Problem 3 - Applied Calculus

Question

Find the general solution to the differential equation $$\frac{d y}{d t}=2 t$$

Step-by-Step Solution

Answer

The general solution is \(y(t) = t^2 + C\).

Step 1: Understand the Problem

We are given the differential equation \(\frac{d y}{d t} = 2t\). This is a first-order differential equation, where \(y\) is the dependent variable and \(t\) is the independent variable. The objective is to find a general solution for \(y\).

Step 2: Separate Variables

Even though separation of variables isn't necessary here, for clarity, let's write the equation in a form that emphasizes derivatives. The equation \(\frac{d y}{d t} = 2t\) suggests that if we integrate both sides with respect to \(t\), we will find \(y(t)\).

Step 3: Integrate Both Sides

Integrate both sides of the equation with respect to \(t\):\[\int \frac{d y}{d t} \, dt = \int 2t \, dt\].The left side, \( \int \frac{d y}{d t} \, dt \), simplifies to \(y\) since the derivative and the integral cancel each other out. For the right side, use the power rule of integration to find \(\int 2t \, dt = t^2 + C\), where \(C\) is the constant of integration.

Step 4: Write the General Solution

Based on the integration in Step 3, the general solution to the differential equation is:\[y(t) = t^2 + C\].This equation represents the general solution, with \(C\) being an arbitrary constant that can take any real value.

Key Concepts

First-Order Differential Equation

A first-order differential equation involves the first derivative of a function, but no higher derivatives. In our original problem, the equation \( \frac{d y}{d t} = 2t \) is a clear example of such an equation. Here, \( \frac{d y}{d t} \) signifies the first derivative of \( y \) with respect to \( t \), showing how \( y \) changes as \( t \) changes.
These equations often describe real-world phenomena where the rate of change is directly proportional to some variable. The primary goal is to determine a function \( y(t) \) whose derivative matches the equation you're given.
First-order differential equations are essential in various fields, ranging from physics to finance, because they can model simple exponential growth or decay, population dynamics, or even heat transfer.

Integration

Integration is a core mathematical concept that allows us to find a function when we know its derivative. It is essentially the reverse process of differentiation. In our problem, integrating both sides of \( \frac{d y}{d t} = 2t \) with respect to \( t \) helps us determine the function \( y(t) \).
Here's the approach:

• On the left, \( \int \frac{d y}{d t} \, dt = y \) simplifies the process because integration and differentiation are inverse processes.
• On the right, \( \int 2t \, dt \) requires careful attention to find the original function whose derivative is \( 2t \).

Integration is crucial in calculus as it allows us to solve differential equations, calculate areas, and model various scenarios with given rates of change.

Power Rule

The power rule is a straightforward method for integrating functions of the form \( t^n \). In our situation, to solve \( \int 2t \, dt \), the power rule is applied as follows:

• First, identify the function in the form \( t^n \). Here, \( n = 1 \) because we have \( 2t \).
• Integrate using the formula \( \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \).
• For \( 2t \), applying the rule gives \( \int 2t \, dt = 2 \times \frac{t^{1+1}}{1+1} = t^2 + C \).

The power rule simplifies integration significantly, making it easier to handle simple polynomial expressions. Understanding this rule allows students to integrate a wide variety of functions quickly.

Constant of Integration

The constant of integration, represented by \( C \), is a pivotal concept in integration. It arises because the process of differentiation eliminates constants, meaning when we integrate, we cannot determine a single unique function without additional information.
In our equation \( y(t) = t^2 + C \), \( C \) represents an arbitrary constant. It signifies that there are infinitely many functions \( y(t) \) that satisfy the original differential equation, each parallel to another but vertically shifted.
The constant becomes critical when solving specific real-world problems where initial conditions or additional data may be provided to uniquely determine \( C \) and consequently the exact function \( y(t) \).
Remember, integration would not fully capture the solution space without acknowledging this constant, allowing the general solution to include all possible scenarios.