Chapter 15: Problem 1
Solve the following first-order linear differential equations; if an initial condition is given, definitize the arbitrary constant: $$\frac{d y}{d t}+5 y=15$$
Short Answer
Expert verified
The general solution is \( y(t) = 3 + Ce^{-5t} \).
Step by step solution
01
Identify the standard form
The given differential equation \( \frac{d y}{d t} + 5y = 15 \) is already in the standard linear form, which is \( \frac{d y}{d t} + P(t)y = Q(t) \). Here, \( P(t) = 5 \) and \( Q(t) = 15 \).
02
Find the integrating factor
The integrating factor \( \mu(t) \) is given by \( e^{\int P(t) \, dt} \). For the equation \( \frac{d y}{d t} + 5y = 15 \), calculate \( \mu(t) = e^{\int 5 \, dt} = e^{5t} \).
03
Multiply the equation by the integrating factor
Multiply the entire differential equation by the integrating factor \( e^{5t} \): \( e^{5t} \frac{d y}{d t} + 5e^{5t}y = 15e^{5t} \).
04
Recognize the left-hand side as a derivative
Notice that the left-hand side of the equation can be written as the derivative of \( (e^{5t} y) \): \( \frac{d}{dt} (e^{5t} y) = 15e^{5t} \).
05
Integrate both sides
Integrate both sides with respect to \( t \): \( \int \frac{d}{dt} (e^{5t} y) \, dt = \int 15e^{5t} \, dt \). The left side becomes \( e^{5t}y \) and the right side becomes \( 3e^{5t} + C \).
06
Solve for y
To isolate \( y \), divide both sides by \( e^{5t} \): \( y = 3 + Ce^{-5t} \).
07
Apply initial conditions (if given)
If an initial condition is given, such as \( y(t_0) = y_0 \), substitute \( t_0 \) and \( y_0 \) into the equation \( y = 3 + Ce^{-5t} \) to solve for \( C \).
08
Write the solution
The general solution to the differential equation is \( y(t) = 3 + Ce^{-5t} \). If an initial condition was provided, we substitute \( C \) from Step 7 to find the particular solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Differential Equations
A linear differential equation is an equation involving a dependent variable (for example, \( y \)) and its derivatives. It is termed "linear" because the equation and its derivatives appear linearly and not in any other form (like squared or cubed). A general first-order linear differential equation can be written as: \[ \frac{dy}{dt} + P(t)y = Q(t) \] where \( P(t) \) and \( Q(t) \) are known functions of \( t \). It's important for students to recognize this form because it guides the method of solution. In our given equation \( \frac{d y}{d t} + 5y = 15 \), \( P(t) = 5 \) and \( Q(t) = 15 \), both constant functions of \( t \). By understanding these components, you can proceed to solve the equation systematically.
Integrating Factor
The integrating factor is a crucial tool when solving first-order linear differential equations. It transforms the differential equation into a form that is easier to integrate. The integrating factor, \( \mu(t) \), is defined as: \[ \mu(t) = e^{\int P(t) \, dt} \] For our equation \( \frac{d y}{d t} + 5y = 15 \), the integrating factor becomes \( e^{\int 5 \, dt} = e^{5t} \). By multiplying the entire equation by this \( e^{5t} \), the equation is adjusted to a format where the left-hand side becomes the derivative of a product of functions. This simplification enables straightforward integration of both sides, leading toward the solution.
Initial Condition
An initial condition provides a specific value of the dependent variable at a particular point, typically used to calculate the constant of integration in a differential equation. Suppose an initial condition is given with \( y(t_0) = y_0 \); it means that when \( t = t_0 \), \( y = y_0 \). This allows you to find the unique value of the constant \( C \) in the solution of the differential equation. Once the general solution, \( y(t) = 3 + Ce^{-5t} \), is found, substitute \( t_0 \) and \( y_0 \) into the equation to solve for \( C \). This step is crucial because it narrows down the many possible solutions to a specific one that satisfies the initial condition, ensuring the problem is completely solved.
First-Order Differential Equations
First-order differential equations are equations involving the first derivative of the unknown function, without involving higher order derivatives. These types of equations take the general form: \[ \frac{dy}{dt} = g(t, y) \] where \( g(t, y) \) is some function involving \( t \) and \( y \). The linear first-order differential equations form a special subset of these wherein the right side is a linear function of \( y \). Solving these equations involves techniques like integrating factors, which help convert the equations into a form where they can be integrated directly. These kinds of differential equations are common in real-world applications, so mastering them builds a foundational skill in solving more complex problems in mathematics and engineering.