Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) Show that \(P=C e^{2 t}\) (where \(C\) is any constant) is a solution to the differential equation \(\frac{d P}{d t}=2 P .\) That is, show that if you compute \(\frac{d P}{d t}\), you get \(2 P\). (b) Show that \(P=e^{2 t}+C\) is not a solution to the differential equation \(\frac{d P}{d t}=2 P\).

Step-by-Step Solution

Answer

The first function \(P=C e^{2 t}\) is a solution to the differential equation \(\frac{d P}{d t}=2 P\) and the second function \(P=e^{2 t}+C\) is not a solution to the given differential equation.

Step 1: Differential of the First Function

Firstly, take the derivative of the first function \(P=C e^{2 t}\) with respect to \(t\) to get \(\frac{d P}{d t}=2C e^{2t}\). Here, \(C e^{2t}\) can be replaced with \(P\). Hence, we get \(\frac{d P}{d t}=2P\). This proves that the first function is a solution to the given differential equation.

Step 2: Testing the Second Function

Now, let's compute the derivative of the second function \(P=e^{2 t}+C\). The derivative is \(\frac{d P}{d t}=2 e^{2t}\).

Step 3: Validation of the Second Function

The obtained derivative doesn't equal to \(2P\) after replacing \(e^{2t}+C\) with \(P\) i.e. \(\frac{d P}{d t}\) is not equal to \(2P\) but is equal to \(2(e^{2t}+C)=2P+2C\), which proves that the second function is not a solution to the given differential equation.

Key Concepts

Derivative Calculation

When dealing with differential equations, understanding how to compute derivatives is crucial. A derivative represents the rate at which a function changes at any given point, and it's a fundamental concept in calculus. In our example, we calculate the derivative of
\(P = Ce^{2t}\) with respect to time \(t\). To do this, we use the rule of derivatives for exponential functions: the derivative of \(e^{kt}\), where \(k\) is a constant, is \(ke^{kt}\).
Thus, the derivative of \(P\) becomes \(\frac{dP}{dt} = 2Ce^{2t}\). Since \(Ce^{2t} = P\), this simplifies to \(\frac{dP}{dt} = 2P\), confirming that \(P\) is indeed a solution to the differential equation \(\frac{dP}{dt} = 2P\). This step-by-step derivative calculation is essential for understanding why the proposed functions either satisfy or do not satisfy the differential equation.

Exponential Functions

Exponential functions, denoted as \(e^x\) where \(e\) is the base of natural logarithms, are widely used in mathematics to model growth or decay processes, such as population growth, radioactive decay, and interest compounding. In our exercise, the function \(P = Ce^{2t}\) showcases an exponential growth pattern, which is characterized by a rate of change proportional to its current value.

In the context of differential equations, exponential functions are often solutions to constant coefficient linear equations. Such functions reveal a direct relationship between the value of \(P\) and its rate of change over time, which is key to solving many real-world problems.

Constant Coefficient Differential Equations

Differential equations with constant coefficients, such as \(\frac{dP}{dt} = 2P\) from our exercise, are a special class of linear differential equations. They are termed 'constant coefficient' because they involve derivatives of the function and the function itself multiplied by constants.
These equations often arise in modeling naturally occurring phenomena where the change rate is proportional to the function's current state. The general solution to a first-order constant coefficient differential equation \(\frac{dx}{dt} = kx\) is given by \(x = Ce^{kt}\) where \(C\) is an arbitrary constant and \(k\) is the coefficient from the equation.

Thus, in our case, the function \(P = Ce^{2t}\) perfectly fits this mold and indeed constitutes a valid solution to the differential equation. The solution method we apply here is powerful and extends to more complex equations in the study of differential equations.