Express the solutions of the initial value problems in Exercises 35 and 36 in terms of integrals.

dy/dx = sin x/x , y(5) = -3

검증된 단계별 안내

Identify the given differential equation and initial condition: \( \frac{dy}{dx} = \frac{\sin x}{x} \) with \( y(5) = -3 \).

Recognize that this is a first-order differential equation where \( \frac{dy}{dx} \) is given explicitly as a function of \( x \).

To find \( y(x) \), integrate both sides with respect to \( x \): \[ y(x) = \int \frac{\sin x}{x} \, dx + C \], where \( C \) is the constant of integration.

Use the initial condition \( y(5) = -3 \) to solve for \( C \): \[ -3 = \int_{}^{5} \frac{\sin t}{t} \, dt + C \]. Note that the variable of integration is changed to \( t \) to avoid confusion.

Express the solution explicitly in terms of an integral from 5 to \( x \): \[ y(x) = -3 + \int_5^{x} \frac{\sin t}{t} \, dt \]. This represents the solution in integral form.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Initial Value Problems (IVPs)

An initial value problem involves a differential equation along with a specified value of the unknown function at a given point. This initial condition allows us to find a unique solution that satisfies both the differential equation and the initial value.

Separable Differential Equations and Integration

When a differential equation is given in the form dy/dx = f(x), the solution can be found by integrating f(x) with respect to x. This process involves expressing y as an integral of the function plus a constant determined by the initial condition.

Definite Integrals and Applying Initial Conditions

To express the solution in terms of integrals, we use definite integrals with limits corresponding to the initial condition. This approach incorporates the initial value directly, allowing the constant of integration to be evaluated and the solution to be written explicitly.

Express the solutions of the initial value problems in Exercises 35 and 36 in terms of integrals. dy/dx = sin x/x , y(5) = -3