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Chapter 10: Problem 25

Solve the initial-value problem. $$y^{\prime}+y=e^{2 t}, y(0)=-1$$

Short Answer

Expert verified
The solution is \( y = \frac{e^{2t}}{3} - \frac{4e^{-t}}{3} \).

Step by step solution

01

Identify the problem

The problem is an initial-value problem for a first-order linear differential equation: \[ y^{\text{'} } + y = e^{2t} \] with an initial condition \( y(0) = -1 \).
02

Find the integrating factor

The integrating factor \( \mu(t) \) is given by \( \mu(t) = e^{\int 1 \, dt} \). Simplify the integral to get the integrating factor: \( \mu(t) = e^t \).
03

Multiply the differential equation by the integrating factor

Multiply both sides of the differential equation by \( e^t \): \[ e^t y^{\text{'} } + e^t y = e^{2t} e^t \] which simplifies to \[ e^t y^{\text{'}} + e^t y = e^{3t} \].
04

Recognize the left-hand side as a derivative

Notice that the left-hand side of the equation is the derivative of \( e^t y \): \[ \frac{d}{dt}(e^t y) = e^{3t} \].
05

Integrate both sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt} (e^t y) \, dt = \int e^{3t} \, dt \]. The left-hand side simplifies to \( e^t y \), and the right-hand side becomes \( \frac{e^{3t}}{3} + C \). So we have: \[ e^t y = \frac{e^{3t}}{3} + C \].
06

Solve for \( y \)

Divide both sides by \( e^t \) to solve for \( y \): \[ y = \frac{e^{3t}}{3e^t} + \frac{C}{e^t} \]. This simplifies to \[ y = \frac{e^{2t}}{3} + Ce^{-t} \].
07

Apply the initial condition

Use the initial condition \( y(0) = -1 \) to find \( C \): \[ -1 = \frac{e^{2 \times 0}}{3} + C e^{0} \]. This simplifies to: \[ -1 = \frac{1}{3} + C \]. Solve for \( C \): \[ C = -1 - \frac{1}{3} = -\frac{4}{3} \].
08

Write the final solution

Substitute \( C \) back into the general solution: \[ y = \frac{e^{2t}}{3} - \frac{4e^{-t}}{3} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial-Value Problem
An initial-value problem involves solving a differential equation with a given initial condition. In simple terms, you start with an equation and a specific starting point. Here, we are given:
  • The first-order linear differential equation: \(\frac{dy}{dt} + y = e^{2t} \)
  • The initial condition: \( y(0) = -1 \)
The goal is to find a function \( y(t) \) that satisfies both the differential equation and the specified initial condition.
Integrating Factor
An integrating factor is a function we multiply both sides of a linear differential equation by to simplify it. For the given differential equation,
  • First, identify the differential equation: \( \frac{dy}{dt} + y = e^{2t} \)

  • The integrating factor \( \mu(t) \) is found using the exponential of the integral of the coefficient of \( y \) in the equation: \[ \mu(t) = e^{\int 1 \, dt} = e^t \]
This integrating factor will help in transforming the left-hand side of the equation.
First-Order Linear Differential Equation
A first-order linear differential equation has the form: \( \frac{dy}{dt} + P(t)y = Q(t) \). To solve the equation:

  • Multiply through by the integrating factor: \( e^t (\frac{dy}{dt} + y) = e^t \, e^{2t} \)
  • This gives: \( e^t y' + e^t y = e^{3t} \)
  • Notice that the left-hand side is the derivative of the product:\[ \frac{d}{dt}(e^t y) = e^{3t} \]
recognizing this helps us integrate both sides more easily.
Integration
To find the solution, integrate both sides of the transformed equation with respect to \( t \):

  • \( \int \frac{d}{dt}(e^t y) \, dt = \int e^{3t} \, dt \)

  • The left-hand side simplifies to: \( e^t y \)
  • The right-hand side integrates as: \( \frac{e^{3t}}{3} + C \)
  • So, we get: \( e^t y = \frac{e^{3t}}{3} + C \)

Finally, solve for \( y \) by dividing through by \( e^t \):\[ y = \frac{e^{2t}}{3} + Ce^{-t} \]
Initial Condition
Initial conditions are specific values for the function and its derivatives at a particular point. They help to determine the unique solution of the differential equation. In our case:
  • The initial condition given is: \( y(0) = -1 \)
  • Substitute this into the general solution: \( -1 = \frac{e^{2 \times 0}}{3} + C e^{0} \)
  • This simplifies to: \( -1 = \frac{1}{3} + C \)
  • Solving for \( C \) gives: \( C = -1 - \frac{1}{3} = -\frac{4}{3} \)

Incorporate \( C \) back into the general solution to obtain the unique solution: \[ y = \frac{e^{2t}}{3} - \frac{4e^{-t}}{3} \].

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Most popular questions from this chapter

For information being spread by mass media, rather than through individual contact, the rate of spread of the information at any time is proportional to the percentage of the population not having the information at that time. Give the differential equation that is satisfied by \(y=f(t),\) the percentage of the population having the information at time \(t\) Assume that \(f(0)=1 .\) Sketch the solution.

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