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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 9.2.32

Solve the Bernoulli equations in Exercises 29–32.


x²y' + 2xy = y³

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1
Rewrite the given differential equation \(x^{2}y' + 2xy = y^{3}\) in the standard Bernoulli form. First, isolate \(y'\) by dividing the entire equation by \(x^{2}\): \[y' + \frac{2}{x}y = \frac{y^{3}}{x^{2}}\]
Identify the Bernoulli equation parameters: it has the form \[y' + P(x)y = Q(x)y^{n}\] where \(P(x) = \frac{2}{x}\), \(Q(x) = \frac{1}{x^{2}}\), and \(n = 3\).
Make the substitution \(v = y^{1-n} = y^{1-3} = y^{-2}\). Then compute \(v'\) in terms of \(y\) and \(y'\): \[v = y^{-2} \implies v' = -2y^{-3}y'\]
Express \(y'\) from the substitution: \[y' = -\frac{1}{2} y^{3} v'\] Substitute \(y'\) and \(y^{3}\) back into the original equation rewritten in step 1 to get a linear differential equation in terms of \(v\).
After substitution, simplify the resulting equation to the form \[v' + p(x)v = q(x)\] where \(p(x)\) and \(q(x)\) are functions of \(x\). Then solve this linear first-order ODE 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) = exp(∫P(x) dx). Multiplying through by μ(x) simplifies the equation to an exact derivative, enabling integration and solution for v, and subsequently y.
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