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Chapter 4: Problem 87

Solve the initial value problems in Exercises \(71-90\) $$\frac{d^{3} y}{d x^{3}}=6 ; \quad y^{\prime \prime}(0)=-8, \quad y^{\prime}(0)=0, \quad y(0)=5$$

Short Answer

Expert verified
The solution is \( y = x^3 - 4x^2 + 5 \).

Step by step solution

01

Integrate the Differential Equation

Start with the given third-order differential equation: \[ \frac{d^{3} y}{d x^{3}} = 6 \] Integrating with respect to \(x\) gives: \[ \frac{d^{2} y}{d x^{2}} = 6x + C_1 \] where \(C_1\) is the constant of integration.
02

Apply the First Initial Condition

Use the initial condition \(y''(0) = -8\) to find \(C_1\). Substitute \(x = 0\) into the equation from Step 1:\[ \frac{d^{2} y}{d x^{2}}(0) = 6(0) + C_1 = -8 \] This simplifies to:\[ C_1 = -8 \] Thus, the equation becomes:\[ \frac{d^{2} y}{d x^{2}} = 6x - 8 \]
03

Integrate the Second Equation

Integrate the second-order equation with respect to \(x\):\[ \frac{dy}{dx} = \int (6x - 8) \, dx = 3x^2 - 8x + C_2 \] where \(C_2\) is another constant of integration.
04

Apply the Second Initial Condition

Use the initial condition \(y'(0) = 0\) to find \(C_2\). Substitute \(x = 0\) into the equation from Step 3:\[ \frac{dy}{dx}(0) = 3(0)^2 - 8(0) + C_2 = 0 \] This simplifies to:\[ C_2 = 0 \] So, the first-order equation becomes:\[ \frac{dy}{dx} = 3x^2 - 8x \]
05

Integrate the First Equation

Integrate the first-order equation:\[ y = \int (3x^2 - 8x) \, dx = x^3 - 4x^2 + C_3 \] where \(C_3\) is the final constant of integration.
06

Apply the Final Initial Condition

Use the initial condition \(y(0) = 5\) to determine \(C_3\). Substitute \(x = 0\) into the equation:\[ y(0) = (0)^3 - 4(0)^2 + C_3 = 5 \] This simplifies to:\[ C_3 = 5 \] So, the function \(y\) becomes:\[ y = x^3 - 4x^2 + 5 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are mathematical equations that involve functions and their derivatives. They help describe the relationship between a function and its rate of change. In the given exercise, we started with a third-order differential equation \(\frac{d^{3} y}{d x^{3}} = 6\). The order of the differential equation refers to the highest derivative present in the equation. Here, since we know the third derivative, our goal is to find the original function \(y\) by reversing the process of differentiation, known as integration. Understanding how to work through differential equations is crucial in many fields such as physics, engineering, and economics.
Integration
Integration is a mathematical operation that allows us to find the original function given its derivative. In our problem, we integrated the third-order derivative to successively find the lower-order derivatives and eventually the function \(y\). Let's break down the integration process used:
  • We started with the third-order derivative, \(\frac{d^{3} y}{d x^{3}} = 6\), integrating it gave us \(\frac{d^{2} y}{d x^{2}} = 6x + C_1\).
  • Subsequent integration of the second-order derivative \(\frac{d^{2} y}{d x^{2}} = 6x - 8\) resulted in the first derivative \(\frac{dy}{dx} = 3x^2 - 8x + C_2\).
  • Finally, integrating \(\frac{dy}{dx} = 3x^2 - 8x\) led us to the original function \(y = x^3 - 4x^2 + C_3\).
These integrations help decipher the underlying function that fulfills the conditions provided in the problem.
Boundary Conditions
Boundary conditions, also known as initial conditions in the context of differential equations, are values given for the function or its derivatives at specific points. They help us determine the constants of integration that appear when we reverse differentiation through integration. In the exercise, we had three boundary conditions:
  • \(y''(0) = -8\) helped find \(C_1\).
  • \(y'(0) = 0\) was used to determine \(C_2\).
  • Finally, \(y(0) = 5\) allowed us to compute \(C_3\).
Using these boundary conditions ensures the solution meets specific requirements, confirming it is the correct representation of the original problem.
Constant of Integration
When we integrate a function, an arbitrary constant, known as the constant of integration, is added because the derivative of any constant is zero. Without any initial conditions, there are infinitely many solutions as this constant could be any number. In the step-by-step solution:
  • \(C_1\) appeared when we integrated the third-order derivative. Using \(y''(0)=-8\) allowed us to find \(C_1 = -8\).
  • After further integration, another constant \(C_2\) emerged and was determined to be \(0\) using \(y'(0)=0\).
  • Finally, \(C_3\) was introduced after integrating the first-order derivative and set to \(5\) using \(y(0)=5\).
These constants tailor the general solution to the specific conditions of the problem, ensuring accuracy and precision in modeling real-world scenarios.

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Most popular questions from this chapter

Projectile motion The range \(R\) of a projectile fired from the origin over horizontal ground is the distance from the origin to the point of impact. If the projectile is fired with an initial velocity \(v_{0}\) at an angle \(\alpha\) with the horizontal, then in Chapter 13 we find that \begin{equation}R=\frac{v_{0}^{2}}{g} \sin 2 \alpha,\end{equation} where \(g\) is the downward acceleration due to gravity. Find the angle \(\alpha\) for which the range \(R\) is the largest possible.

Solve the initial value problems in Exercises \(71-90\) $$\frac{d^{3} \theta}{d t^{3}}=0 ; \quad \theta^{\prime \prime}(0)=-2, \quad \theta^{\prime}(0)=-\frac{1}{2}, \quad \theta(0)=\sqrt{2}$$

Right, or wrong? Say which for each formula and give a brief reason for each answer. \begin{equation}\begin{array}{l}{\text { a. }{\int x \sin x d x=\frac{x^{2}}{2} \sin x+C}} \\ {\text { b. } \int x \sin x d x=-x \cos x+C} \\\ {\text { c. } \int x \sin x d x=-x \cos x+\sin x+C}\end{array}\end{equation}

Find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function's first and second derivatives. How are the values at which these graphs intersect the \(x\)-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? \begin{equation}y=x^{5}-5 x^{4}-240\end{equation}

Solve the initial value problems in Exercises \(71-90\) . $$\frac{d y}{d x}=\frac{1}{x^{2}}+x, \quad x>0 ; \quad y(2)=1$$
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