Problem 8 - Calculus for Biology and Medicine

Question

\text { In Problems , solve each pure-time differential equation. }$$ \frac{d h}{d t}=4-16 t^{2}, \text { where } h(1)=0 $$

Step-by-Step Solution

Answer

The solution is \( h(t) = 4t - \frac{16}{3}t^3 + \frac{4}{3} \).

Step 1: Set Up the Integral

To solve the differential equation \( \frac{d h}{d t}=4-16 t^{2} \), we integrate both sides with respect to \( t \). The left side becomes \( \int \frac{d h}{d t} \, dt = h(t) \), and the right side becomes \( \int (4-16 t^{2}) \, dt \).

Step 2: Integrate the Right Side

Now, integrate \( 4 - 16t^2 \) with respect to \( t \). This gives: \( 4t - \frac{16}{3}t^3 + C \), where \( C \) is the integration constant.

Step 3: Solve for the Constant with Initial Condition

Use the initial condition \( h(1) = 0 \) to solve for \( C \). Substitute \( t = 1 \) and \( h = 0 \) into the integrated function: \( 0 = 4(1) - \frac{16}{3}(1)^3 + C \). Simplifying gives \( 0 = 4 - \frac{16}{3} + C \).

Step 4: Calculate the Constant

Continue simplifying: \( 0 = 4 - \frac{16}{3} + C \) becomes \( 0 = \frac{12}{3} - \frac{16}{3} + C = \frac{-4}{3} + C \). Solving for \( C \), we find \( C = \frac{4}{3} \).

Step 5: Write the General Solution

Substitute \( C = \frac{4}{3} \) back into the integrated function to get the general solution: \( h(t) = 4t - \frac{16}{3}t^3 + \frac{4}{3} \).

Key Concepts

Understanding Integration

Integration is like the reverse process of differentiation. In the context of solving differential equations, we use integration to find an unknown function from its derivative. Let's dive deeper into the exercise. We started with the differential equation \( \frac{d h}{d t}=4-16 t^{2} \). To solve it, we integrate both sides. When we integrate the left side, which is the derivative of \( h(t) \), we simply get \( h(t) \). This happens because integration and differentiation are inverse operations.

For the right side, we integrate the function \( 4-16t^2 \). This involves finding a function whose derivative gives us \( 4-16t^2 \). The result is \( 4t - \frac{16}{3}t^3 + C \), where \( C \) is a constant of integration. This constant is crucial as it accounts for any initial values or specific conditions the problem might have.

Importance of Initial Conditions

In differential equations, initial conditions are used to find specific solutions tailored to a problem's constraints. It's like having extra information to pinpoint one solution among infinitely many possibilities. For our exercise, the initial condition given is \( h(1) = 0 \).

This means when \( t = 1 \), the value of \( h(t) \) has to be \( 0 \). With this condition, we substitute \( t = 1 \) into the integrated function \( h(t) = 4t - \frac{16}{3}t^3 + C \). Doing this allows us to solve for \( C \), the integration constant. By performing the calculation: \( 0 = 4(1) - \frac{16}{3}(1)^3 + C \), we find \( C \) to be \( \frac{4}{3} \).

Ultimately, this specific \( C \) value helps construct the unique solution that satisfies both the differential equation and initial condition.

Exploring Pure-Time Differential Equations

A pure-time differential equation is a type of ordinary differential equation (ODE) where the rate of change of a variable depends solely on time. The equation \( \frac{d h}{d t}=4-16 t^{2} \) is an example. In this scenario, the change in \( h \) is governed only by \( t \) and not by \( h \) itself.

What makes pure-time differential equations straightforward is their dependency exclusively on time, eliminating any potential complexities introduced by additional variables. This aspect simplifies the integration process because we only need to focus on the time variable \( t \).

Solving such equations involves integrating the time-dependent expression, just like we did with \( 4 - 16t^2 \). This facilitates deriving all potential solutions dependent on the time variable, guided by any initial conditions present. They are a great stepping stone as you explore more complex differential equations involving multiple variables.