Chapter 31: Problem 36
Find the particular solution of the given differential equation for the indicated values. $$\frac{d s}{d t}=\sec s ; \quad t=0 \text { when } s=0$$
Short Answer
Expert verified
The particular solution is \( \sin s = t \).
Step by step solution
01
Recognize the Differential Equation Type
The given differential equation is \( \frac{ds}{dt} = \sec s \). This is a first-order separable differential equation, which means we can separate the variables \( s \) and \( t \) to solve it.
02
Separate Variables
Rearrange the equation to separate the variables. We get:\[ \cos s \, ds = dt \]This allows us to integrate both sides with respect to their respective variables.
03
Integrate Both Sides
Integrate both sides of the separated equation:\[ \int \cos s \, ds = \int dt \]The integrals will be:\[ \sin s = t + C \]where \( C \) is the constant of integration.
04
Apply Initial Conditions
Use the given initial conditions to find the value of the constant \( C \).\[ t = 0 \text{ when } s = 0 \]Substitute these values into the equation:\[ \sin(0) = 0 + C \]This simplifies to:\[ 0 = C \]Thus, the constant \( C = 0 \).
05
Write the Particular Solution
Substitute \( C = 0 \) back into the integrated equation to get the particular solution:\[ \sin s = t \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Differential Equations
First-order differential equations are equations that involve the first derivative of a function. These types of equations generally appear in the form \( \frac{dy}{dx} = f(x, y) \) where we need to find the function \( y(x) \).
These equations are fundamental in many scientific fields because they model the rate of change of systems. A common technique for solving first-order equations is separation of variables, especially when they are separable. This means we can rearrange the equation to get all \( y \) terms on one side and all \( x \) terms on the other.
In the given exercise \( \frac{ds}{dt} = \sec s \), it is a first-order separable differential equation. We separate the variables by rearranging it as \( \cos s \, ds = dt \). Integrating both sides then helps us find the function that satisfies this equation.
Key points to remember include:
These equations are fundamental in many scientific fields because they model the rate of change of systems. A common technique for solving first-order equations is separation of variables, especially when they are separable. This means we can rearrange the equation to get all \( y \) terms on one side and all \( x \) terms on the other.
In the given exercise \( \frac{ds}{dt} = \sec s \), it is a first-order separable differential equation. We separate the variables by rearranging it as \( \cos s \, ds = dt \). Integrating both sides then helps us find the function that satisfies this equation.
Key points to remember include:
- First-order equations involve only the first derivative.
- They often require techniques like separation of variables for solutions.
- Understanding the behavior of the solution depends heavily on the integrating process.
Particular Solution
A particular solution to a differential equation is a solution that fits a specific condition or set of conditions, often called initial conditions.
For the differential equation \( \frac{ds}{dt} = \sec s \) in the exercise, we aim to find a solution that not only satisfies the equation itself but also the given condition \( t = 0 \) when \( s = 0 \).
To achieve this, we integrate the separated equation \( \cos s \, ds = dt \) which leads us to \( \sin s = t + C \), introducing a constant of integration \( C \).
By applying the initial conditions, we determine \( C \) to obtain the particular solution \( \sin s = t \).
Features of the particular solution:
For the differential equation \( \frac{ds}{dt} = \sec s \) in the exercise, we aim to find a solution that not only satisfies the equation itself but also the given condition \( t = 0 \) when \( s = 0 \).
To achieve this, we integrate the separated equation \( \cos s \, ds = dt \) which leads us to \( \sin s = t + C \), introducing a constant of integration \( C \).
By applying the initial conditions, we determine \( C \) to obtain the particular solution \( \sin s = t \).
Features of the particular solution:
- The solution specifically satisfies the initial condition.
- It offers a precise answer rather than a family of solutions.
- Particular solutions are crucial for accurately modeling systems in real-life situations.
Initial Conditions
Initial conditions serve as the starting values for determining specific solutions to differential equations. By providing specific values for variables at a given point, they help in narrowing down the general solutions to particular ones.
In our given problem, the initial condition is \( t = 0 \) when \( s = 0 \).
Applying this condition to the integrated result \( \sin s = t + C \), we substitute \( 0 \) for both \( s \) and \( t \), thus finding that \( C = 0 \).
This step is crucial as it allows us to finalize and verify the particular solution \( \sin s = t \).
Significance of initial conditions:
In our given problem, the initial condition is \( t = 0 \) when \( s = 0 \).
Applying this condition to the integrated result \( \sin s = t + C \), we substitute \( 0 \) for both \( s \) and \( t \), thus finding that \( C = 0 \).
This step is crucial as it allows us to finalize and verify the particular solution \( \sin s = t \).
Significance of initial conditions:
- They provide a unique solution to the differential equation among infinite possibilities.
- Ensure the solution aligns with real-world constraints or observations.
- Initial conditions are often derived from physical laws or empirical data.