Problem 21 - Calculus: Early Transcendentals

Question

Find the solution for the mass-spring equation \(y^{\prime \prime}+4 y^{\prime}+29 y=689 \cos (2 t)\).

Step-by-Step Solution

Answer

The solution is \( y(t) = e^{-2t}(C_1 \cos(5t) + C_2 \sin(5t)) + 25 \cos(2t) - 8 \sin(2t) \).

Step 1: Identify the Type of Equation

The given equation is \( y^{\prime \prime} + 4y^{\prime} + 29y = 689 \cos(2t) \), which is a second-order linear non-homogeneous differential equation with constant coefficients.

Step 2: Write the Associated Homogeneous Equation

The homogeneous form of the given equation is \( y^{\prime \prime} + 4y^{\prime} + 29y = 0 \).

Step 3: Solve the Characteristic Equation

The characteristic equation for the homogeneous equation is \( r^2 + 4r + 29 = 0 \). Using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 4, \) and \( c = 29 \). The discriminant is \( 16 - 116 = -100 \). This implies complex roots.

Step 4: Find the Roots and Form the General Solution

The roots are \( r = -2 \pm i5 \). These complex roots are in the form \( \alpha \pm i\beta \). The general solution of the homogeneous equation is \( y_h = e^{-2t}(C_1 \cos(5t) + C_2 \sin(5t)) \).

Step 5: Formulate the Particular Solution for the Non-Homogeneous Part

Assume a particular solution of the form \( y_p = A\cos(2t) + B\sin(2t) \). Substitute \( y_p \) and its derivatives into the non-homogeneous equation to determine \( A \) and \( B \).

Step 6: Calculate Derivatives for the Particular Solution

The derivatives are \( y_p' = -2A \sin(2t) + 2B \cos(2t) \) and \( y_p'' = -4A \cos(2t) - 4B \sin(2t) \).

Step 7: Substitute and Solve for Coefficients

Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the original equation to find \( A \) and \( B \). Simplifying, we have: \( (-4A - 8B + 29A) \cos(2t) + (-4B + 8A + 29B) \sin(2t) = 689 \cos(2t) \).

Step 8: Equate Coefficients

Equating coefficients, we get two equations: \( 25A - 8B = 689 \) for \( \cos(2t) \) and \( 25B + 8A = 0 \) for \( \sin(2t) \).

Step 9: Solve the System of Equations

Solving these equations, we find \( A = 25 \) and \( B = -8 \).

Step 10: Write the General Solution of the Non-Homogeneous Equation

The complete solution is the sum of the homogeneous and particular solutions: \( y(t) = e^{-2t}(C_1 \cos(5t) + C_2 \sin(5t)) + 25 \cos(2t) - 8 \sin(2t) \).

Key Concepts

Second-Order Linear Differential Equation

A second-order linear differential equation involves the second derivative of a function, often representing systems undergoing forces or motions, like in a mass-spring system.
It is called "linear" because the terms depend linearly on the function and its derivatives.
In our example, the equation is expressed as:

• \( y'' + 4y' + 29y = 689 \cos(2t) \).
• The order is determined by the highest derivative, which here is the second derivative \( y'' \).

Such equations frequently come in the form \[ ay'' + by' + cy = g(t) \], where \( g(t) \) is any function of \( t \) (in our case, \( 689 \cos(2t) \)).
The linearity ensures that any function solutions can be superimposed, or added together, to find the general solution.
This specific case is also non-homogeneous due to the presence of a non-zero function \( g(t) \), leading to a slightly more complex solution process.

Characteristic Equation

The characteristic equation is key to solving the homogeneous part of a second-order linear differential equation.
For the mass-spring problem, we first focus on the homogeneous equation:

• \( y'' + 4y' + 29y = 0 \)

Here, the characteristic equation is derived by substituting the equation with \( y = e^{rt} \), which simplifies to:

• \( r^2 + 4r + 29 = 0 \)

This is a quadratic equation, and solving it gives the roots of the form \( r = -2 \pm i5 \). A quadratic solution often uses the quadratic formula:

• \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The roots reveal critical information about the system's behavior, especially when they are complex, indicating oscillating behavior in the solution due to the imaginary part of the roots.

Particular Solution

A particular solution addresses the non-homogeneous component of the equation.
To find it, we guess a solution form resembling the non-homogeneous term. In this case:

• \( y_{\text{p}} = A \cos(2t) + B \sin(2t) \)

We differentiate this guess and substitute back into the original differential equation to solve for the constants \( A \) and \( B \).
By substituting and equating coefficients for like terms (cosine and sine), we obtain a system of equations:

• \( 25A - 8B = 689 \)
• \( 25B + 8A = 0 \)

Solving these equations, we find \( A = 25 \) and \( B = -8 \). The particular solution thus models the specific response of the system to the non-homogeneous driving force.

Homogeneous Solution

The homogeneous solution captures the natural response of the system, in absence of external forces, based on the homogeneous differential equation.
By solving the characteristic equation, we obtain complex roots \( r = -2 \pm i5 \). These lead to a general form:

• \[ y_{\text{h}} = e^{-2t} (C_1 \cos(5t) + C_2 \sin(5t)) \]

The negative real part \(-2t\) indicates an exponential decay, and the imaginary component results in oscillations, modeled by cosine and sine functions.
This forms a damped oscillatory pattern, typical in mass-spring systems with friction or damping. This portion of the solution decays over time due to the exponential term, symbolizing energy loss in the physical system.
The constants \( C_1 \) and \( C_2 \) are determined by initial conditions, unique to each specific scenario in practice.