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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 9.2.6
Chapter 9, Problem 9.2.6

First-Order Linear Equations
Solve the differential equations in Exercises 1–14.


(1+x)y' + y = √x

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1
Rewrite the given differential equation \((1+x)y' + y = \sqrt{x}\) in the standard linear form \(y' + P(x)y = Q(x)\) by dividing both sides by \((1+x)\), resulting in \(y' + \frac{1}{1+x}y = \frac{\sqrt{x}}{1+x}\).
Identify the integrating factor \(\mu(x)\), which is given by \(\mu(x) = e^{\int P(x)\,dx}\). Here, \(P(x) = \frac{1}{1+x}\), so compute \(\mu(x) = e^{\int \frac{1}{1+x} dx}\).
Calculate the integral \(\int \frac{1}{1+x} dx\), which is \(\ln|1+x|\), and thus the integrating factor becomes \(\mu(x) = e^{\ln|1+x|} = |1+x|\). Since \(1+x > 0\) for the domain of interest, we can write \(\mu(x) = 1+x\).
Multiply the entire differential equation by the integrating factor \(\mu(x) = 1+x\) to get \((1+x)y' + y = \sqrt{x}\) multiplied by \(1+x\), which simplifies back to the left side being the derivative of \(\mu(x)y\), i.e., \(\frac{d}{dx}[(1+x)y] = (1+x) \cdot \frac{\sqrt{x}}{1+x} = \sqrt{x}\).
Integrate both sides with respect to \(x\): \(\int \frac{d}{dx}[(1+x)y] dx = \int \sqrt{x} dx\). This gives \((1+x)y = \int x^{1/2} dx + C\). Then solve for \(y\) by dividing both sides by \((1+x)\).

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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). It involves the first derivative of the unknown function and can be solved using an integrating factor. Recognizing this form is essential to apply the correct solution method.
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Classifying Differential Equations

Integrating Factor Method

The integrating factor is a function, usually denoted μ(x), used to multiply both sides of a linear differential equation to make the left side an exact derivative. It is found by μ(x) = e^(∫P(x)dx), enabling straightforward integration to solve for y.
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Euler's Method

Solving for y after Integration

After multiplying by the integrating factor, the equation becomes d/dx[μ(x)y] = μ(x)Q(x). Integrating both sides with respect to x and then isolating y gives the general solution. This step requires careful integration and algebraic manipulation.
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Related Practice
Textbook Question

Integral Equations

In Exercises 7–12, write an equivalent first-order differential equation

and initial condition for y.


y = ∫₁ ͯ 1/t dt

Textbook Question

First-Order Linear Equations

Solve the differential equations in Exercises 1–14.


y' + (tanx)y = cos²x, -π/2 < x < π/2

Textbook Question

Integral Equations

In Exercises 7–12, write an equivalent first-order differential equation

and initial condition for y.


y = 1 + ∫₀ ͯ y(t) dt

Textbook Question

First-Order Linear Equations

Solve the differential equations in Exercises 1–14.


(t-1)³ ds/dt + 4(t-1)²s = t+1, t >1

Textbook Question

Using Euler’s Method

In Exercises 15–20, use Euler’s method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places.


y' = 2xy + 2y, y(0) = 3, dx = 0.2

Textbook Question

Solving Initial Value Problems

Solve the initial value problems in Exercises 15–20.


θ dy/dθ + y = sin θ, θ > 0, y(π/2) = 1