Chapter 4: Problem 4
Solve the given differential equation subject to the given condition. Note that \(y(a)\) denotes the value of \(y\) at \(t=a\). $$ \frac{d y}{d t}=-0.003 y, y(-2)=3 $$
Short Answer
Expert verified
The solution is \( y(t) = 3 e^{-0.003(t+2)} \).
Step by step solution
01
Identify the Type of Differential Equation
This is a first-order linear ordinary differential equation of the form \( \frac{dy}{dt} = -0.003y \), which is separable.
02
Separate Variables
Separate variables by rewriting the equation as \( \frac{dy}{y} = -0.003 dt \).
03
Integrate Both Sides
Integrate both sides: \( \int \frac{dy}{y} = \int -0.003 dt \). This yields \( \ln|y| = -0.003t + C \), where \( C \) is the constant of integration.
04
Solve for y
Exponentiate both sides to solve for \( y \): \( |y| = e^{-0.003t + C} \). Let \( C_1 = e^C \), and since \( y \) is positive, the solution is \( y = C_1 e^{-0.003t} \).
05
Apply the Initial Condition
Use the initial condition \( y(-2) = 3 \) to find \( C_1 \). Substitute into the equation: \( 3 = C_1 e^{0.006} \).
06
Solve for Constant C_1
From \( 3 = C_1 e^{0.006} \), solve for \( C_1 \): \( C_1 = \frac{3}{e^{0.006}} \).
07
Finalize the Particular Solution
Substitute \( C_1 \) back into the solution: \( y = \frac{3}{e^{0.006}} e^{-0.003t} = 3 e^{-0.003(t+2)} \). Thus, the particular solution is \( y(t) = 3 e^{-0.003(t+2)} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equation
A first-order linear differential equation is a common type of differential equation that involves variables and their rates of change. In the equation, the highest derivative is the first derivative, indicated by \( \frac{dy}{dt} \). First-order linear differential equations take the general form:
\[\frac{dy}{dt} + P(t)y = Q(t)\]
In our exercise, the equation is \( \frac{dy}{dt}=-0.003y \). Notice it fits as a first-order linear equation since the structure holds the main variable, \( y \), and its derivative. The key step in solving such equations is to isolate the derivative, making the equation easier to manipulate through methods like integration.
\[\frac{dy}{dt} + P(t)y = Q(t)\]
In our exercise, the equation is \( \frac{dy}{dt}=-0.003y \). Notice it fits as a first-order linear equation since the structure holds the main variable, \( y \), and its derivative. The key step in solving such equations is to isolate the derivative, making the equation easier to manipulate through methods like integration.
Separable Differential Equations
Separable differential equations are a subtype of first-order differential equations, allowing for variables to be separated on different sides of the equation. After separation, each side can be integrated independently. In our example \( \frac{dy}{dt} = -0.003y \), we separate the variables to get:
\[\frac{dy}{y} = -0.003 \, dt\]
Separating variables is a powerful method as it converts a differential equation into a simpler algebraic form. Once variables are separated, we integrate both sides, which creates a solvable equation for \( y \). This approach simplifies the process of finding a general solution to the differential equation.
\[\frac{dy}{y} = -0.003 \, dt\]
Separating variables is a powerful method as it converts a differential equation into a simpler algebraic form. Once variables are separated, we integrate both sides, which creates a solvable equation for \( y \). This approach simplifies the process of finding a general solution to the differential equation.
Initial Condition Problem
An initial condition problem involves finding a particular solution of a differential equation that satisfies a given condition. Typically, this condition provides the value of the solution at a specific point, often denoted as \( y(a) = c \). For our problem, the initial condition given is \( y(-2) = 3 \).
Once a general solution is found, the initial condition allows us to determine the constant of integration, which makes the solution unique to this specific situation. By substituting the initial values into the solution, we solve for the constant. This step ensures the solution accurately reflects the initial terms in the relevant context, providing insights into real-world applications, like predicting future behavior from known starting points.
Once a general solution is found, the initial condition allows us to determine the constant of integration, which makes the solution unique to this specific situation. By substituting the initial values into the solution, we solve for the constant. This step ensures the solution accurately reflects the initial terms in the relevant context, providing insights into real-world applications, like predicting future behavior from known starting points.
Ordinary Differential Equations
An ordinary differential equation (ODE) involves functions of only one independent variable and their derivatives. The equations model many essential phenomena, ranging from simple motion to complex biological systems. The differential equation considered here is an example of an ODE because it focuses on how the rate of change of \( y \) depends solely on the variable \( t \), without referencing other variables.
Ordinary differential equations differ from partial differential equations that involve multiple independent variables. They can be "linear" or "non-linear," and "separable" when variables can be separated and handled individually. Understanding ODEs allows us to predict and understand behaviors in various scientific fields, supporting robust problem-solving and analytical thinking.
Ordinary differential equations differ from partial differential equations that involve multiple independent variables. They can be "linear" or "non-linear," and "separable" when variables can be separated and handled individually. Understanding ODEs allows us to predict and understand behaviors in various scientific fields, supporting robust problem-solving and analytical thinking.