Question

Transform the equation $$ y^{\prime \prime}+4 y^{\prime}+40 y=0 $$ into a system of equations and solve. Apply the initial conditions \(y(0)=1, y^{\prime}(0)=0\)

Short answer

Question: Determine the specific solution to the given second-order linear differential equation with the specified initial conditions: \(y''+4y'+40y=0\), \(y(0)=1\), and \(y'(0)=0\). Answer: The specific solution to the given differential equation with the specified initial conditions is: \(y(t)=e^{-2t}\cos(6t)+\frac{1}{3}e^{-2t}\sin(6t)\).

Step-by-step solution

Transform the given equation into a system of first-order ODEs

To transform the given equation into a system of first-order ODEs, let \(u_1 = y\) and \(u_2 = y'\). Therefore, we can write the equation as follows: $$u_2'=y''=-4y'-40y=-4u_2 - 40u_1$$ Now, we have a system of linear first-order ODEs: $$ \begin{cases} u_1'=u_2 \\ u_2'=-4u_2-40u_1 \end{cases} $$

Write down the transformation matrix

We can rewrite the system of equations in matrix form as: $$\begin{bmatrix} u_1' \\ u_2' \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -40 & -4 \end{bmatrix}\begin{bmatrix} u_1\\ u_2\end{bmatrix}$$ So, the transformation matrix is: $$A=\begin{bmatrix}0 & 1 \\ -40 & -4 \end{bmatrix}$$

Write down the initial conditions

The initial conditions are given by: \(y(0)=1\) and \(y'(0)=0\). Since we have defined \(u_1=y\) and \(u_2=y'\), our initial conditions can be rewritten as \(u_1(0)=1\) and \(u_2(0)=0\).

Find the eigenvalues and eigenvectors of the matrix A

To find the eigenvalues, we need to solve the characteristic equation: $$|A-\lambda I|=0$$ Plugging in the values from matrix A, we get: $$\begin{vmatrix} -\lambda & 1 \\ -40 & -4 - \lambda \end{vmatrix}=0$$ Calculating the determinant, we have: $$\lambda^2+4\lambda+40=0$$ Using the quadratic formula to solve for \(\lambda\): $$\lambda=(-4\pm\sqrt{4^2-4(1)(40)})/(2)= -2\pm 6i$$ So, the eigenvalues are \(\lambda_1=-2+6i\) and \(\lambda_2=-2-6i\). Now, we will find the corresponding eigenvectors. For \(\lambda_1=-2+6i\), the equation is: $$(A-\lambda_1I)X_1=0$$ which gives us: $$\begin{bmatrix} 2-6i & 1 \\ -40 & -2-6i \end{bmatrix}\begin{bmatrix} x_{11} \\ x_{12} \end{bmatrix}=0$$ By solving the augmented matrix, we find that the eigenvector for \(\lambda_1\) is: $$X_1=\begin{bmatrix} 1 \\ 2-6i \end{bmatrix}$$ Similarly, for \(\lambda_2=-2-6i\), the equation is: $$(A-\lambda_2I)X_2=0$$ which gives us: $$\begin{bmatrix} 2+6i & 1 \\ -40 & -2+6i \end{bmatrix}\begin{bmatrix} x_{21} \\ x_{22} \end{bmatrix}=0$$ By solving the augmented matrix, we find that the eigenvector for \(\lambda_2\) is: $$X_2=\begin{bmatrix} 1 \\ 2+6i \end{bmatrix}$$

Formulate the general solution

Using the eigenvalues and eigenvectors, we can now formulate the general solution for the system of equations as: $$U(t)=c_1e^{(-2+6i)t}\begin{bmatrix} 1 \\ 2-6i \end{bmatrix}+c_2e^{(-2-6i)t}\begin{bmatrix} 1 \\ 2+6i \end{bmatrix}$$ Since we are only interested in the real part \(u_1\), we can write the general solution for \(y(t)\) as: $$y(t)=c_1e^{-2t}\cos(6t)+c_2e^{-2t}\sin(6t)$$

