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 9.4.40

Chapter 9, Problem 9.4.40

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

Verified step by step guidance

Recognize that the given differential equation is of the form \(a(t)y'(t) + a'(t)y(t) = f(t)\), where \(a(t) = t^3\). This means the left side can be expressed as the derivative of the product \(a(t)y(t)\), i.e., \(\frac{d}{dt}(t^3 y(t))\).

Rewrite the equation using this product rule form: \(\frac{d}{dt}(t^3 y(t)) = \frac{1 + t}{t}\).

Integrate both sides with respect to \(t\) to find \(t^3 y(t)\): \(\int \frac{d}{dt}(t^3 y(t)) \, dt = \int \frac{1 + t}{t} \, dt\).

Simplify the integral on the right side by splitting the fraction: \(\int \left( \frac{1}{t} + 1 \right) dt = \int \frac{1}{t} dt + \int 1 dt\).

After integrating, solve for \(y(t)\) by dividing both sides by \(t^3\), then use the initial condition \(y(1) = 6\) to find the constant of integration.

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Key Concepts

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

First-Order Linear Differential Equations

These are differential equations of the form y' + p(t)y = q(t), where p and q are functions of t. They can often be solved using integrating factors or by recognizing patterns that simplify the equation. Understanding their structure is essential for applying appropriate solution methods.

Product Rule and Recognizing Derivatives of Products

The product rule states that d/dt [a(t)y(t)] = a(t)y'(t) + a'(t)y(t). Recognizing when the left side of a differential equation matches this derivative allows rewriting the equation in a simpler form, facilitating direct integration to find solutions.

Initial Value Problems and Integration

An initial value problem specifies the value of the solution at a particular point, enabling determination of the integration constant after solving the differential equation. Integrating both sides with respect to t and applying the initial condition yields the unique solution.

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Initial Value Problems

Related Practice

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

33–42. Solving initial value problems Solve the following initial value problems.

y'(t) = 1 + eᵗ, y(0) = 4

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

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

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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