Skip to main content
Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 9, Problem 9.2.15

Solving Initial Value Problems
Solve the initial value problems in Exercises 15–20.
dy/dt + 2y = 3, y(0) = 1

Verified step by step guidance
1
Identify the type of differential equation. The given equation is a first-order linear ordinary differential equation of the form \(\frac{dy}{dt} + P(t)y = Q(t)\), where \(P(t) = 2\) and \(Q(t) = 3\).
Find the integrating factor \(\mu(t)\) using the formula \(\mu(t) = e^{\int P(t)\,dt}\). In this case, calculate \(\mu(t) = e^{\int 2\,dt}\).
Multiply both sides of the differential equation by the integrating factor \(\mu(t)\) to rewrite the left side as the derivative of a product: \(\frac{d}{dt}[\mu(t) y] = \mu(t) Q(t)\).
Integrate both sides with respect to \(t\) to find \(\mu(t) y = \int \mu(t) Q(t)\, dt + C\), where \(C\) is the constant of integration.
Use the initial condition \(y(0) = 1\) to solve for the constant \(C\), then solve for \(y(t)\) by dividing both sides by \(\mu(t)\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m
Was this helpful?

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), where the solution involves finding an integrating factor to simplify the equation. Recognizing this form allows the use of systematic methods to solve for y(t).
Recommended video:
07:39
Classifying Differential Equations

Integrating Factor Method

This technique involves multiplying the entire differential equation by an integrating factor, usually e^(∫P(t)dt), to rewrite the left side as a derivative of a product. This simplifies solving the equation by enabling direct integration.
Recommended video:
07:33
Euler's Method

Initial Value Problems (IVP)

An IVP specifies the value of the unknown function at a particular point, such as y(0) = 1. This condition is used to find the unique solution to the differential equation that satisfies the initial condition.
Recommended video:
05:03
Initial Value Problems