2}\left[\begin{array}{c}e^{-t / 2}+e^{3 t / 2} \\\ -e^{-t / 2}+e^{3 t / 2}\end{array}\right] .\) (a) Write a program that applies the Runge-Kutta method (12) to this problem. (b) Run your program on the interval \(0 \leq t \leq 1\), using step size \(h=0.01\). (c) Run your program on the interval \(0 \leq t \leq 1\), using step size \(h=0.005\). (d) Let \(\mathbf{y}_{100}\) and \(\mathbf{y}_{200}\) denote the numerical approximations to \(\mathbf{y}(1)\) computed in parts (b) and (c), respectively. Compute the error vectors \(\mathbf{y}(1)-\mathbf{y}_{100}\) and \(\mathbf{y}(1)-\overline{\mathbf{y}}_{200}\). By roughly what fractional amount is the error reduced when the step size is halved?

In each exercise, determine all equilibrium solutions (if any).\(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}1 & 1 & 0 \\ 0 & -1 & 2 \\\ 0 & 0 & 1\end{array}\right] \mathbf{y}+\left[\begin{array}{l}2 \\ 3 \\\ 2\end{array}\right]\)

In each exercise, find the general solution of the homogeneous linear system and then solve the given initial value problem. $$ \mathbf{y}^{r}=\left[\begin{array}{rr} -5 & -2 \\ 12 & 5 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(1)=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] $$
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