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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.5.68
Chapter 8, Problem 8.5.68

Solve the initial value problems in Exercises 67–70 for x as a function of t.
(3t⁴ + 4t² + 1) (dx/dt) = 2√3, x(1) = -π√3/4

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Start by isolating the derivative \( \frac{dx}{dt} \) in the given differential equation: \( (3t^{4} + 4t^{2} + 1) \frac{dx}{dt} = 2\sqrt{3} \). Divide both sides by \( 3t^{4} + 4t^{2} + 1 \) to get \( \frac{dx}{dt} = \frac{2\sqrt{3}}{3t^{4} + 4t^{2} + 1} \).
Rewrite the equation as \( dx = \frac{2\sqrt{3}}{3t^{4} + 4t^{2} + 1} dt \) to prepare for integration with respect to \( t \).
Integrate both sides: \( \int dx = \int \frac{2\sqrt{3}}{3t^{4} + 4t^{2} + 1} dt \). The left side integrates to \( x(t) + C \), where \( C \) is the constant of integration.
Focus on the integral \( \int \frac{2\sqrt{3}}{3t^{4} + 4t^{2} + 1} dt \). Consider factoring or substituting to simplify the denominator \( 3t^{4} + 4t^{2} + 1 \) if possible, or use partial fractions if applicable.
After finding the antiderivative, apply the initial condition \( x(1) = -\frac{\pi \sqrt{3}}{4} \) to solve for the constant \( C \). Substitute \( t = 1 \) and \( x = -\frac{\pi \sqrt{3}}{4} \) into your expression for \( x(t) \) and solve for \( C \).

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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 so that all terms involving one variable are on one side and all terms involving the other variable are on the opposite side. This allows integration of both sides separately to find the solution. Recognizing and rearranging the given equation into separable form is essential for solving it.
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Initial Value Problems (IVP)

An initial value problem specifies the value of the unknown function at a particular point, which helps determine the constant of integration after solving the differential equation. Using the given initial condition x(1) = -π√3/4 ensures the solution is unique and fits the problem's context.
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Initial Value Problems

Integration Techniques

Solving the differential equation involves integrating expressions with respect to t. Familiarity with integrating polynomials and constants, as well as handling square roots and constants, is necessary to find the explicit form of x(t). Proper integration leads to the general solution before applying initial conditions.
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Integration by Parts for Definite Integrals
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