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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.AAE.5
Chapter 9, Problem 9.AAE.5

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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1
Identify the given differential equation: \( (x^{2} + y^{2}) \, dx + x y \, dy = 0 \). Our goal is to rewrite it in the form \( M(x,y) \, dx + N(x,y) \, dy = 0 \) and check if it is homogeneous.
Recall that a function \(f(x,y)\) is homogeneous of degree \(n\) if \(f(tx, ty) = t^{n} f(x,y)\) for all \(t\). Check the degrees of \(M(x,y) = x^{2} + y^{2}\) and \(N(x,y) = x y\) to confirm homogeneity.
Since \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions of degree 2, the differential equation is homogeneous. Next, use the substitution \(v = \frac{y}{x}\) to reduce the equation to a separable form.
Express \(y\) as \(v x\), then compute \(dy = v \, dx + x \, dv\). Substitute \(y\) and \(dy\) into the original equation to rewrite it entirely in terms of \(x\), \(v\), and \(dv\).
After substitution, simplify the resulting expression and separate variables to obtain an equation involving \(v\) and \(x\) that can be integrated to find the implicit solution.

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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 all terms are of the same degree when variables are scaled. Typically, it can be written as M(x,y)dx + N(x,y)dy = 0 where M and N are homogeneous functions of the same degree. Recognizing this form allows the use of substitution methods to simplify the equation.
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Substitution Method (y = vx or x = vy)

To solve homogeneous equations, the substitution y = vx (or x = vy) is used to reduce the equation to a separable form. This substitution leverages the homogeneity by expressing y in terms of x and a new variable v, simplifying the differential equation into one involving v and x only, which is easier to solve.
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Separable Differential Equations

After substitution, the resulting equation often becomes separable, meaning it can be written as g(v) dv = h(x) dx. This allows integration of each side independently. Understanding how to separate variables and integrate is essential to find the general solution of the original differential equation.
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Related Practice
Textbook Question

In Exercises 1–22, solve the differential equation.

y' = xeʸ√(x-2)

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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.


y' = y/x + cos ((y-x)/x)

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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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Textbook Question

In Exercises 1–22, solve the differential equation.


sec x dy + x cos² y dx = 0

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