Textbook Question
A family of exponential functions
b. Verify that the arc length of the curve y=f(x) on the interval [0, ln 2] is A(2^a-1) - 1/4a²A (2^-a - 1).
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Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 9.3.21
17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y'(t) = yeᵗ, y(0) = −1
Verified step by step guidance1
First, identify if the differential equation is separable. The given equation is \(y'(t) = y e^{t}\). Since the right side can be written as a product of a function of \(y\) and a function of \(t\), it is separable.
Rewrite the differential equation using Leibniz notation: \(\frac{dy}{dt} = y e^{t}\). Then separate variables by dividing both sides by \(y\) and multiplying both sides by \(dt\): \(\frac{1}{y} dy = e^{t} dt\).
Integrate both sides: \(\int \frac{1}{y} dy = \int e^{t} dt\). This will give you the antiderivatives on each side.
After integrating, you will have \(\ln|y| = e^{t} + C\), where \(C\) is the constant of integration. Solve for \(y\) by exponentiating both sides: \(y = \pm e^{e^{t} + C} = A e^{e^{t}}\), where \(A = \pm e^{C}\).
Use the initial condition \(y(0) = -1\) to find the constant \(A\). Substitute \(t=0\) and \(y=-1\) into the solution and solve for \(A\). Then write the particular solution.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Separable Differential Equations
A differential equation is separable if it can be written as a product of a function of y and a function of t, allowing the variables to be separated on opposite sides of the equation. This form enables integration with respect to each variable independently to find the solution.
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Solving Separable Differential Equations
Initial Value Problems (IVP)
An initial value problem involves solving a differential equation with a given initial condition, such as y(t₀) = y₀. The initial condition helps determine the specific solution curve among the family of solutions by fixing the constant of integration.
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Integration Techniques for Solving ODEs
Once variables are separated, integration is used to solve each side with respect to its variable. This often involves integrating exponential functions or polynomials, and applying the initial condition to solve for the integration constant to find the explicit solution.
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06:18Integration by Parts for Definite Integrals
Related Practice
Textbook Question
23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.
B′(t) = 0.004B − 800, B(0) = 40,000
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Textbook Question
Does the function y(t) = 2t satisfy the differential equation y'''(t) + y'(t) = 2?
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Textbook Question
11–16. Initial value problems Solve the following initial value problems.
y'(x) = −y + 2, y(0) = −2
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