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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 9.1.27

21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.
u''(x) = 55x⁹ + 36x⁷ - 21x⁵ + 10x⁻³

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Recognize that the given differential equation is a second-order ordinary differential equation of the form \(u''(x) = f(x)\), where \(f(x) = 55x^{9} + 36x^{7} - 21x^{5} + 10x^{-3}\).
To find the general solution \(u(x)\), integrate the right-hand side function \(f(x)\) twice with respect to \(x\). The first integration will give \(u'(x)\), and the second integration will give \(u(x)\).
Perform the first integration: calculate \(u'(x) = \int (55x^{9} + 36x^{7} - 21x^{5} + 10x^{-3}) \, dx\). Integrate each term separately using the power rule for integration, which states \(\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C\) for \(n \neq -1\).
After finding \(u'(x)\), perform the second integration: calculate \(u(x) = \int u'(x) \, dx\). Again, integrate each term separately and add a new arbitrary constant of integration.
Combine the results to write the general solution as \(u(x) = \) (the expression from the second integration) \(+ C_1 x + C_2\), where \(C_1\) and \(C_2\) are arbitrary constants representing the general solution's family.

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Key Concepts

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

Second-Order Differential Equations

A second-order differential equation involves the second derivative of an unknown function. Solving such equations means finding a function whose second derivative satisfies the given equation. The general solution includes all possible functions that fit the equation, often expressed with arbitrary constants.
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Integration to Find General Solutions

To solve u''(x) = f(x), integrate the right-hand side twice with respect to x. Each integration introduces an arbitrary constant, reflecting the family of solutions. This process transforms the differential equation into an explicit formula for u(x).
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Handling Polynomial and Negative Powers in Integration

When integrating terms like x⁹ or x⁻³, apply the power rule: ∫x^n dx = x^(n+1)/(n+1) for n ≠ -1. For negative powers, ensure the integral is defined and carefully add constants. This technique is essential for integrating the given right-hand side accurately.
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Related Practice
Textbook Question

33–38. {Use of Tech} Solutions in implicit form Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one function, be sure to indicate which function corresponds to the solution of the initial value problem.

yy'(x) = 2x/(2 + y)², y(1) = −1

Textbook Question

Solution of the logistic equation Use separation of variables to show that the solution of the initial value problem

P'(t) = rP (1-P/K), P(0) = P₀

is P(t) = K/((K/P₀ − 1)e⁻ʳᵗ + 1)

Textbook Question

9–14. Growth rate functions Make a sketch of the population function P (as a function of time) that results from the following growth rate functions. Assume the population at time t = 0 begins at some positive value.


Textbook Question

23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.


B′(t) = 0.005B − 500, B(0) = 50,000

Textbook Question

20–22. {Use of Tech} Solving the Gompertz equation Solve the Gompertz equation in Exercise 19 with the given values of r, K, and M₀. Then graph the solution to be sure that M(0) and lim(t→∞) M(t) are correct.


r = 0.05, K = 1200, M₀ = 90

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Textbook Question

39–42. Special equations A special class of first-order linear equations have the form a(t)y'(t)+a'(t)y(t)=f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form

a(t)y'(t) + a'(t)y(t) = d/dt (a(t)y(t)) = f(t). 

Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following initial value problems. 


t³y′(t) + 3t²y = (1 + t)/t, y(1) = 6

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