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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 9 - First-Order Differential EquationsProblem 9.1.25
Chapter 9, Problem 9.1.25

Show that the solution of the initial value problem
y' = x + y, y(x₀) = y₀
is
y = -1 -x + (1 + x₀ + y₀) exp(x-x₀).

Verified step by step guidance
1
Recognize that the given differential equation is a first-order linear ordinary differential equation of the form \(y' - y = x\).
Identify the integrating factor \(\mu(x)\), which is given by \(\mu(x) = e^{\int -1 \, dx} = e^{-x}\).
Multiply both sides of the differential equation by the integrating factor to get \(e^{-x} y' - e^{-x} y = x e^{-x}\), which simplifies to \(\frac{d}{dx}(e^{-x} y) = x e^{-x}\).
Integrate both sides with respect to \(x\): \(\int \frac{d}{dx}(e^{-x} y) \, dx = \int x e^{-x} \, dx\), leading to \(e^{-x} y = \int x e^{-x} \, dx + C\).
Solve the integral on the right side using integration by parts, then multiply both sides by \(e^{x}\) to isolate \(y\). Finally, apply the initial condition \(y(x_0) = y_0\) to solve for the constant \(C\) and write the solution in the form \(y = -1 - x + (1 + x_0 + y_0) e^{x - x_0}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First-Order Linear Differential Equations

A first-order linear differential equation has the form y' + p(x)y = q(x). Such equations can be solved using an integrating factor, which simplifies the equation into an exact derivative, allowing integration to find the general solution.
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Classifying Differential Equations

Integrating Factor Method

The integrating factor is a function, usually denoted μ(x) = exp(∫p(x) dx), used to multiply both sides of a linear differential equation. This transforms the left side into the derivative of (μ(x)y), enabling straightforward integration to solve for y.
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Euler's Method

Initial Value Problem (IVP)

An initial value problem specifies the value of the solution at a particular point, y(x₀) = y₀. After finding the general solution, the initial condition is used to determine the constant of integration, yielding the unique solution that satisfies the problem.
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Related Practice
Textbook Question

Use Euler’s method with dx = 0.2 to estimate y(2) if y′ = y/x and y(1) = 2. What is the exact value of y(2)?

Textbook Question

Solving Initial Value Problems

Solve the initial value problems in Exercises 15–20.


(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x > -1, y(0) = 5

Textbook Question

Solving Initial Value Problems

Solve the initial value problems in Exercises 15–20.


t dy/dt + 2y = t³, t > 0, y(2) = 1

Textbook Question

Solve the Bernoulli equations in Exercises 29–32.


x²y' + 2xy = y³

Textbook Question

Use Euler’s method with dx = 0.5 to estimate y(5) if y′ = y²/√x and y(1) = −1. What is the exact value of y(5)?

Textbook Question

First-Order Linear Equations

Solve the differential equations in Exercises 1–14.


xdy/dx + y = e ͯ, x > 0