Initial Value Problems

Solve the initial value problems in Exercises 71–90.

dy/dx = 3x⁻²ᐟ³, y(−1) = −5

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Identify the given differential equation and initial condition: \( \frac{dy}{dx} = 3x^{-\frac{2}{3}} \) with \( y(-1) = -5 \).

Rewrite the differential equation in integral form: \( dy = 3x^{-\frac{2}{3}} dx \).

Integrate both sides with respect to \( x \): \( y = \int 3x^{-\frac{2}{3}} dx + C \), where \( C \) is the constant of integration.

Use the power rule for integration: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \), making sure to add 1 to the exponent \( -\frac{2}{3} + 1 = \frac{1}{3} \).

Apply the initial condition \( y(-1) = -5 \) to solve for \( C \) by substituting \( x = -1 \) and \( y = -5 \) into the integrated expression.

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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 Power Functions

Integrating power functions involves applying the power rule for integration, which states ∫x^n dx = (x^(n+1))/(n+1) + C, for n ≠ -1. Understanding how to handle fractional and negative exponents is essential for solving the given differential equation.

Initial Conditions and Particular Solutions

Initial conditions specify the value of the solution at a particular point, allowing determination of the constant of integration. This transforms the general solution of a differential equation into a unique particular solution that satisfies the given condition.

Initial Value ProblemsSolve the initial value problems in Exercises 71–90.dy/dx = 3x⁻²ᐟ³, y(−1) = −5