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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 9.1.18
Chapter 9, Problem 9.1.18

17–20. Verifying solutions of initial value problems Verify that the given function y is a solution of the initial value problem that follows it.
y(t) = 8t⁶ - 3; ty'(t) - 6y(t) = 18, y(1) = 5

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1
Start by identifying the given function and the initial value problem. The function is \(y(t) = 8t^{6} - 3\), and the differential equation is \(t y'(t) - 6 y(t) = 18\) with the initial condition \(y(1) = 5\).
Find the derivative of the function \(y(t)\) with respect to \(t\). Use the power rule to differentiate \(y(t) = 8t^{6} - 3\), which gives \(y'(t) = 48t^{5}\).
Substitute \(y(t)\) and \(y'(t)\) into the differential equation \(t y'(t) - 6 y(t) = 18\). This means replacing \(y'(t)\) with \$48t^{5}\( and \)y(t)$ with \(8t^{6} - 3\).
Simplify the left-hand side expression after substitution: calculate \(t imes 48t^{5} - 6 imes (8t^{6} - 3)\) and simplify the terms to check if it equals 18 for all \(t\).
Verify the initial condition by substituting \(t = 1\) into the original function \(y(t) = 8t^{6} - 3\) and check if \(y(1) = 5\) holds true.

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Key Concepts

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

Initial Value Problem (IVP)

An initial value problem consists of a differential equation along with a specified value of the unknown function at a particular point. Solving an IVP means finding a function that satisfies both the differential equation and the initial condition, ensuring the solution is unique and fits the given starting value.
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Verification of a Solution

To verify a solution to a differential equation, substitute the proposed function and its derivative into the equation to check if the equality holds. Additionally, confirm that the function satisfies the initial condition by evaluating it at the given point.
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Differentiation of Polynomial Functions

Differentiation involves finding the derivative of a function, which represents its rate of change. For polynomial functions like y(t) = 8t⁶ - 3, apply the power rule by multiplying the coefficient by the exponent and reducing the exponent by one to find y'(t).
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Related Practice
Textbook Question

17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.

y'(t) = cos² y, y(1) = π/4

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

5–10. First-order linear equations Find the general solution of the following equations.


v'(y) − v/2 = 14

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Textbook Question

Orthogonal trajectories Use the method in Exercise 44 to find the orthogonal trajectories for the family of circles x² + y² = a²

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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′(x) = 4x csc y, y(0) = π/2

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.