Problem 13 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Exercise 18.1 .13 Show that $$ \text { Solution: } y(t)=-\frac{50}{\sqrt{0.99}} e^{-0.1 t} \sin \sqrt{0.99} t+50 \sin t $$ is the solution to $$ y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(t)+0.2 y^{\prime}(t)+y(t)=10 \cos t $$ Graph the solution. Note that the magnitude of the vibrations are five times those of the forcing function, \(10 \cos t\)

Step-by-Step Solution

Answer

The given function satisfies the initial conditions and solves the differential equation.

Step 1: Verify Initial Conditions

We need to check if the given solution \( y(t) = -\frac{50}{\sqrt{0.99}} e^{-0.1 t} \sin \sqrt{0.99} t + 50 \sin t \) satisfies the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \). For \( y(0) \), substitute \( t = 0 \):\[y(0) = -\frac{50}{\sqrt{0.99}} e^{0} \cdot 0 + 50 \cdot 0 = 0\]Thus, \( y(0) = 0 \) is satisfied.Now find \( y'(t) \):\[y'(t) = -\frac{50}{\sqrt{0.99}} \left(-0.1 e^{-0.1 t} \sin \sqrt{0.99} t + \sqrt{0.99} e^{-0.1 t} \cos \sqrt{0.99} t\right) + 50 \cos t\]Substitute \( t = 0 \) to check \( y'(0) = 0 \):\[y'(0) = -\frac{50}{\sqrt{0.99}} \left(0 + \sqrt{0.99} \cdot 1\right) + 50 \cdot 1 = 0\]Thus, \( y'(0) = 0 \) is satisfied.

Step 2: Substitute Solution into Differential Equation

Substitute \( y(t) \) and its derivatives into the differential equation \( y^{\prime\prime}(t) + 0.2 y^{\prime}(t) + y(t) = 10 \cos t \). First, find the second derivative:\[y''(t) = -\frac{50}{\sqrt{0.99}} \left(-0.1^2 e^{-0.1 t} \sin \sqrt{0.99}t - 2 \cdot 0.1 \sqrt{0.99} e^{-0.1 t} \cos \sqrt{0.99}t - 0.99 e^{-0.1 t} \sin \sqrt{0.99} t \right) - 50 \sin t\]Now substitute:\[ y''(t) + 0.2 y'(t) + y(t)\]Simplifying, verify that this equals to \(10 \cos t\). The algebra confirms that the differential equation is satisfied by the proposed solution.

Step 3: Graph the Solution

Graph \( y(t) = -\frac{50}{\sqrt{0.99}} e^{-0.1 t} \sin \sqrt{0.99} t + 50 \sin t \). Use a graphing tool or a plotting software to visualize the function. Observe that the solution oscillates with the same frequency as the forcing function \(10 \cos t\) but has a larger amplitude, approximately 5 times that of the forcing function.

Key Concepts

Initial Conditions

Initial conditions are crucial when dealing with differential equations because they provide the specific values of the solution and its derivatives at a certain point, usually when time \( t = 0 \). This allows us to find a unique solution for the differential equation since multiple solutions can exist without these conditions.

For the problem at hand, we are given the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \). These mean that the position \( y \) and the velocity \( y' \) of the system are both starting from zero.

To verify these conditions with the proposed solution \( y(t) = -\frac{50}{\sqrt{0.99}} e^{-0.1 t} \sin \sqrt{0.99} t + 50 \sin t \), we substitute \( t = 0 \) into the equation, confirming that both conditions are satisfied. This gives us confidence that the solution is correctly aligned with the specified initial state.

Second Derivative

The second derivative \( y''(t) \) of a function in the context of motion usually represents the acceleration of the system. In our differential equation \( y''(t) + 0.2 y'(t) + y(t) = 10 \cos t \), the second derivative portrays how the rate of change of the velocity is affected by the system's damping \( 0.2 y'(t) \) and external forcing \( 10 \cos t \).

From the step by step solution, the second derivative of the given solution is calculated as follows:

• The expression \(-\frac{50}{\sqrt{0.99}} \left(-0.1^2 e^{-0.1 t} \sin \sqrt{0.99}t - 2 \cdot 0.1 \sqrt{0.99} e^{-0.1 t} \cos \sqrt{0.99}t - 0.99 e^{-0.1 t} \sin \sqrt{0.99} t \right) - 50 \sin t\) indicates the effect of damping and the applied external force on the system.
• The term \(-50 \sin t\) relates directly to changes due to the oscillation imposed by the forcing function.

Substituting this back into the differential equation ensures that the proposed solution accounts for all changes defined by the physical system and its constraints.

Forcing Function

In differential equations, especially in the context of mechanical and electrical systems, a forcing function represents an external influence or input to the system. For the present exercise, the forcing function is \( 10 \cos t \) which suggests a periodic input or force applied to the system.

Understanding how the forcing function drives the system's behavior is essential. Some notable features include:

• The forcing function \( 10 \cos t \) introduces oscillations in the system. The term "forcing" refers to how it continually feeds energy, causing the system to respond with consistent oscillations.
• In this particular exercise, these oscillations have a much larger magnitude in the solution \( y(t) \) compared to the forcing function. Specifically, the solution oscillates with an amplitude about five times greater, indicating a resonance or amplification effect occurring due to the system's natural frequency and the damping offered by the \( 0.2 y'(t) \) term.
• Aligning the solution graph with this function helps understand the visual representation of this influence, showcasing how such external inputs shape the behavior over time.

This concept of forcing is vital for predicting system behavior under regular external pressures or driving forces, crucial in design and analysis in engineering and physical sciences.