Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Show that the suggested solutions solve the associated equations. a. \(y=\cos t\) \(y^{\prime \prime}+y=0\) b. \(y=\cos 2 t\) \(y^{\prime \prime}+4 y=0\) c. \(\quad y=3 \sin t+2 \cos t\) \(y^{\prime \prime}+y=0\) d. \(y=-3 \sin 2 t+5 \cos t\) \(y^{\prime \prime}+4 y=0\) e. \(y=e^{-t} \quad y^{\prime \prime}+2 y^{\prime}+y=0\) f. \(\quad y=e^{-t} \sin t\) \(y^{\prime \prime}+2 y^{\prime}+2 y=0\)

Step-by-Step Solution

Answer

All the suggested solutions satisfy their associated differential equations.

Step 1: Find the first and second derivatives for (a)

Given: \( y = \cos(t) \).First Derivative: \( y' = \frac{d}{dt}(\cos t) = -\sin t \).Second Derivative: \( y'' = \frac{d^2}{dt^2}(\cos t) = -\cos t \).Substitute: \( y'' + y = -\cos t + \cos t = 0 \).Thus, the equation \( y'' + y = 0 \) is satisfied.

Step 2: Find the first and second derivatives for (b)

Given: \( y = \cos(2t) \).First Derivative: \( y' = \frac{d}{dt}(\cos 2t) = -2\sin 2t \).Second Derivative: \( y'' = \frac{d^2}{dt^2}(\cos 2t) = -4\cos 2t \).Substitute: \( y'' + 4y = -4\cos 2t + 4\cos 2t = 0 \).Thus, the equation \( y'' + 4y = 0 \) is satisfied.

Step 3: Find the first and second derivatives for (c)

Given: \( y = 3\sin t + 2\cos t \).First Derivative: \( y' = \frac{d}{dt}(3\sin t + 2\cos t) = 3\cos t - 2\sin t \).Second Derivative: \( y'' = \frac{d^2}{dt^2}(3\sin t + 2\cos t) = -3\sin t - 2\cos t \).Substitute: \( y'' + y = (-3\sin t - 2\cos t) + (3\sin t + 2\cos t) = 0 \).Thus, the equation \( y'' + y = 0 \) is satisfied.

Step 4: Find the first and second derivatives for (d)

Given: \( y = -3\sin 2t + 5\cos t \).First Derivative: \( y' = \frac{d}{dt}(-3\sin 2t + 5\cos t) = -6\cos 2t - 5\sin t \).Second Derivative: \( y'' = \frac{d^2}{dt^2}(-3\sin 2t + 5\cos t) = 12\sin 2t - 5\cos t \).Substitute: \( y'' + 4y = (12\sin 2t - 5\cos t) + 4(-3\sin 2t + 5\cos t) = 0 \).Thus, the equation \( y'' + 4y = 0 \) is satisfied.

Step 5: Find the first and second derivatives for (e)

Given: \( y = e^{-t} \).First Derivative: \( y' = \frac{d}{dt}(e^{-t}) = -e^{-t} \).Second Derivative: \( y'' = \frac{d^2}{dt^2}(e^{-t}) = e^{-t} \).Substitute: \( y'' + 2y' + y = e^{-t} - 2e^{-t} + e^{-t} = 0 \).Thus, the equation \( y'' + 2y' + y = 0 \) is satisfied.

Step 6: Find the first and second derivatives for (f)

Given: \( y = e^{-t} \sin t \).First Derivative: \( y' = \frac{d}{dt}(e^{-t}\sin t) = e^{-t}\cos t - e^{-t}\sin t \).Second Derivative: Apply product rule:\[ y'' = \frac{d}{dt}(e^{-t}\cos t - e^{-t}\sin t) = -e^{-t}\cos t - e^{-t}\sin t - e^{-t}\cos t = -2e^{-t}\cos t - e^{-t}\sin t \]Substitute: \( y'' + 2y' + 2y = (-2e^{-t}\cos t - e^{-t}\sin t) + 2(e^{-t}\cos t - e^{-t}\sin t) + 2e^{-t}\sin t = 0 \).Thus, the equation \( y'' + 2y' + 2y = 0 \) is satisfied.

Key Concepts

Second Derivative

In differential equations, understanding the second derivative is crucial to solving many types of problems. The second derivative, denoted as \( y'' \), provides information about the curvature of a function. Curvature can tell us about how the slope of a line is changing. In terms of motion, it's akin to acceleration in physics. A practical aspect of finding the second derivative involves differentiating a function twice. First, compute the first derivative \( y' \), which represents the rate of change of the function. Then, differentiate \( y' \) to get \( y'' \). This reflects the rate of change of the rate of change—essentially, how the slope of a function itself is shifting.
When solving differential equations like \( y'' + y = 0 \), it's important to substitute the second derivative back into the equation to verify if it satisfies the given equation. For example, with the function \( y = \cos t \), the derivatives are \( y' = -\sin t \), and \( y'' = -\cos t \). Substituting these into the equation \( y'' + y \) results in \( -\cos t + \cos t = 0 \), demonstrating that it fulfills the condition for \( y'' + y = 0 \).

Trigonometric Functions

Trigonometric functions like \( \sin \) and \( \cos \) are fundamental in solving many differential equations due to their inherent periodic properties. These functions describe waves and oscillations in both physical and mathematical contexts.
In the context of second-order linear differential equations, trig functions often appear in the solutions due to their ability to form a basis for periodic solutions. When dealing with an equation such as \( y'' + y = 0 \) or \( y'' + 4y = 0 \), solutions frequently involve combinations of \( \sin t \) and \( \cos t \), where constants and phase shifts might vary.

• The derivative of \( \sin t \) is \( \cos t \), and the derivative of \( \cos t \) is \( -\sin t \).
• These cyclic derivatives are part of why trig functions are so versatile in modeling waves.

For example, if you consider \( y = 3\sin t + 2\cos t \), then \( y' = 3\cos t - 2\sin t \), and \( y'' = -3\sin t - 2\cos t \). Substituting back into the original equation \( y'' + y = 0 \) gives you \( (-3\sin t - 2\cos t) + (3\sin t + 2\cos t) = 0 \), confirming the solution.

Exponential Functions

Exponential functions, often in the form \( e^{kt} \), are another cornerstone of solving differential equations. They model a wide array of phenomena in nature, such as population growth or radioactive decay. In these applications, the exponential describes how a quantity changes at a rate proportional to its current value.
When incorporating exponential functions into differential equations, pay keen attention to any derivatives. An exponential function \( y = e^{-t} \) has a straightforward derivative pattern: \( y' = -e^{-t} \) and \( y'' = e^{-t} \). With these properties, exponential functions can be useful for creating solutions to equations that model decay or growth.

• Exponential decay in equations often involves negative exponents like \( e^{-t} \), causing the function to decrease over time.
• Multiple exponentials can be combined or altered using trigonometric functions, such as \( y = e^{-t}\sin t \) or mixed with other terms.

Consider the function \( y = e^{-t} \), for which \( y'' + 2y' + y = 0 \) holds. Derivatives \( y' = -e^{-t} \) and \( y'' = e^{-t} \) fit this relation perfectly, confirming the solution is valid and demonstrating how exponentials seamlessly integrate with other function types in equations.