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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 4, Problem 4.7.73

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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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 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.
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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.
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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.
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Initial Value Problems