Textbook Question
In Exercises 1–22, solve the differential equation.
y' = xeʸ√(x-2)
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Ch. 9 - First-Order Differential Equations
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 9, Problem 9.AAE.9
Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.
y' = y/x + cos ((y-x)/x)
Verified step by step guidance1
Rewrite the given differential equation: \(y' = \frac{y}{x} + \cos\left(\frac{y - x}{x}\right)\).
Recognize that the equation involves expressions of \(y\) and \(x\) in the form \(\frac{y}{x}\) and \(\frac{y - x}{x}\), which suggests a substitution using \(v = \frac{y}{x}\) to transform it into a homogeneous equation.
Express \(y\) in terms of \(v\) and \(x\): \(y = vx\). Then, differentiate both sides with respect to \(x\) using the product rule: \(y' = v + x \frac{dv}{dx}\).
Substitute \(y = vx\) and \(y' = v + x \frac{dv}{dx}\) back into the original equation to rewrite it entirely in terms of \(v\) and \(x\): \(v + x \frac{dv}{dx} = v + \cos(v - 1)\).
Simplify the equation by canceling terms and isolate \(\frac{dv}{dx}\) to get a separable differential equation in \(v\) and \(x\), which can then be solved by integration.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Homogeneous Differential Equations
A differential equation is homogeneous if it can be expressed so that the right-hand side is a function of the ratio y/x alone. This allows substitution like v = y/x to simplify the equation into a separable form, making it easier to solve.
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Classifying Differential Equations
Substitution Method (v = y/x)
By substituting v = y/x, the original variables y and x are transformed into a single variable v, reducing the differential equation to one involving v and x. This substitution leverages the homogeneity property to simplify the equation.
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Separable Differential Equations
Once the equation is rewritten in terms of v and x, it often becomes separable, meaning it can be expressed as a product of a function of v and a function of x. This allows integration on both sides to find the general solution.
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06:06Solving Separable Differential Equations
Related Practice
Textbook Question
In Exercises 39–42, use Euler’s method with the specified step size to estimate the value of the solution at the given point x*. Find the value of the exact solution at x*.
y' = 2y²(x-1), y(2) = -1/2, dx = 0.1, x* = 3
Textbook Question
Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.
(x²+y²)dx + xy dy = 0
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Textbook Question
Show that (0, 0) and (c/d, a/b) are equilibrium points. Explain the meaning of each of these points.
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Textbook Question
Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.
(x sin y/x - y cos y/x)dx + (x cos y/x) dy = 0
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Textbook Question
Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.
(x.exp(y/x) + y)dx - x dy = 0
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