Chapter 3: Problem 39
Find the particular solution to the differential equation \(\frac{d y}{d t}=e^{(t+y)}\) that passes through \((1,0)\), given that \(y=-\ln \left(C-e^{t}\right)\) is a general solution.
Short Answer
Expert verified
The particular solution is \( y = -\ln(1 + e - e^{t}) \).
Step by step solution
01
Understand the General Solution
The general solution of the differential equation is given as \( y = -\ln(C - e^{t}) \). This equation contains a constant \( C \) which we need to determine using the initial condition.
02
Apply the Initial Condition
Use the point \((1, 0)\) to find the value of \( C \). Substitute \( t = 1 \) and \( y = 0 \) into the general solution: \[0 = -\ln(C - e^{1})\] Simplifying, we get \[\ln(C - e) = 0.\] Therefore,\[C - e = 1.\] Hence, \[ C = e + 1. \]
03
Substitute Back Into the General Solution
Now substitute \( C = e + 1 \) back into the general solution:\[y = -\ln((e + 1) - e^{t}).\]Simplify the expression within the logarithm:\[y = -\ln(1 + e - e^{t}).\]
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.
Particular Solution
In solving differential equations, we often seek the particular solution that satisfies specific conditions. A particular solution is a specific instance of the broader solution set that satisfies both the differential equation and some predefined initial condition.
In this exercise, we start with the general solution: \( y = -\ln(C - e^{t}) \).
To find the particular solution, we apply the initial condition provided: the solution passes through the point \((1, 0)\).
By using this specific condition, we are able to pinpoint which solution from the infinite possibilities of the general solution truly fits. The process involves replacing \( t \) and \( y \) in the general equation with 1 and 0, respectively, then solving for the constant \( C \).
This way, we narrow down from the generic to the specific, capturing the particular solution that describes not just any path on the differential equation's curve, but the exact path that includes our known point. Once we establish \( C = e + 1 \), we substitute it back into the general equation, yielding the particular solution: \( y = -\ln((e + 1) - e^{t}) \).
In this exercise, we start with the general solution: \( y = -\ln(C - e^{t}) \).
To find the particular solution, we apply the initial condition provided: the solution passes through the point \((1, 0)\).
By using this specific condition, we are able to pinpoint which solution from the infinite possibilities of the general solution truly fits. The process involves replacing \( t \) and \( y \) in the general equation with 1 and 0, respectively, then solving for the constant \( C \).
This way, we narrow down from the generic to the specific, capturing the particular solution that describes not just any path on the differential equation's curve, but the exact path that includes our known point. Once we establish \( C = e + 1 \), we substitute it back into the general equation, yielding the particular solution: \( y = -\ln((e + 1) - e^{t}) \).
Initial Condition
The concept of an "initial condition" is fundamental in differential equations. It's essentially a piece of additional information that specifies the value of the unknown function at a particular point. This information is crucial for finding a unique solution from a general family of solutions, making it possible to homing in on a singular path that precisely matches the scenario in question.
In our original exercise, the initial condition is that the solution curve must pass through the point \((1, 0)\). This means that when \( t = 1 \), the output of the function \( y \) should be 0.
Using the initial condition in the context of our general solution \( y = -\ln(C - e^{t}) \), substituting in the given \( t \) and \( y \) values helps us solve for the constant \( C \). By executing the substitution and solving through this lens, we find \( C \) that aligns the curve to pass exactly through \((1, 0)\). Ultimately, this unique solution is what we're after, as it serves the essential role of describing a specific physical or geometrical condition.
In our original exercise, the initial condition is that the solution curve must pass through the point \((1, 0)\). This means that when \( t = 1 \), the output of the function \( y \) should be 0.
Using the initial condition in the context of our general solution \( y = -\ln(C - e^{t}) \), substituting in the given \( t \) and \( y \) values helps us solve for the constant \( C \). By executing the substitution and solving through this lens, we find \( C \) that aligns the curve to pass exactly through \((1, 0)\). Ultimately, this unique solution is what we're after, as it serves the essential role of describing a specific physical or geometrical condition.
General Solution
A general solution represents a set of solutions to a differential equation that includes all potential paths the function \( y \) might take in relation to its independent variable (often \( t \) or \( x \)).
For many differential equations, not just one but infinitely many functions can satisfy the equation. Hence, annotations like the constant \( C \) appear in the general solution.
In our case, the general solution given is \( y = -\ln(C - e^{t}) \). The family of solutions described by this equation contains an element of flexibility: different values of \( C \) characterize different curves.
However, to solve a practical problem, we need more than the possibilities offered by the general solution. That's where particular solutions, picked using an initial condition, come into play. These general solutions lay the groundwork on which the real-world constraint—our initial condition—is applied.
For many differential equations, not just one but infinitely many functions can satisfy the equation. Hence, annotations like the constant \( C \) appear in the general solution.
In our case, the general solution given is \( y = -\ln(C - e^{t}) \). The family of solutions described by this equation contains an element of flexibility: different values of \( C \) characterize different curves.
However, to solve a practical problem, we need more than the possibilities offered by the general solution. That's where particular solutions, picked using an initial condition, come into play. These general solutions lay the groundwork on which the real-world constraint—our initial condition—is applied.
Exponential Functions
Exponential functions are a key component of calculus and differential equations. They involve variables appearing as exponents and are critical in describing processes with constant relative growth rates, such as population growth or radioactive decay.
In our exercise, the variable \( e^{t} \) within the differential equation \( \frac{d y}{d t}=e^{(t+y)} \) plays a crucial role. Exponential functions like these are particularly interesting because they grow rapidly, and their properties often significantly simplify both calculus operations and their practical applications.
In the general solution \( y = -\ln(C - e^{t}) \), the simplicity and power of the exponential function \( e^{t} \) is utilized, showing how exponential growth or decay integrates smoothly with other functions, like the natural logarithm, within solutions to differential equations.
Understanding exponential growth enables one to appreciate its impact on not only mathematics but various real-life phenomena modeled by such equations.
In our exercise, the variable \( e^{t} \) within the differential equation \( \frac{d y}{d t}=e^{(t+y)} \) plays a crucial role. Exponential functions like these are particularly interesting because they grow rapidly, and their properties often significantly simplify both calculus operations and their practical applications.
In the general solution \( y = -\ln(C - e^{t}) \), the simplicity and power of the exponential function \( e^{t} \) is utilized, showing how exponential growth or decay integrates smoothly with other functions, like the natural logarithm, within solutions to differential equations.
Understanding exponential growth enables one to appreciate its impact on not only mathematics but various real-life phenomena modeled by such equations.
