Problem 7 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) Is \(y=e^{t}+\ln t\) a solution to the differential equation \(\frac{d y}{d t}=y-\frac{y}{t}\) ? (b) Is \(y=t e^{t}\) a solution to the differential equation \(\frac{d y}{d t}=y-\frac{y}{t}\) ?

Step-by-Step Solution

Answer

The provided functions should be evaluated by differentiating and substituting into the differential equation. The value of each \(y'\) and \(y\) should be substituted to the differential equation. If, after simplification, the left hand side equals right hand side, then the function is a solution. Otherwise, it is not.

Step 1 - Differentiate the function for part (a)

Differentiate \(y=e^{t}+\ln t\) with respect to \(t\) to get \(y' = e^{t}+\frac{1}{t}\)

Step 2 - Substitute into the differential equation for part (a)

Substitute \(y' = e^{t}+\frac{1}{t}\) and \(y=e^{t}+\ln t\) into the given differential equation \(\frac{d y}{d t} = y - \frac{y}{t}\). Simplify to see if the equation holds true.

Step 3 - Evaluate function for part(a)

After substitution and simplification, if the left hand side equals the right hand side, then \(y = e^{t}+\ln t\) is a solution to the differential equation. Otherwise, it isn't a solution.

Step 4 - Differentiate the function for part (b)

Differentiate \(y = t e^{t}\) with respect to \(t\) to get \(y' = t e^{t}+e^{t}\)

Step 5 - Substitute into the differential equation for part (b)

Substitute \(y' = t e^{t}+e^{t}\) and \(y = t e^{t}\) into the given differential equation \(\frac{d y}{d t} = y - \frac{y}{t}\). Simplify to see if the equation holds true.

Step 6 - Evaluate function for part (b)

After substitution and simplification, if the left hand side equals the right hand side, then \(y= t e^{t}\) is a solution to the differential equation. Otherwise, it isn't a solution.

Key Concepts

Calculus

Understanding calculus is essential when solving differential equations. Calculus deals with concepts such as derivatives and integrals to study how things change. It’s like having a toolbox that helps you find rates of change or total amounts. In calculus:

• Derivatives tell you how fast things are changing at a specific instant. They are used in exercises to find solutions relevant to real-world problems, like motion or growth.
• Integration involves finding accumulation or total values by inverse operations of derivatives. It’s like folding up small changes to see the bigger picture.

When faced with a differential equation like \[\frac{d y}{d t} = y - \frac{y}{t}\]understanding these fundamental calculus concepts breaks down the process into manageable parts, allowing you to identify whether given functions like \(y=e^{t} + \ln t\) or \(y=te^{t}\) satisfy the equation.
This allows you to decide the effectiveness of particular solutions. Mastering calculus provides the foundation necessary for engaging with differential equations.

Derivatives

Derivatives are a key focus in calculus as they describe the rate at which a quantity changes. In the given problem, you need derivatives to verify solutions to the differential equation. By differentiating given functions, such as \(y = e^{t} + \ln t\), you find the derivative:\[y' = e^{t} + \frac{1}{t}\].

• To differentiate \(e^{t}\), recognize it as a basic exponential function. Its rate of change is proportional to itself, reflected as \(e^{t}\).
• For \(\ln t\), use the fundamental derivative rule \(\frac{d}{dt}[\ln t] = \frac{1}{t}\).

This derivative process helps determine if these derivatives match the conditions laid out in the differential equation. By ensuring that the derivative you calculate aligns with the original differential equation, you assess whether the proposed function is a genuine solution.
It transforms abstract differential problems into logical sequences of operations.

Exponential Functions

Exponential functions appear frequently in differential equations, often as part of natural processes like growth or decay. An exponential function of the form \(y = e^{t}\) grows at a rate proportional to its current value.

• Its intuitive structure allows it to model dynamic systems, such as population growth or radioactive decay effectively.
• In our exercise, for \(y = te^{t}\), the derivative involves the product rule since it involves a product of functions.

To solve the differential equation, you calculate:\[y' = t e^{t} + e^{t}\]using the product rule and verify this against the differential equation conditions. Understanding how exponential expressions behave under differentiation is central to mastering solutions for differential equations. Practice with exponential functions demystifies their nature and builds confidence handling them in calculus problems.

Logarithmic Functions

Logarithmic functions, like \(\ln t\), are inverses of exponential functions. They are used to solve for time in exponential growth scenarios or in calculus problems where scales of magnitude are in play.

• Derivatives of logarithmic functions help understand the behavior of composite functions such as \(y = e^{t} + \ln t\).
• In the derivative of \(\ln t\), \(\frac{1}{t}\) indicates how the rate of change scales inversely with \(t\).

Substituting and simplifying derivatives involving \(\ln t\) in the differential equation problem verifies whether the proposed function is valid. Logarithmic nuances add depth to calculus concepts and offer powerful tools for reconfiguring and analyzing complex equations.
Embracing these concepts strengthens your problem-solving abilities in advanced calculus contexts, equipping you to tackle challenges with varied mathematical phenomena.