Ch. 9 - First-Order Differential Equations

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 9 - First-Order Differential Equations

Problem 9.2.16

Chapter 9, Problem 9.2.16

Solving Initial Value Problems
Solve the initial value problems in Exercises 15–20.

t dy/dt + 2y = t³, t > 0, y(2) = 1

Verified step by step guidance

Rewrite the given differential equation in standard linear form by dividing both sides by \(t\) (since \(t > 0\)): \[\frac{dy}{dt} + \frac{2}{t} y = t^2\]

Identify the integrating factor \(\mu(t)\), which is given by: \[\mu(t) = e^{\int P(t) \, dt} = e^{\int \frac{2}{t} \, dt}\]

Calculate the integrating factor \(\mu(t)\) explicitly by evaluating the integral: \[\mu(t) = e^{2 \ln |t|} = e^{\ln |t|^2} = t^2\] (since \(t > 0\), absolute value can be dropped)

Multiply both sides of the differential equation by the integrating factor \(t^2\) to write the left side as a derivative of a product: \[t^2 \frac{dy}{dt} + 2t y = t^4\] which can be expressed as: \[\frac{d}{dt} (t^2 y) = t^4\]

Integrate both sides with respect to \(t\): \[\int \frac{d}{dt} (t^2 y) \, dt = \int t^4 \, dt\] This gives: \[t^2 y = \frac{t^5}{5} + C\] where \(C\) is the constant of integration. Use the initial condition \(y(2) = 1\) to solve for \(C\).

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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 dy/dt + P(t)y = Q(t). They can be solved using an integrating factor, which simplifies the equation into an exact derivative, allowing integration on both sides to find the general solution.

Integrating Factor Method

The integrating factor is a function, usually denoted μ(t) = e^(∫P(t) dt), used to multiply both sides of a linear differential equation. This transforms the left side into the derivative of (μ(t)y), making it easier to integrate and solve for y.

Initial Value Problems (IVP)

An IVP specifies the value of the unknown function at a particular point, such as y(2) = 1. After finding the general solution, the initial condition is used to determine the specific constant, yielding a unique solution to the differential equation.

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Initial Value Problems

Related Practice

Textbook Question

First-Order Linear Equations

Solve the differential equations in Exercises 1–14.

(t-1)³ ds/dt + 4(t-1)²s = t+1, t >1

Textbook Question

Using Euler’s Method

In Exercises 15–20, use Euler’s method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places.

y' = 2xy + 2y, y(0) = 3, dx = 0.2

Textbook Question

Solving Initial Value Problems

Solve the initial value problems in Exercises 15–20.

(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x > -1, y(0) = 5

Textbook Question

Solve the Bernoulli equations in Exercises 29–32.

x²y' + 2xy = y³

Textbook Question

Show that the solution of the initial value problem

y' = x + y, y(x₀) = y₀

y = -1 -x + (1 + x₀ + y₀) exp(x-x₀).