Question

Solve \(y^{\prime}=e^{k t}\) with the initial condition \(y(0)=0\) and solve \(y^{\prime}=1\) with the same initial condition. As \(k\) approaches 0, what do you notice?

Short answer

As \( k \to 0 \), \( y(t) = \frac{e^{kt} - 1}{k} \) approaches \( y(t) = t \).

Step-by-step solution

Understanding the Differential Equation

We begin by identifying the form of the given differential equation, \( y' = e^{kt} \). This equation is separable as the right-hand side is solely a function of \( t \). We will integrate with respect to \( t \) to solve for \( y(t) \).

Integrating the Differential Equation

Integrate \( y' = e^{kt} \) with respect to \( t \):\[ y(t) = \int e^{kt} \; dt = \frac{e^{kt}}{k} + C \]where \( C \) is the constant of integration.

Applying the Initial Condition

Apply the initial condition \( y(0) = 0 \) to find \( C \):\[ 0 = \frac{e^{k \cdot 0}}{k} + C \Rightarrow 0 = \frac{1}{k} + C \]Solving for \( C \), we get \( C = -\frac{1}{k} \).

Solution for the First Equation

Substitute \( C = -\frac{1}{k} \) back into the expression for \( y(t) \):\[ y(t) = \frac{e^{kt}}{k} - \frac{1}{k} = \frac{e^{kt} - 1}{k} \]

Solving the Second Differential Equation

For the equation \( y' = 1 \), integrate with respect to \( t \):\[ y(t) = \int 1 \; dt = t + C \]

Applying Initial Condition to Second Equation

Apply the initial condition \( y(0) = 0 \) to find \( C \):\[ 0 = 0 + C \Rightarrow C = 0 \]Thus, \( y(t) = t \).

Examining Behavior as \( k \to 0 \)

As \( k \) approaches 0, consider the limit: \[ \lim_{k \to 0} \frac{e^{kt} - 1}{k} \]Using L'Hôpital's Rule, since the limit has the form \( \frac{0}{0} \), \[ \lim_{k \to 0} \frac{e^{kt} - 1}{k} = \lim_{k \to 0} \frac{t e^{kt}}{1} = t \]This matches the solution \( y(t) = t \) obtained for the second equation when \( k = 0 \).

Conclusion

Both equations lead to the same solution \( y(t) = t \) as \( k \to 0 \), illustrating that the exponential growth term effectively becomes linear growth.

Key concepts

Separable Equations

A separable differential equation is a type of differential equation that can be expressed in a form where each variable can be separated onto opposite sides of the equation. In mathematical terms, a separable equation is one that can be rewritten in the form:
\[ \frac{dy}{dt} = g(t) \cdot h(y) \]This allows us to isolate variables and integrals with respect to one another. For example, if we have \( y' = e^{kt} \), it is separable because the right-hand side depends only on \( t \). This means we can integrate with respect to \( t \) to solve for \( y(t) \).

• The technique involves rearranging the equation to isolate \( dy \) and \( dt \) on opposite sides.
• Complete the integration for each part separately.
• This separation is what makes it easier to integrate and solve for the function \( y \).

Understanding separable equations is crucial as it provides a systematic method to solve complex differential equations by breaking them down into more manageable parts.

Integration

Integration is a mathematical process of finding the antiderivative of a function. In the context of solving differential equations, it involves integrating both sides of the equation to express a function. For instance, when faced with a separable differential equation like \( y' = e^{kt} \), we integrate the right-hand side to find \( y(t) \).

• Start by rewriting the equation so one side is in terms of \(dy\) and the other in terms of \(dt\).
• Proceed with integrating both sides separately to solve for \( y(t) \).
• Remember to add a constant of integration, denoted as \( C \), after integrating.

After integrating \( y' = e^{kt} \), you will find \( y(t) = \frac{e^{kt}}{k} + C \). This expression shows the role of integration in solving the equation. The constant reflects the range of solutions depending on initial conditions.

Initial Conditions

Initial conditions are essential in solving differential equations because they determine the specific solution from a family of solutions. Often represented as \( y(t_0) = y_0 \), initial conditions help in calculating the constant of integration, \( C \). For example, using the condition \( y(0) = 0 \), we can find \( C \) in our exercise.

• Initial conditions are usually given as a point through which the function passes, like \( y(0) = 0 \).
• These conditions help to "pin" the solution in real-world applications, making it unique.
• They also allow us to make predictions and understand the behavior of the system modeled by the differential equation.

In our example, applying \( y(0) = 0 \) to the solution \( y(t) = \frac{e^{kt} - 1}{k} \) helped in conclusively finding that \( C = -\frac{1}{k} \). This exercise shows how initial conditions refine the general solution into one specific path or outcome, especially when analyzing limits like \( k \to 0 \).