Skip to main content
Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
9장, 문제 9.2.30

Solve the Bernoulli equations in Exercises 29–32.


y' - y = xy²

검증된 단계별 안내
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.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
8m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
07:39
Classifying Differential Equations

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.
추천 영상:
07:33
Euler's Method

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.
추천 영상:
06:06
Solving Separable Differential Equations