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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 9.1.23
Chapter 9, Problem 9.1.23

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)?

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Identify the differential equation and initial condition: \(\frac{dy}{dx} = \frac{y^2}{\sqrt{x}}\) with \(y(1) = -1\).
Set the step size \(\Delta x = 0.5\) and determine the points at which to approximate \(y\): starting from \(x=1\), the points are \$1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5$.
Apply Euler's method iteratively using the formula \(y_{n+1} = y_n + \Delta x \cdot f(x_n, y_n)\), where \(f(x,y) = \frac{y^2}{\sqrt{x}}\). For each step, calculate the slope \(f(x_n, y_n)\) and update \(y\) accordingly.
Continue this process until you reach \(x=5\), which will give you the Euler's method estimate for \(y(5)\).
To find the exact value of \(y(5)\), solve the differential equation analytically by separating variables and integrating both sides, then apply the initial condition to find the constant of integration and evaluate \(y\) at \(x=5\).

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Key Concepts

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

Euler's Method

Euler's method is a numerical technique to approximate solutions of first-order differential equations. It uses a step size (dx) to incrementally estimate the value of the function by applying the slope (derivative) at the current point. This method is especially useful when an exact solution is difficult to find.
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Euler's Method

Solving First-Order Differential Equations

A first-order differential equation relates a function and its derivative. To find the exact solution, one often separates variables or uses an integrating factor. Understanding how to manipulate and solve these equations is essential to find the exact value of y(5) in this problem.
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Solving Separable Differential Equations

Initial Value Problems (IVP)

An initial value problem specifies the value of the function at a starting point, here y(1) = -1. This condition allows for a unique solution to the differential equation and is crucial for both numerical methods like Euler's and for finding the exact solution.
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Initial Value Problems
Related Practice
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

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).

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

Textbook Question

Solving Initial Value Problems

Solve the initial value problems in Exercises 15–20.

dy/dt + 2y = 3, y(0) = 1