Problem 4 - Calculus: Early Transcendentals

Question

Find the general solution of the equation. $$y^{\prime}+e^{t} y=-2 e^{t}$$

Step-by-Step Solution

Answer

The general solution is \( y(t) = -e^{2t-e^t} + Ce^{-e^t} \).

Step 1: Identify the type of equation

This is a first-order linear ordinary differential equation of the form \( y' + p(t)y = g(t) \), where \( p(t) = e^t \) and \( g(t) = -2e^t \).

Step 2: Compute the Integrating Factor

The integrating factor \( \mu(t) \) is given by \( e^{\int p(t) dt} \). Here, \( \int e^t \, dt = e^t \), thus \( \mu(t) = e^{e^t} \).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \mu(t) = e^{e^t} \). The equation becomes \[ e^{e^t} y' + e^{e^t + t} y = -2 e^{e^t + t} \].

Step 4: Recognize the Left-Hand Side as a Derivative

Notice that the left-hand side can be written as the derivative of the product \( (e^{e^t} y)' \). So we have \[ (e^{e^t} y)' = -2 e^{e^t + t} \].

Step 5: Integrate Both Sides

Integrate both sides with respect to \( t \): \[ \int (e^{e^t} y)' \, dt = \int -2 e^{e^t + t} \, dt \]. The left side integrates to \( e^{e^t} y \). For the right side, use substitution \( u = e^t \), thus \( du = e^t \, dt \) to get:\[ \int -2 u \, du = -u^2 + C \]. Replace back \( u = e^t \), \[ -e^{2t} + C \].

Step 6: Solve for y

The equation from the integration is \[ e^{e^t} y = -e^{2t} + C \]. Solve for \( y \) by dividing both sides by \( e^{e^t} \):\[ y = -e^{2t - e^t} + Ce^{-e^t} \].

Step 7: Present the General Solution

The general solution is \[ y(t) = -e^{2t-e^t} + Ce^{-e^t} \], where \( C \) is an arbitrary constant.

Key Concepts

Integrating Factor

In solving first-order linear ordinary differential equations, the integrating factor plays a vital role. It is a tool for transforming a non-exact differential equation into an exact one, simplifying the process of finding a solution. The integrating factor is typically denoted by \( \mu(t) \) and is calculated using the formula \( e^{\int p(t) \, dt} \), where \( p(t) \) is the coefficient of \( y \) in the differential equation.

The rationale for using an integrating factor is to multiply the entire differential equation by \( \mu(t) \), thereby restructuring it such that the left-hand side becomes a derivative of the product of the integrating factor and \( y(t) \). This clever manipulation allows us to integrate both sides of the equation easily.

• Identify \( p(t) \) in the differential equation.
• Compute the integrating factor by finding \( e^{\int p(t) \, dt} \).
• Multiply through by this factor to make the equation exact.

In our exercise, \( \mu(t) = e^{e^t} \), making the left-hand side the derivative \( (e^{e^t} y)' \). This greatly eases the process of integrating and solving the equation.

Ordinary Differential Equation

Ordinary differential equations (ODEs) are equations that involve functions of one independent variable and their derivatives. Specifically, first-order linear ordinary differential equations are simplified forms where the highest order derivative present is of the first order.

In the context of the given problem, the differential equation is \( y' + e^{t} y = -2 e^{t} \). Here:

• \( y' \) is the first derivative of unknown function \( y(t) \).
• \( e^{t} \) is the coefficient function \( p(t) \).
• \( -2 e^{t} \) acts as the non-homogeneous part \( g(t) \).

To solve this ODE, identifying its components—\( p(t) \) and \( g(t) \)—is crucial. The structure is defined generally as \( y' + p(t)y = g(t) \). Mastery of recognizing and manipulating these equations comes from understanding their inherent simplicity and the strategies like integrating factors that resolve them deftly.

General Solution

Finding the general solution of a first-order linear differential equation like the one presented involves using the integrating factor and integrating the equation to express \( y(t) \) in terms of \( t \) and an arbitrary constant. This constant represents the family of solutions that satisfies the equation across different initial conditions.

To derive the general solution, execute the following steps:

• Integrate both sides after employing the integrating factor.
• Express the derivative in a simplified form after integrating to isolate \( y \).
• Solve for \( y \) by dividing each term by the integrating factor.

The result in our exercise yielded a general solution of \( y(t) = -e^{2t-e^t} + Ce^{-e^t} \), where \( C \) is an arbitrary constant. This illustrates how the solution depends not solely on the function \( g(t) \) but also on the specific conditions we've yet to apply, making it versatile for various scenarios.