Problem 1 - Calculus for Biology and Medicine

Question

\text { In Problems 1-8, solve each pure-time differential equation. } \frac{d y}{d t}=t+\sin t, \text { where } y(0)=0

Step-by-Step Solution

Answer

The solution is \( y(t) = \frac{t^2}{2} - \cos t + 1 \).

Step 1: Identify the Differential Equation Type

The differential equation given is \( \frac{dy}{dt} = t + \sin t \). This is a first-order linear differential equation, a function of time \( t \) only, with no explicit dependence on \( y \). It can be integrated directly as it equates the derivative of \( y \) with an expression in terms of \( t \) alone.

Step 2: Integrate the Differential Equation

Separate the variables to integrate with respect to time: \[ y(t) = \int (t + \sin t) \ dt. \]We can split this into two integrals:\[ y(t) = \int t \ dt + \int \sin t \ dt. \]

Step 3: Solve the First Integral

Calculate \( \int t \ dt \):\[ \int t \ dt = \frac{t^2}{2} + C_1, \]where \( C_1 \) is the integration constant.

Step 4: Solve the Second Integral

Calculate \( \int \sin t \ dt \):\[ \int \sin t \ dt = -\cos t + C_2, \]where \( C_2 \) is another integration constant.

Step 5: Combine the Results

Combine both integrals and simplify:\[ y(t) = \frac{t^2}{2} - \cos t + C, \]where \( C = C_1 + C_2 \) represents the combined integration constant.

Step 6: Apply Initial Condition

Use the initial condition \( y(0) = 0 \) to solve for \( C \):\[ 0 = \frac{0^2}{2} - \cos(0) + C \]\[ 0 = 0 - 1 + C \]\( C = 1 \).

Step 7: Write the Final Solution

Substitute \( C = 1 \) back into the equation:\[ y(t) = \frac{t^2}{2} - \cos t + 1. \]This is the particular solution that satisfies the original differential equation with the given initial condition.

Key Concepts

Understanding First-Order Linear Differential Equations

The differential equation given in the problem, \( \frac{dy}{dt} = t + \sin t \), is known as a first-order linear differential equation. These equations are characterized by involving the first derivative of the unknown function, here \( y \), with respect to an independent variable, in this case \( t \). The distinguishing feature is that the equation can be expressed in the standard form \( \frac{dy}{dt} + P(t)y = Q(t) \), where \( P(t) \) and \( Q(t) \) are functions of \( t \).

In our exercise, \( P(t) = 0 \) because there is no function of \( y \) in the equation, simplifying it to \( \frac{dy}{dt} = Q(t) = t + \sin t \). This form confirms it's purely based on \( t \), making solutions straightforward by integration.

These equations appear often in a variety of fields such as physics and engineering, where they are used to model phenomena that change over time. Understanding their structure is key to solving them effectively.

The Role of Integration in Solving Differential Equations

Integration is a crucial process when solving first-order linear differential equations like \( \frac{dy}{dt} = t + \sin t \). This method is used to reverse what differentiation does, enabling us to find the original function \( y(t) \) given its rate of change.

In the provided solution, the integration is performed by:

• Separating the expression into simpler parts: \( \int (t + \sin t) \ dt = \int t \ dt + \int \sin t \ dt \).
• Solving each integral individually, \( \int t \ dt \) and \( \int \sin t \ dt \), to find the functions \( \frac{t^2}{2} \) and \(-\cos t \) respectively.
• Combining these results and adding an integration constant \( C \), as unknowns are common when solving indefinite integrals.

This step uncovers the function or family of functions that satisfies the differential equation. It's important to handle each integral carefully to ensure the correct result, reflecting the original conditions of the problem.

Applying the Initial Condition for Particular Solutions

Initial conditions provide specific information about the solution at a certain point, helping us find a particular, rather than general, solution to the differential equation. In the problem, \( y(0) = 0 \) is the initial condition applied to our integrated solution to determine the integration constant \( C \).

By substituting \( t=0 \) and \( y=0 \) into the equation \( y(t) = \frac{t^2}{2} - \cos t + C \), we arrive at:

• \( 0 = \frac{0^2}{2} - \cos(0) + C \)
• \( 0 = 0 - 1 + C \)

Solving this gives \( C = 1 \). This particular value of \( C \) ensures that the solution curve passes through the given initial point.

This process is essential in problems modeled by differential equations, as initial conditions provide the specificity needed to describe unique real-world scenarios accurately.