Apply the initial conditions

To find the specific solution, apply the initial conditions: \(y(0)=1\) and \(y'(0)=0\). Plugging them into the general solution, we get: $$y(0) = c_1\underline{e^{0}\cos(0)} + c_2\underline{e^{0}\sin(0)}$$ Hence, \(c_1 = 1\). To find \(c_2\), we take the derivative of \(y(t)\) with respect to \(t\): $$y'(t)=-2c_1e^{-2t}\cos(6t)-6c_1e^{-2t}\sin(6t)+2c_2e^{-2t}\sin(6t)-6c_2e^{-2t}\cos(6t)$$ And then apply the initial condition \(y'(0)=0\): $$y'(0) = -2\underline{c_1e^{0}\cos(0)}-6\underline{c_1e^{0}\sin(0)}+2\underline{c_2e^{0}\sin(0)}\underline{-6c_2e^{0}\cos(0)}$$ Which gives \(c_2 = \frac{1}{3}\).

Write the specific solution

Substituting the values of \(c_1\) and \(c_2\) in the general solution, we get the specific solution: $$y(t)=e^{-2t}\cos(6t)+\frac{1}{3}e^{-2t}\sin(6t)$$ This is the specific solution to the given differential equation with the specified initial conditions.

Key concepts

System of First-Order ODEs

Understanding the structure of a system of first-order ordinary differential equations (ODEs) is pivotal when dealing with complex equations involving higher-order derivatives. Such systems can be visualized as multiple equations, each representing a change in a specific variable with respect to another (often time). In the context of our exercise, the transformation of a second-order ODE into a system of first-order ODEs is achieved by introducing a set of substitution variables, like \(u_1 = y\) and \(u_2 = y'\).
This systematisation process is not just a mathematical trick; it lays the foundation for applying powerful linear algebra tools to solve the equations. When the system is written in matrix form, it elegantly encapsulates the relationships between the different variables and their rates of change, allowing for a more systematic approach to finding the solution. This matrix form leverages the ability to use computational methods to find solutions, which is especially beneficial for more complex systems that are otherwise difficult to solve analytically. When initially learning about this process, it’s helpful to:

• Clearly define each substitution variable.
• Understand how each equation in the system corresponds to a derivative of the original variable.
• Recognize the structure of the transformation matrix and its role in the system.

By mastering this initial transformation, students can pave the way towards a deeper comprehension of how various physical systems evolve over time, since many such systems are inherently described by first-order ODEs.

Eigenvalues and Eigenvectors

When dealing with systems of linear first-order ODEs written in matrix form, the concepts of eigenvalues and eigenvectors are not mere abstractions but practical tools offering valuable insights into the system's behavior. An eigenvalue represents the factor by which an eigenvector is scaled when a linear transformation is applied. To find these, one must delve into the characteristic equation of the matrix representing the system.
For students, especially those in disciplines such as engineering and physics, understanding these concepts is crucial. They describe important properties of linear transformations that are applicable in various scenarios, including stability analysis and natural frequency determination of mechanical systems. Key steps in grasping the concept include:

• Comprehending how to derive the characteristic equation from the matrix.
• Understanding the significance of real vs complex eigenvalues and their implications.
• Learning to calculate eigenvectors corresponding to each eigenvalue, as they point in the direction of the eigenvalues' effect.

In the given exercise, the eigenvalues turn out to be complex, indicating oscillatory motion when examining the dynamic system's solution. This insight is pivotal as it uncovers the natural modes in which the system prefers to behave.

Initial Value Problem

An initial value problem in the context of differential equations is one where the solution is required to satisfy certain conditions at a specific point, usually the start of the interval of interest. By providing these 'initial conditions,' one can pin down a specific solution from a family of potential solutions to the differential equation.
For students, this can be likened to having a snapshot at the beginning of a story and then following how the plot unfolds uniquely due to that specific starting point. Solving initial value problems requires integrating the initial conditions into the general solution:

• Identification and correct substitution of the initial conditions into the general solution of the differential equation.
• Deriving additional equations by differentiating the general solution (if required) to apply further initial conditions.
• Solving the resulting set of equations to find the specific constants that satisfy the initial conditions.

In our exercise, applying the initial condition pertains to a physical starting position and an initial velocity, guiding us to a unique solution that describes the motion of the system over time. By understanding and applying initial conditions, students can effectively navigate through complex differential equations to find meaningful and applicable solutions that directly correspond to physical phenomena.