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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.7.87
Chapter 4, Problem 4.7.87

Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5

Verified step by step guidance
1
Recognize that the given differential equation is a third-order ordinary differential equation: \(\frac{d^{3}y}{dx^{3}} = 6\). Our goal is to find the function \(y(x)\) that satisfies this equation along with the initial conditions.
Integrate the differential equation once with respect to \(x\) to find the second derivative \(y''(x)\). Since \(\frac{d^{3}y}{dx^{3}} = 6\), integrating gives \(y''(x) = 6x + C_1\), where \(C_1\) is an integration constant.
Use the initial condition \(y''(0) = -8\) to solve for \(C_1\). Substitute \(x=0\) into \(y''(x) = 6x + C_1\) and set it equal to \(-8\) to find \(C_1\).
Integrate \(y''(x)\) to find the first derivative \(y'(x)\). This gives \(y'(x) = \int y''(x) \, dx = \int (6x + C_1) \, dx = 3x^{2} + C_1 x + C_2\), where \(C_2\) is another integration constant.
Use the initial condition \(y'(0) = 0\) to solve for \(C_2\). Substitute \(x=0\) into \(y'(x)\) and set it equal to \(0\) to find \(C_2\). Then integrate \(y'(x)\) once more to find \(y(x)\), and use the initial condition \(y(0) = 5\) to solve for the last constant.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Higher-Order Derivatives and Integration

This involves understanding derivatives beyond the first order, such as the third derivative given here. To solve the differential equation, you integrate the third derivative step-by-step to find the second, first derivatives, and finally the original function, adding constants of integration at each step.
Recommended video:
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Higher Order Derivatives

Initial Conditions in Differential Equations

Initial conditions specify the values of the function and its derivatives at a particular point, allowing us to determine the constants of integration uniquely. Here, values for y(0), y′(0), and y″(0) are given, which help solve for the constants after integration.
Recommended video:
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Solutions to Basic Differential Equations

Solving Initial Value Problems (IVPs)

An IVP requires finding a function that satisfies a differential equation and meets given initial conditions. The process involves integrating the differential equation and applying the initial values to find the specific solution that fits the problem.
Recommended video:
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Initial Value Problems
Related Practice
Textbook Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

71. y' = x(x² - 12)

Textbook Question

Finding Extrema from Graphs


In Exercises 1–6, determine from the graph whether the function has any absolute extreme values on [a, b]. Then explain how your answer is consistent with Theorem 1.


Textbook Question

Root Finding

2. Use Newton's method to estimate the one real solution of x^3 +3x + 1 = 0. Start with x_0 = 0 and then find x_2.

Textbook Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

82. y' = sin t, for 0 ≤ t ≤ 2π

Textbook Question

109. Suppose the derivative of the function y = f(x) is

y'=(x-1)^2(x-2).

At what points, if any, does the graph of f have a local minimum, local maximum, or

point of inflection? (Hint: Draw the sign pattern for y'.)

Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(1 − cot²x) dx