Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 9 - Differential Equations

Problem 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

Verified step by step guidance

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.

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.

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

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.

y(t) = C₁ sin4t + C₂ cos4t; y''(t) + 16y(t) = 0

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Textbook Question

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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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

110

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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

What is the equilibrium solution of the equation y'(t) = 3y − 9? Is it stable or unstable?

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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

Verified video answer for a similar problem:

Key Concepts

Initial Value Problem (IVP)

Verification of a Solution

Differentiation of Polynomial Functions

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