Question

Solve the generic problem \(y^{\prime}=a y+b\) with initial condition \(y(0)=c\).

Short answer

The solution is \( y(t) = (c + \frac{b}{a}) e^{at} - \frac{b}{a} \).

Step-by-step solution

Identify the Type of Differential Equation

The given problem is a first-order linear differential equation of the form \( y' = ay + b \), where \( y' \) is the derivative of \( y \) with respect to \( t \), \( a \) and \( b \) are constants, and \( y(0) = c \) is the initial condition.

Write the Standard Form

We can express the equation in the standard linear differential equation form as \( \frac{dy}{dt} - ay = b \). The goal is to solve for \( y(t) \).

Calculate the Integrating Factor

The integrating factor \( \mu(t) \) for a linear differential equation \( \frac{dy}{dt} - ay = b \) is calculated as \( \mu(t) = e^{\int -a \, dt} = e^{-at} \).

Multiply by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{-at} \): \( e^{-at} \frac{dy}{dt} - ae^{-at} y = be^{-at} \).

Solve the Left-Hand Side

Recognize that the left-hand side is the derivative of \( y e^{-at} \). Therefore, rewrite as \( \frac{d}{dt} (y e^{-at}) = be^{-at} \).

Integrate Both Sides

Integrate both sides with respect to \( t \). The integral of the left side is \( y e^{-at} \), and on the right side, we have \( \int be^{-at} \, dt = -\frac{b}{a} e^{-at} + C \), where \( C \) is the constant of integration.

Solve for y(t)

Solve for \( y(t) \) by multiplying both sides by \( e^{at} \): \( y(t) = (c + \frac{b}{a}) e^{at} - \frac{b}{a} \), using the initial condition \( y(0) = c \) to find \( C \).

Apply Initial Condition

When \( t = 0 \), \( y(0) = c = (c + \frac{b}{a}) e^{0} - \frac{b}{a} \). Simplifying gives us \( C = c + \frac{b}{a} \). Substitute \( C \) back to get \( y(t) = (c + \frac{b}{a}) e^{at} - \frac{b}{a} \).

Key concepts

Integrating Factor

In solving first-order linear differential equations, the integrating factor method is a powerful tool that simplifies the equation into a form that is easier to solve. The main idea is to multiply the original differential equation by a cleverly chosen function, known as the integrating factor. This makes the left-hand side of the equation an exact derivative, which can be easily integrated.

- The standard form of the first-order linear differential equation is usually written as \( \frac{dy}{dt} + P(t)y = Q(t) \).
- To find the integrating factor, \( \mu(t) \), we first identify \( P(t) \), the coefficient of \( y \) in the equation.
- The integrating factor is given by \( \mu(t) = e^{\int P(t) \, dt} \).

In the specific exercise \( y^{\prime} = ay + b \), we rewrite it in the form \( \frac{dy}{dt} - ay = b \), where \( P(t) = -a \). Therefore, the integrating factor becomes \( \mu(t) = e^{-at} \).

This transforms the differential equation and plays a crucial role in finding its solution.

Initial Conditions

Initial conditions help us find a specific solution from the general solution of a differential equation. These conditions provide the necessary information to determine unique constants that appear after integration.

- For any differential equation, applying the initial conditions means substituting the known values into the solution.
- This often involves using a known value of the function such as \( y(t_0) = c \), where \( t_0 \) is a specified point, as seen in our exercise where \( y(0) = c \).

In our exercise, the initial condition \( y(0) = c \) is used to solve for the constant \( C \). By substituting \( t = 0 \) and \( y = c \) into the equation, we solve for \( C \), ensuring the solution satisfies this specific condition.

Initial conditions ensure that the solution is not just general, but tailor-fitted to the problem, making it highly practical and relevant to specific scenarios.

Solution of Differential Equations

The solution of a differential equation represents the function that satisfies the equation for given conditions. Solving first-order linear differential equations typically involves finding a particular solution that adheres to any initial conditions specified.

- The general solution involves integrating the differential equation to find \( y(t) \).
- Initial conditions are then applied to solve for any constants of integration, tailoring the general solution to fit initial parameters.

In our exercise, after applying the integrating factor and simplifying, the differential equation is integrated to find \( y(t) = (c + \frac{b}{a}) e^{at} - \frac{b}{a} \). This expression represents the particular solution that fits the initial condition \( y(0) = c \).

Solutions to differential equations can offer insights into dynamic systems, predicting behavior over time and under various conditions, making them invaluable in fields like physics, biology, and engineering.