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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주요 개념

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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.

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.

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.

Solve the initial value problems in Exercises 55–58.55. dy/dt = e^t sin(e^t − 2),y(ln 2) = 0