Initial Value Problems

Solve the initial value problems in Exercises 71–90.

dy/dx = 1/x² + x, x > 0; y(2) = 1

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Identify the given differential equation and initial condition: \( \frac{dy}{dx} = \frac{1}{x^{2}} + x \), with \( y(2) = 1 \).

Rewrite the differential equation to express \( dy \) in terms of \( dx \): \( dy = \left( \frac{1}{x^{2}} + x \right) dx \).

Integrate both sides with respect to \( x \) to find the general solution: \( y = \int \left( \frac{1}{x^{2}} + x \right) dx + C \).

Compute the integral by splitting it into two parts: \( \int \frac{1}{x^{2}} dx \) and \( \int x \, dx \), then combine the results.

Use the initial condition \( y(2) = 1 \) to solve for the constant of integration \( C \) by substituting \( x = 2 \) and \( y = 1 \) into the general solution.

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주요 개념

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

Separable Differential Equations

A separable differential equation can be written as a product of a function of x and a function of y, allowing the variables to be separated on opposite sides of the equation. This technique simplifies solving by integrating each side independently.

Integration of Functions

Solving dy/dx = f(x) involves integrating the right-hand side with respect to x to find the general solution y(x). Understanding how to integrate functions like 1/x² and x is essential for finding the antiderivative.

Initial Conditions and Particular Solutions

An initial value problem includes a condition like y(2) = 1, which allows determination of the constant of integration after finding the general solution. This yields a unique particular solution satisfying the given initial condition.

Initial Value ProblemsSolve the initial value problems in Exercises 71–90.dy/dx = 1/x² + x, x > 0; y(2) = 1