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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
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.1.15
Chapter 9, Problem 9.1.15

7–16. Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume C, C1, C2 and C3 are arbitrary constants.
u(t) = C₁t⁵ + C₂t⁻⁴ - t³; t²u''(t) - 20u(t) = 14t³

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Identify the given function and the differential equation: the function is \(u(t) = C_1 t^5 + C_2 t^{-4} - t^3\), and the differential equation is \(t^2 u''(t) - 20 u(t) = 14 t^3\).
Compute the first derivative \(u'(t)\) by differentiating each term of \(u(t)\) with respect to \(t\): use the power rule for derivatives.
Compute the second derivative \(u''(t)\) by differentiating \(u'(t)\) again with respect to \(t\).
Substitute \(u(t)\) and \(u''(t)\) into the left-hand side of the differential equation: calculate \(t^2 u''(t) - 20 u(t)\).
Simplify the expression obtained after substitution and verify if it equals the right-hand side \$14 t^3\( for all \)t\(, confirming that \)u(t)$ is a solution.

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Key Concepts

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

General Solution of Differential Equations

A general solution includes arbitrary constants representing all possible solutions to a differential equation. Verifying a solution involves substituting it into the equation to check if it satisfies the relationship for all values in the domain.
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Solutions to Basic Differential Equations

Second Derivative and Differentiation

The second derivative measures the rate of change of the first derivative and is essential in second-order differential equations. Calculating u''(t) accurately is crucial for substituting into the given differential equation.
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The Second Derivative Test: Finding Local Extrema

Substitution and Simplification in Differential Equations

To verify a solution, substitute the function and its derivatives into the differential equation. Simplify the resulting expression to confirm if both sides of the equation are equal, ensuring the function satisfies the equation.
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Related Practice
Textbook Question

12–16. Sketching direction fields Use the window [-2, 2] x [-2, 2] to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed.

y(x) = sin y, y(−2) = 1/2

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Textbook Question

Logistic growth in India The population of India was 435 million in 1960 (t=0) and 487 million in 1965 (t=5). The projected population for 2050 is 1.57 billion.

e. Discuss some possible shortcomings of this model. Why might the carrying capacity be either greater than or less than the value predicted by the model?

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Textbook Question

Direction fields Consider the direction field for the equation y′=y(2−y) shown in the figure and initial conditions of the form y(0)=A.

d. For what values of A are the corresponding solutions decreasing, for t≥0

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Textbook Question

11–18. Solving initial value problems Use the method of your choice to find the solution of the following initial value problems.

y′(t) = -3y + 9, y(0) = 4

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Textbook Question

What are the assumptions underlying the predator-prey model discussed in this section?

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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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