교과서 질문
Evaluate the integrals in Exercises 77–90.
79. ∫(from -1 to 0)6dt/√(3-2t-t²)
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Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- 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
7장, 문제 7.4.5
In Exercises 5–8, show that each function is a solution of the given initial value problem.
5. Differential Equation: 2y + y' = 4x + 2
Initial condition: y(-1) = e² - 2
Solution candidate: y = e^(-2x) + 2x
검증된 단계별 안내1
Identify the given differential equation: \(2y + y' = 4x + 2\) and the initial condition \(y(-1) = e^{2} - 2\).
Calculate the derivative of the candidate solution \(y = e^{-2x} + 2x\). Use the chain rule for the exponential term: \(y' = \frac{d}{dx}(e^{-2x}) + \frac{d}{dx}(2x) = -2e^{-2x} + 2\).
Substitute the candidate solution \(y\) and its derivative \(y'\) into the left side of the differential equation: \(2y + y' = 2(e^{-2x} + 2x) + (-2e^{-2x} + 2)\).
Simplify the expression from the substitution to verify if it equals the right side of the differential equation \(4x + 2\). This confirms whether the candidate function satisfies the differential equation.
Check the initial condition by substituting \(x = -1\) into the candidate solution \(y = e^{-2(-1)} + 2(-1)\) and verify if it equals \(e^{2} - 2\). This confirms the solution meets the initial value condition.
비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
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주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
Initial Value Problem (IVP)
An initial value problem consists of a differential equation paired with a specific condition that the solution must satisfy at a given point. This condition, called the initial condition, ensures a unique solution by specifying the value of the function at a particular input, such as y(-1) = e² - 2.
추천 영상:
05:03
Initial Value Problems
Verification of a Solution to a Differential Equation
To verify a candidate solution, substitute it and its derivative into the differential equation to check if the equation holds true for all x in the domain. This process confirms whether the proposed function satisfies the relationship defined by the differential equation.
추천 영상:
04:00Solutions to Basic Differential Equations
Differentiation of Exponential and Polynomial Functions
Differentiation rules for exponential functions like e^(-2x) and polynomial terms such as 2x are essential to find y'. The derivative of e^(-2x) is -2e^(-2x), and the derivative of 2x is 2, which are used to substitute into the differential equation for verification.
추천 영상:
6:13Exponential Functions
관련 실천
교과서 질문
Solve the initial value problems in Exercises 115–120.
117. dy/dx = 1/(x√(x² - 1)), x > 1; y(2) = π
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교과서 질문
Use l’Hôpital’s rule to find the limits in Exercises 7–52.
24. lim (x → π/2) (ln(csc x)) / (x - (π/2))²
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교과서 질문
Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76.
Try it—you just keep on cycling. Find the limits some other way.
73. lim (x → ∞) (2^x - 3^x) / (3^x + 4^x)
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교과서 질문
In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = e^(5-7x)
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교과서 질문
In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
29. y = (1 - t)coth⁻¹(√t)
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