교과서 질문
Integral Equations
In Exercises 7–12, write an equivalent first-order differential equation
and initial condition for y.
y = ln x + ∫ₓᵉ √ (t² + (y(t))²) dt
16
views
Ch. 9 - First-Order Differential Equations
Hass15th EditionThomas' CalculusISBN: 9780137616077
당신이 사용하는 게 아니라요?교과서 변경다른 교과서를 선택하세요
TOC.tableOfContents
- 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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
9장, 문제 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)
검증된 단계별 안내1
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.
비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
6m도움이 되었나요?
주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
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.
추천 영상:
07:39
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.
추천 영상:
06:30Disk Method Using y-Axis
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.
추천 영상:
06:06Solving Separable Differential Equations
관련 실천
교과서 질문
Write the formula for a logistic function that has values between y = 0 and y = 1, crosses the line y = 1/2 at x = 0, and has slope 5 at this point.
15
views
교과서 질문
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
35
views
교과서 질문
Show that (0, 0) and (c/d, a/b) are equilibrium points. Explain the meaning of each of these points.
26
views
교과서 질문
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
24
views
교과서 질문
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
40
views