Problem 66 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(63-66\). $$ \frac{d^{2} y}{d t^{2}}=1-e^{2 t}, \quad y(1)=-1 \quad \text { and } \quad y^{\prime}(1)=0 $$

Step-by-Step Solution

Answer

The solution is \( y(t) = \frac{t^2}{2} - \frac{1}{4} e^{2t} + \frac{1}{2} e^{2} t - t + 1 - \frac{1}{4} e^{2} \).

Step 1: Analyze Given Information

We are given a second-order differential equation: \( \frac{d^{2} y}{d t^{2}} = 1 - e^{2t} \), with initial conditions \( y(1) = -1 \) and \( y'(1) = 0 \). Our task is to find the function \( y(t) \) that satisfies both the differential equation and these initial conditions.

Step 2: Integrate to Find First Derivative

Starting with the equation \( \frac{d^{2} y}{d t^{2}} = 1 - e^{2t} \), we integrate with respect to \( t \) to determine the first derivative, \( \frac{d y}{d t} = \int (1 - e^{2t}) \, dt = t - \frac{1}{2} e^{2t} + C_1 \).

Step 3: Apply Initial Condition for First Derivative

To find \( C_1 \), use the initial condition \( y'(1) = 0 \): \[ y'(t) = t - \frac{1}{2} e^{2t} + C_1 \] \[ 0 = 1 - \frac{1}{2} e^{2} + C_1 \] \[ C_1 = \frac{1}{2} e^{2} - 1 \]. Thus, \[ y'(t) = t - \frac{1}{2} e^{2t} + \frac{1}{2} e^{2} - 1 \].

Step 4: Integrate to Find General Solution for y(t)

Integrate \( y'(t) = t - \frac{1}{2} e^{2t} + \frac{1}{2} e^{2} - 1 \) to find \( y(t) \):\[ y(t) = \int \left(t - \frac{1}{2} e^{2t} + \frac{1}{2} e^{2} - 1\right) dt \] \[ = \frac{t^2}{2} - \frac{1}{4} e^{2t} + \frac{1}{2} e^{2} t - t + C_2 \].

Step 5: Apply Initial Condition for y(t)

Use the initial condition \( y(1) = -1 \) to find \( C_2 \):\[ -1 = \frac{1}{2} - \frac{1}{4} e^{2} + \frac{1}{2} e^{2} - 1 + C_2 \] \[ C_2 = 1 - \frac{1}{4} e^{2} \]. Thus, \[ y(t) = \frac{t^2}{2} - \frac{1}{4} e^{2t} + \frac{1}{2} e^{2} t - t + 1 - \frac{1}{4} e^{2} \].

Step 6: Final Solution

The solution to the initial value problem is \( y(t) = \frac{t^2}{2} - \frac{1}{4} e^{2t} + \frac{1}{2} e^{2} t - t + 1 - \frac{1}{4} e^{2} \). This function satisfies both the differential equation and the given initial conditions.

Key Concepts

Initial Value Problems

Initial value problems in differential equations involve finding a function that satisfies a given differential equation and specific conditions at a particular point, known as initial conditions. This helps determine a unique solution to the differential equation.

• Initial conditions are specified values for the function and its derivatives at a particular point.
• In our problem, the initial conditions are given as: \( y(1) = -1 \) and \( y'(1) = 0 \).
• These conditions are crucial as they guide us in computing the constants of integration that arise when solving differential equations.

By applying these initial conditions, you ensure that your solution is not just any function that satisfies the differential equation, but the one and only function that also adheres to these conditions.

Second-Order Differential Equations

A second-order differential equation involves the second derivative of the function. It is more complex than a first-order equation, requiring additional integration steps for a complete solution.The given exercise is of the form \( \frac{d^{2} y}{d t^{2}} = 1 - e^{2t} \). Here's how to tackle it:

• The equation includes a second derivative \( \frac{d^{2} y}{d t^{2}} \) indicating it's second-order.
• Our goal is to find the function \( y(t) \) and its derivative \( y'(t) \) based on integrating the given equation.

Handling second-order equations typically requires two integrations, as both the first and second derivatives need to be addressed. After the first integration, you obtain the first derivative; a second integration delivers the original function.

Integration Techniques

Integration is a fundamental technique in solving differential equations, especially in reducing higher-order derivatives. Let's see how it's applied step-by-step:**First Integration**Given the equation \( \frac{d^{2} y}{d t^{2}} = 1 - e^{2t} \), start by integrating to find the first derivative:

• Apply the integral to both sides: \( \frac{d y}{d t} = \int (1 - e^{2t}) \, dt \).
• This results in \( \frac{d y}{d t} = t - \frac{1}{2} e^{2t} + C_1 \), where \( C_1 \) is a constant.

**Second Integration**Integrate again to find \( y(t) \):

• Apply the integral to the first derivative: \( y(t) = \int \left(t - \frac{1}{2} e^{2t} + C_1\right) dt \).
• You find \( y(t) = \frac{t^2}{2} - \frac{1}{4} e^{2t} + C_1 t + C_2 \).

Integration allows us to reverse the differentiation process, moving from the derivative back to the original function. The constants of integration, \( C_1 \) and \( C_2 \), are determined using the initial conditions provided.

Calculus

Calculus forms the backbone of solving differential equations. It encompasses both differentiation and integration, which are used to solve and manipulate equations.**Differentiation**- Differentiation involves finding the derivative of a function, which shows how the function changes at any point.- In our context, the second derivative \( \frac{d^{2} y}{d t^{2}} \) is provided, indicating the rate of change of the rate of change of \( y \).**Integration**- Integration is applied to "undo" differentiation, helping to find an original function from its derivative.- The integration process results in additive constants, which are crucial in determining a unique function that satisfies initial conditions.Calculus allows us to explore the behavior of functions and solve complex differential equations by breaking them down into more manageable parts. Mastery of both these tools provides the power to address a wide range of mathematical problems effectively.