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Magazine Chapter 10: Problem 5
Find particular solutions \(\frac{d Q}{d t}=0.3 Q-120, \quad Q=50\) when \(t=0\)
Short Answer
Expert verified
The particular solution is \( Q(t) = -350e^{0.3t} + 400 \).
Step by step solution
01
Recognize the Differential Equation Type
The given equation \( \frac{dQ}{dt} = 0.3Q - 120 \) is a first-order linear ordinary differential equation, and it can be rewritten as \( \frac{dQ}{dt} = aQ + b \), where \( a = 0.3 \) and \( b = -120 \).
02
Find the Homogeneous Solution
To find the homogeneous solution, consider the equation \( \frac{dQ}{dt} = 0.3Q \). This is a separable differential equation, and its solution is \( Q_h(t) = Ce^{0.3t} \), where \( C \) is an arbitrary constant.
03
Find the Particular Solution
We need a particular solution to \( \frac{dQ}{dt} = 0.3Q - 120 \). Assume a constant solution of the form \( Q = K \). Substituting into the differential equation, we get \( 0 = 0.3K - 120 \). Solving for \( K \), we have \( K = 400 \). Thus, \( Q_p = 400 \) is a particular solution.
04
Write the General Solution
The general solution is the sum of the homogeneous and particular solutions: \( Q(t) = Ce^{0.3t} + 400 \).
05
Apply the Initial Condition
Use the initial condition \( Q(0) = 50 \) to find \( C \). Substitute \( t = 0 \) and \( Q = 50 \) into the general solution: \( 50 = Ce^{0.3\times0} + 400 \). This simplifies to \( 50 = C + 400 \), leading to \( C = -350 \).
06
Write the Particular Solution with Initial Condition
Substitute \( C = -350 \) back into the general solution to obtain the particular solution: \( Q(t) = -350e^{0.3t} + 400 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order Linear Differential Equations
In calculus, first-order linear differential equations are essential for understanding change and growth over time. The general form is given by:
\[ \frac{dy}{dt} + P(t)y = Q(t) \]where \( P(t) \) and \( Q(t) \) are functions of \( t \), and \( y \) is the function we want to solve for.
These equations are called 'first-order' because they involve only the first derivative (rate of change) of the function \( y \).
Key aspects to remember:
\[ \frac{dy}{dt} + P(t)y = Q(t) \]where \( P(t) \) and \( Q(t) \) are functions of \( t \), and \( y \) is the function we want to solve for.
These equations are called 'first-order' because they involve only the first derivative (rate of change) of the function \( y \).
Key aspects to remember:
- The coefficients (\( a \) and \( b \)), which are constants in many problems, define the behavior of the solution.
- An exponential component often arises in solutions, addressing growth or decay processes.
Initial Conditions
Initial conditions in differential equations specify the value of the solution at a specific point, often at the start of the time period being considered. For example, in our problem:
\( Q(0) = 50 \)
This tells us where the process begins; it is also crucial for determining specific solutions from general solutions.
These conditions are essential for applying particular scenarios and ensuring solutions reflect real-world situations precisely. Without them, we'd only have a general family of solutions.
\( Q(0) = 50 \)
This tells us where the process begins; it is also crucial for determining specific solutions from general solutions.
These conditions are essential for applying particular scenarios and ensuring solutions reflect real-world situations precisely. Without them, we'd only have a general family of solutions.
- Initial conditions are used to uniquely determine the constant in the general solution of a differential equation.
- They align the mathematical solution with practical real-life conditions.
Particular Solutions
A particular solution to a differential equation is a solution that satisfies the differential equation and any given initial conditions. It represents one specific instance among the family of possible solutions.
In our problem:
The particular solution arises from assuming the form \( Q = K \) and solving for \( K \).
This is often a constant solution used to adjust the general solution.
In our problem:
The particular solution arises from assuming the form \( Q = K \) and solving for \( K \).
This is often a constant solution used to adjust the general solution.
- Particular solutions can be constants, functions, or combinations thereof.
- They help 'anchor' the equation to specific situations or points.
Homogeneous Solutions
Homogeneous solutions address the part of the differential equation without the external, non-homogeneous component (in our case, the \(-120\)). By setting the external influence to zero, for instance:
\[ \frac{dQ}{dt} = 0.3Q \],
we solve the equation as if it's purely driven by its own dynamics.
The solution, typically in the form \( Q_h(t) = Ce^{at} \), arises from natural growth or decay principles.
\[ \frac{dQ}{dt} = 0.3Q \],
we solve the equation as if it's purely driven by its own dynamics.
The solution, typically in the form \( Q_h(t) = Ce^{at} \), arises from natural growth or decay principles.
- This form helps understand the natural or intrinsic behavior of the system.
- It allows the understanding of transient states before the external factors act.
