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Magazine Chapter 9: Problem 24
Find the particular solution determined by the initial condition. $$y^{\prime}-y=e^{2 x}, \quad y(1)=1$$
Short Answer
Expert verified
The particular solution determined by the initial condition is \(y(x) = e^{2x} + e^{x}(e^{-1} - e)\).
Step by step solution
01
Find the Integrating Factor (IF)
To find the IF, we need to consider the coefficient of y in the differential equation. In our case, the coefficient is -1. We will now find the IF using the formula \(\mbox{IF} = e^{\int (-1) dx}\).
\[\mbox{IF} = e^{\int (-1) dx} = e^{-x}\]
02
Multiply the equation by the Integrating Factor and integrate both sides
Now we will multiply the given ODE by the IF:
\(e^{-x}(y'-y) = e^{-x} \cdot e^{2x}\)
Simplify the equation:
\(e^{-x}y'-e^{-x}y = e^{x}\)
Now, notice that the left-hand side of the equation is a perfect derivative:
\[\frac{d}{dx}(ye^{-x}) = e^{-x}y' - e^{-x}y\]
So we can rewrite our equation as:
\[\frac{d}{dx}(ye^{-x}) = e^{x}\]
Now integrate both sides with respect to x:
\[\int \frac{d}{dx}(ye^{-x}) dx = \int e^{x} dx\]
\[ye^{-x} = e^{x} + C\]
03
Apply the initial condition and solve for the constant of integration
Given the initial condition, \(y(1)=1\), we will plug in x=1 and y=1 into the equation we found above:
\[1 \cdot e^{-1} = e^{1} + C\]
Now, solve for C:
\[C = e^{-1} - e\]
04
Write down the particular solution
Finally, plug the value of C back into the equation and multiply both sides by \(e^x\) to obtain the particular solution:
\[y = e^{x}(e^{x} + e^{-1} - e)\]
Therefore, the particular solution determined by the initial condition is:
\[y(x) = e^{2x} + e^{x}(e^{-1} - e)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
In the world of solving differential equations, an Integrating Factor (IF) becomes a valuable tool, especially for linear first-order Ordinary Differential Equations (ODEs). When you have a linear ODE of the form \( y' + P(x)y = Q(x) \), the IF helps simplify the process of solving it. The idea is to multiply the entire differential equation by this factor to turn the left-hand side into the derivative of a product, making the equation easier to solve.
- The formula to find the IF is \( e^{\int P(x) \, dx} \).
- In this case, for the problem \( y' - y = e^{2x} \), the coefficient \( P(x) \) is \(-1\).
- Therefore, the IF is \( e^{\int -1 \, dx} = e^{-x} \).
Initial Value Problem
An Initial Value Problem (IVP) is a type of differential equation problem where you are given a differential equation along with particular values of the variable and its derivatives at a certain point. In an IVP, you attempt to find a specific solution that not only satisfies the differential equation but also meets these initial conditions.
- For our given problem, the initial condition is \( y(1) = 1 \).
- This means that at \( x = 1 \), the value of \( y \) must be 1, which provides a boundary to our solution.
- The initial condition is crucial as it helps determine the specific or `particular' solution from the general solution of the differential equation.
Particular Solution
After obtaining a general solution from integrating the ODE, we use the initial condition to find the particular solution. This process ensures that the solution is not only applicable for a wide range of scenarios but satisfies the given initial values.
- When the general solution is found, it typically includes an integration constant \( C \).
- The initial condition helps us solve for this \( C \), determining the specific path of the solution curve.
- In our problem, substituting \( y(1) = 1 \) into the equation helps find \( C = e^{-1} - e \).
ODE (Ordinary Differential Equation)
An Ordinary Differential Equation, or ODE, is an equation involving derivatives of a function with respect to one variable. They are key in expressing relationships between dynamically changing quantities. ODEs come in different orders, typically identified by the highest derivative present.
- Our specific problem is a first-order linear ODE \( y' - y = e^{2x} \).
- Here, the term \( y' \) refers to the first derivative of \( y \) with respect to \( x \), highlighting the simplest form of ODEs.
- These equations can be found everywhere in math and science, modeling phenomena such as population dynamics, heat transfer, and mechanical motion.
