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 9.4.42

Chapter 9, Problem 9.4.42

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.

(t² + 1)y′(t) + 2ty = 3t², y(2) = 8

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 \(a(t) = t^2 + 1\) and \(f(t) = 3t^2\).

Rewrite the left side as the derivative of a product: \(\frac{d}{dt} \big(a(t) y(t)\big) = f(t)\), which means \(\frac{d}{dt} \big((t^2 + 1) y(t)\big) = 3t^2\).

Integrate both sides with respect to \(t\): \(\int \frac{d}{dt} \big((t^2 + 1) y(t)\big) dt = \int 3t^2 dt\).

After integration, express the solution as \((t^2 + 1) y(t) = \int 3t^2 dt + C\), where \(C\) is the constant of integration.

Use the initial condition \(y(2) = 8\) to substitute \(t=2\) and \(y=8\) into the equation to solve for \(C\), then write the explicit formula for \(y(t)\) by dividing both sides by \((t^2 + 1)\).

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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, such as rewriting the left side as a derivative of a product.

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). Identifying that the left side of the equation matches this derivative allows rewriting the differential equation in a simpler form, facilitating direct integration.

Solving Initial Value Problems by Integration

Once the equation is expressed as d/dt [a(t)y(t)] = f(t), integrating both sides with respect to t yields a(t)y(t) = ∫f(t) dt + C. Applying the initial condition y(t₀) = y₀ helps determine the constant of integration, providing a unique solution.

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Initial Value Problems

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Textbook Question

11–16. Initial value problems Solve the following initial value problems.

y'(t) − 3y = 12, y(1) = 4

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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) = 0.004B − 800, B(0) = 40,000

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Textbook Question

Does the function y(t) = 2t satisfy the differential equation y'''(t) + y'(t) = 2?

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Textbook Question

11–16. Initial value problems Solve the following initial value problems.

y'(x) = −y + 2, y(0) = −2

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Textbook Question

Stability of Euler's method Consider the initial value problem y′(t) = −ay, y(0) = 1 where a > 0; it has the exact solution y(t) = e⁻ᵃᵗ, which is a decreasing function.

a. Show that Euler's method applied to this problem with time step h can be written u₀ = 1, uₖ₊₁ = (1 − ah)uₖ for k = 0, 1, 2, ...

b. Show by substitution that uₖ = (1 − ah)ᵏ is a solution of the equations in part (a), for k = 0, 1, 2, ...

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Textbook Question

Explain how to solve a separable differential equation of the form

g(t)y'(y) = h(t)

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