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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
Chapter 9, Problem 9.2.30

Solve the Bernoulli equations in Exercises 29–32.


y' - y = xy²

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1
Identify the given differential equation as a Bernoulli equation. The general form of a Bernoulli equation is \(y' + P(x)y = Q(x)y^n\). Here, rewrite the equation \(y' - y = xy^2\) as \(y' + (-1) y = x y^2\), so \(P(x) = -1\), \(Q(x) = x\), and \(n = 2\).
Make the substitution \(v = y^{1-n} = y^{1-2} = y^{-1}\). This substitution transforms the nonlinear equation into a linear one in terms of \(v\).
Differentiate \(v = y^{-1}\) with respect to \(x\) to find \(v'\). Using the chain rule, \(v' = -y^{-2} y' = -\frac{y'}{y^2}\).
Rewrite the original equation in terms of \(v\) and \(v'\). From the original equation, express \(y'\) as \(y' = y + x y^2\). Substitute \(y' = -y^2 v'\) and \(y = \frac{1}{v}\) to get an equation involving \(v\) and \(v'\).
Simplify the resulting equation to obtain a linear first-order differential equation in \(v\): \(v' + P(x)(1-n) v = (1-n) Q(x)\). Then solve this linear equation using an integrating factor.

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Key Concepts

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

Bernoulli Differential Equation

A Bernoulli differential equation has the form y' + P(x)y = Q(x)y^n, where n is any real number other than 0 or 1. It is nonlinear due to the y^n term but can be transformed into a linear equation using an appropriate substitution.
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Substitution Method for Bernoulli Equations

To solve a Bernoulli equation, use the substitution v = y^(1-n), which converts the nonlinear equation into a linear differential equation in terms of v. This allows the use of standard methods for linear equations to find the solution.
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Solving Linear First-Order Differential Equations

Once transformed, the equation becomes linear and can be solved using an integrating factor, μ(x) = e^(∫P(x)dx). Multiplying through by μ(x) simplifies the equation, enabling integration and solution for the dependent variable.
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