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Chapter 6: Problem 10

Find the solution of the initial value problem. $$\frac{d y}{d x}=x^{5}+x^{6}, \quad y(1)=2$$

Short Answer

Expert verified
The solution is \( y = \frac{x^6}{6} + \frac{x^7}{7} + \frac{71}{42} \).

Step by step solution

01

Recognize the Form of the Differential Equation

The given differential equation is \( \frac{dy}{dx} = x^5 + x^6 \). This is a straightforward first-order ordinary differential equation where the right side is a polynomial.
02

Integrate Both Sides with Respect to x

To solve \( \frac{dy}{dx} = x^5 + x^6 \), integrate both sides with respect to \( x \). The integral of the left side will just be \( y \), and for the right side, integrate each term separately: \[ \int (x^5 + x^6) \, dx = \int x^5 \, dx + \int x^6 \, dx \].
03

Compute the Integrals

Integrate each term: - \( \int x^5 \, dx = \frac{x^6}{6} + C_1 \)- \( \int x^6 \, dx = \frac{x^7}{7} + C_2 \)Thus, the solution after integration is:\[ y = \frac{x^6}{6} + \frac{x^7}{7} + C \].
04

Apply the Initial Condition to Find C

Use the initial condition \( y(1) = 2 \).Substitute \( x = 1 \) and \( y = 2 \) into the integrated function:\[ 2 = \frac{1^6}{6} + \frac{1^7}{7} + C \].Simplify and solve for \( C \):\[ 2 = \frac{1}{6} + \frac{1}{7} + C \].Calculate \( \frac{1}{6} + \frac{1}{7} = \frac{7}{42} + \frac{6}{42} = \frac{13}{42} \).Thus, \( C = 2 - \frac{13}{42} = \frac{84}{42} - \frac{13}{42} = \frac{71}{42} \).
05

Write the Final Solution

Now substitute \( C \) back into the general solution to get the specific solution satisfying the initial condition:\[ y = \frac{x^6}{6} + \frac{x^7}{7} + \frac{71}{42} \].

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Key Concepts

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

Integration Techniques
Solving first-order ordinary differential equations often requires using integration techniques. In this problem, we are given the differential equation \( \frac{dy}{dx} = x^5 + x^6 \), which we recognize as a polynomial. This identification is crucial because it guides us to integrate each term separately. Here's how we accomplish this:
  • First, rewrite the equation in terms of integrals: \( y = \int (x^5 + x^6) \, dx \).
  • Separate the integrals: \( \int x^5 \, dx + \int x^6 \, dx \).
These steps demonstrate fundamental integration techniques:
  • For \( \int x^5 \, dx \), increase the exponent by one and divide: \( \frac{x^6}{6} \).
  • Similarly, for \( \int x^6 \, dx \), the result is \( \frac{x^7}{7} \).
The integration results give us a general solution: \( y = \frac{x^6}{6} + \frac{x^7}{7} + C \). In this equation, \( C \) represents an unknown constant, which we will determine using other information provided in the problem.
Initial Value Problems
Initial value problems involve a differential equation and conditions specified at a particular point. These problems provide not just the form of the solution, but also let us determine specific constants. In our case, the differential equation is \( \frac{dy}{dx} = x^5 + x^6 \). The initial condition given is \( y(1) = 2 \). This condition means that when \( x = 1 \), \( y \) must equal 2.To apply this:
  • Substitute \( x = 1 \) into the general solution \( y = \frac{x^6}{6} + \frac{x^7}{7} + C \).
  • This gives us the equation: \( 2 = \frac{1^6}{6} + \frac{1^7}{7} + C \).
  • Calculate the values: \( \frac{1}{6} + \frac{1}{7} = \frac{13}{42} \).
  • Resolve for \( C \): \( C = 2 - \frac{13}{42} = \frac{71}{42} \).
This process lets us find the specific constant \( C \), fully determining our solution.
Polynomial Functions
Polynomial functions play a significant role in calculus, especially when dealing with integrations and differential equations. In this problem, the function \( x^5 + x^6 \) is a polynomial. Polynomials are generally in the form of expression consisting of terms with coefficients multiplied by variables raised to a nonnegative integer exponent.Key characteristics of polynomial functions:
  • They are continuous and smooth, which means you don't have any breaks or gaps.
  • When integrating polynomials, you increase each term's exponent by one and divide by the new exponent.
  • Polynomials can be as simple as a linear function or as complex as multi-variable high degree functions.
In differential equations, polynomial right-hand sides are relatively easy to manage, both for integration and differentiation. Recognizing the structure early can help solve differential equations effectively, particularly when it involves initial value problems.

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Most popular questions from this chapter

Explain what is wrong with the statement. $$\frac{d}{d x} \int_{0}^{5} t^{2} d t=x^{2}$$

On the moon, the acceleration due to gravity is about \(1.6 \mathrm{m} / \mathrm{sec}^{2}\) (compared to \(g \approx 9.8 \mathrm{m} / \mathrm{sec}^{2}\) on earth). If you drop a rock on the moon (with initial velocity 0 ), find formulas for: (a) Its velocity, \(v(t),\) at time \(t\). (b) The distance, \(s(t),\) it falls in time \(t\).

Are the statements true or false? Give an explanation for your answer. If \(y=F(x)\) is a solution to the differential equation \(d y / d x=f(x),\) then \(F(x)\) is an antiderivative of \(f(x)\).

An object is shot vertically upward from the ground with an initial velocity of \(160 \mathrm{ft} / \mathrm{sec}\). (a) At what rate is the velocity decreasing? Give units. (b) Explain why the graph of velocity of the object against time (with upward positive) is a line. (c) Using the starting velocity and your answer to part (b), find the time at which the object reaches the highest point. (d) Use your answer to part (c) to decide when the object hits the ground. (e) Graph the velocity against time. Mark on the graph when the object reaches its highest point and when it lands. (f) Find the maximum height reached by the object by considering an area on the graph. (g) Now express velocity as a function of time, and find the greatest height by antidifferentiation.

Use the chain rule to calculate the derivative. $$\frac{d}{d x} \int_{-x^{2}}^{x^{2}} e^{t^{2}} d t$$
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