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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.3.55
Chapter 7, Problem 7.3.55

Solve the initial value problems in Exercises 55–58.
55. dy/dt = e^t sin(e^t − 2),y(ln 2) = 0

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Identify the given differential equation: \(\frac{dy}{dt} = e^{t} \sin\left(e^{t} - 2\right)\) with the initial condition \(y(\ln 2) = 0\).
Recognize that this is a first-order ordinary differential equation that can be solved by direct integration since the right-hand side is expressed explicitly in terms of \(t\).
Set up the integral to find \(y(t)\) by integrating both sides with respect to \(t\): \(y(t) = \int e^{t} \sin\left(e^{t} - 2\right) \, dt + C\).
Use substitution to evaluate the integral: let \(u = e^{t} - 2\), then compute \(du = e^{t} dt\), which allows rewriting the integral as \(\int \sin(u) \, du\).
Integrate \(\sin(u)\) to get \(-\cos(u)\), then substitute back \(u = e^{t} - 2\) to express \(y(t)\), and finally use the initial condition \(y(\ln 2) = 0\) to solve for the constant \(C\).

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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 separable differential equation can be written as dy/dt = g(t)h(y), allowing the variables y and t to be separated on opposite sides of the equation. This enables integration with respect to each variable independently, which is essential for solving the given initial value problem.
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Integration of Composite Functions

Solving dy/dt = e^t sin(e^t − 2) requires integrating a function involving a composite argument, sin(e^t − 2). Recognizing the inner function and applying substitution methods simplifies the integral, making it possible to find the explicit solution.
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Initial Value Problems (IVP)

An initial value problem specifies the value of the solution at a particular point, here y(ln 2) = 0. This condition is used to determine the constant of integration after solving the differential equation, ensuring the solution fits the given initial condition.
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