Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Use logarithmic differentiation to show that \(y=t e^{3 t}\) satisfies \(y^{\prime \prime}-6 y^{\prime}+9 y=0\).

Step-by-Step Solution

Answer

The function \( y = t e^{3t} \) satisfies the given differential equation.

Step 1: Differentiate using Product and Exponential Rules

First, we need to differentiate the function \( y = t e^{3t} \). Using the product rule for differentiation, \((uv)' = u'v + uv'\), where \( u=t \) and \( v=e^{3t} \). We apply the product rule: \( y' = (t)'e^{3t} + t(e^{3t})' \). The derivative of \( t \) is \( 1 \) and the derivative of \( e^{3t} \) is \( 3e^{3t} \) (using the chain rule). So, the first derivative is \( y' = e^{3t} + 3te^{3t} \).

Step 2: Differentiate Again to Find Second Derivative

Now, differentiate \( y' = e^{3t} + 3te^{3t} \) to obtain \( y'' \), the second derivative. Applying the product rule again to the term \( 3te^{3t} \), we get: \( y'' = (e^{3t})' + ((3t)'e^{3t} + 3t(e^{3t})') \). Simplifying each term: the derivative of \( e^{3t} \) is \( 3e^{3t} \), \( (3t)' = 3 \), and \( (e^{3t})' = 3e^{3t} \). Thus, \( y'' = 3e^{3t} + 3e^{3t} + 9te^{3t} = 6e^{3t} + 9te^{3t} \).

Step 3: Substitute Derivatives into Differential Equation

Substitute \( y \), \( y' \), and \( y'' \) into the differential equation \( y'' - 6y' + 9y = 0 \). We have: \( y'' = 6e^{3t} + 9te^{3t} \), \( y' = e^{3t} + 3te^{3t} \), and \( y = te^{3t} \). Substitute to get: \( (6e^{3t} + 9te^{3t}) - 6(e^{3t} + 3te^{3t}) + 9(te^{3t}) = 0 \).

Step 4: Simplify the Expression

Simplify the expression obtained: \( (6e^{3t} + 9te^{3t}) - (6e^{3t} + 18te^{3t}) + 9te^{3t} = 0 \). Let's simplify this: \( 6e^{3t} + 9te^{3t} - 6e^{3t} - 18te^{3t} + 9te^{3t} \). Combine like terms: All terms cancel out, leading to \( 0 = 0 \), verifying the solution.

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used for differentiating products of two functions. When you have a function that is the product of two separate functions, say \( u \) and \( v \), the product rule states that the derivative \((uv)'\) is given by \( u'v + uv' \).

To break it down:

• Differentiate the first function \( u \) and multiply it by the original second function \( v \)
• Then differentiate the second function \( v \) and multiply it by the original first function \( u \)
• Add these two results together to get the derivative of the product

In the context of our exercise, if we consider \( y=t e^{3t} \), \( u=t \) and \( v=e^{3t} \), applying the product rule helps us find the derivative \( y' \). We take the derivative of \( t \), which is 1, and multiply it by \( e^{3t} \). Then, take the derivative of \( e^{3t} \) (using the chain rule) which results in \( 3e^{3t} \), and multiply it by \( t \). Combining these gives \( y' = e^{3t} + 3te^{3t} \).

This demonstrates the power of the product rule to tackle functions that are not straightforward to differentiate by simple means.

Exponential Function

An exponential function is characterized by a constant base raised to a variable exponent, often expressed as \( e^x \) where \( e \) is a mathematical constant approximately equal to 2.71828. This type of function represents growth or decay processes in natural and financial contexts.

In calculus, the differentiation of exponential functions exhibits a unique property: the derivative of \( e^x \) is simply \( e^x \) itself. When dealing with more complex forms like \( e^{3t} \), the chain rule is applied to find the derivative:\( (e^{3t})' = 3e^{3t} \). This adjustment accounts for the variable's coefficient and showcases the consistent nature of exponential derivatives.

Within our problem, the function \( y=t e^{3t} \) includes \( e^{3t} \), demonstrating how exponential functions behave when differentiated and integrated in more complicated expressions, contributing significantly to calculus' study of growth, decay, and changes over time.

Second Derivative

The second derivative of a function gives us valuable insights into the function's concavity and the rate of change of its slope, which is highly useful for understanding the behavior of curves. Essentially, it tells us how the function's rate of change itself is changing.

To find the second derivative, denoted as \( y'' \), we differentiate the first derivative. In our case, starting from \( y' = e^{3t} + 3te^{3t} \), applying the product rule once more, as discussed, we find \( y'' = 6e^{3t} + 9te^{3t} \).

The calculation of the second derivative in this context involves repeated application of differentiation rules, reinforcing the concept that calculus often builds upon itself to solve more complex problems. When analyzing differential equations, the second derivative clarifies how initial changes evolve, shaping our understanding of the equation's solutions.

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. It describes a relationship among changing quantities, presenting a core tool in modeling various natural and engineered systems.

In this particular exercise, we deal with a second-order linear homogeneous differential equation: \(y'' - 6y' + 9y = 0\). Solving such an equation usually requires finding a solution function, \( y(t) \), that changes its derivatives in accordance with the described relationship.

By performing substitution of the computed \( y \), \( y' \), and \( y'' \) into the differential equation, we've shown that \( y(t) = te^{3t} \) satisfies this equation. This substitution confirms that the left-hand side simplifies to zero, indicating that our function rightly adheres to the conditions of the equation.

In practical scenarios, differential equations like these form the backbone for modeling dynamic phenomena, including physics, engineering, economics, and beyond, providing a mathematical avenue for solving real-world problems that involve rates of change.