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Magazine Chapter 24: Problem 70
Let \(y=y(x)\) be the solution of the differential equation \(\frac{d y}{d x}+2 y=f(x)\), where \(f(x)= \begin{cases}1, & x \in[0,1] \\ 0, & \text { otherwise }\end{cases}\) If \(y(0)=0\), then \(y\left(\frac{3}{2}\right)\) is (a) \(\frac{e^{2}-1}{2 e^{3}}\) (b) \(\frac{e^{2}-1}{e^{3}}\) (c) \(\frac{1}{2 e}\) (d) \(\frac{e^{2}+1}{2 e^{4}}\)
Short Answer
Expert verified
Option (a) \( \frac{e^2 - 1}{2e^3} \).
Step by step solution
01
Identify the System Type
This is a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) with \( P(x) = 2 \) and a piecewise function \( f(x) \).
02
Solve the Homogeneous Equation
First, find the solution to the homogeneous equation \( \frac{dy}{dx} + 2y = 0 \). This results in \( y_h = Ce^{-2x} \), where \( C \) is a constant.
03
Determine an Integrating Factor
To solve the non-homogeneous equation, find the integrating factor \( \mu(x) = e^{\int 2 \, dx} = e^{2x} \).
04
Solve the Non-Homogeneous Equation for 0 鈮� x 鈮� 1
For \( x \in [0, 1] \), since \( Q(x) = 1 \), the equation becomes: \[ e^{2x} \frac{d}{dx}(y e^{2x}) = e^{2x} \times 1. \] Integrating both sides gives: \[ y(x)e^{2x} = \int_0^x e^{2u} \, du = \frac{1}{2}(e^{2x} - 1) + C. \] Thus, \( y(x) = \frac{1}{2}(1 - e^{-2x}) + Ce^{-2x} \).
05
Initial Condition and Particular Solution
Apply the initial condition \( y(0) = 0 \). This implies \( \frac{1}{2}(1 - 1) + C \times 1 = 0 \), hence \( C = 0 \). Thus, for \( 0 \leq x \leq 1 \), \( y(x) = \frac{1}{2}(1 - e^{-2x}) \).
06
Solve for x > 1 with New Q(x)
For \( x > 1 \), \( Q(x) = 0 \) since the function is zero in this interval. The differential equation \( \frac{dy}{dx} + 2y = 0 \) again leads us to: \( y(x) = De^{-2x} \).
07
Match Solutions at x = 1
At \( x = 1 \), the solutions must be continuous. For \( 0 \leq x \leq 1 \), \( y(1) = \frac{1}{2}(1 - e^{-2}) \). Set this equal to the solution for \( x > 1 \): \( D e^{-2} = \frac{1}{2}(1 - e^{-2}) \). Solve for \( D \): \( D = \frac{1}{2}e^2(1 - e^{-2}) = \frac{1}{2}(e^2 - 1) \).
08
Evaluate y at x = 3/2
For \( x > 1 \), use \( y(x) = \frac{1}{2}(e^2 - 1)e^{-2x} \). At \( x = \frac{3}{2} \), \( y\left(\frac{3}{2}\right) = \frac{1}{2}(e^2 - 1)e^{-3} \).
09
Choose the Correct Option
The correct value of \( y\left(\frac{3}{2}\right) \) matches option (a) \( \frac{e^2 - 1}{2e^3} \).
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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
A first-order linear differential equation is an equation involving a function and its first derivative. This type of equation is of the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] Here, the goal is to find the function \( y(x) \). Understanding these equations is crucial because they often describe a wide range of physical systems and processes. **Characteristics:**
- Linear in nature because the variable \( y \) and its derivative appear to the power one.
- Includes a term \( P(x)y \) that depends on \( y \) and possibly on the independent variable \( x \).
- \( Q(x) \) represents the non-homogeneous part that can be an external input or forcing function.
Integrating Factor Method
The Integrating Factor Method is a powerful technique used to solve first-order linear differential equations. It simplifies the process, allowing us to transform the equation into an easily integrable form. To apply this method effectively, we first identify the integrating factor \( \mu(x) \), which is computed from the function \( P(x) \) in the differential equation: - For an equation of the form \[ \frac{d y}{d x} + P(x)y = Q(x) \] - The integrating factor is calculated as: \[ \mu(x) = e^{\int P(x) \, dx} \] Once the integrating factor is determined, it is multiplied through the differential equation. This transforms the left-hand side into the derivative of the product of the integrating factor and the function \( y(x) \): \[ \mu(x) \frac{d}{dx}(y e^{\int P(x) \, dx}) = \mu(x) Q(x) \] Next, integrate both sides with respect to \( x \). This yields the solution to the equation after simplification. By applying this method, you ensure that the solution remains continuous and consistent through intervals where \( Q(x) \) changes.
Initial Value Problems
Initial Value Problems (IVPs) are a special category of differential equations where the solution is determined with additional specified values at a particular point. These specific starting details help us determine a unique solution from potentially many general solutions. In first-order differential equations, an IVP typically includes an initial condition such as: - \( y(x_0) = y_0 \) This condition tells us the value of the function \( y \) at a particular value \( x_0 \). It is crucial since it allows mathematicians and scientists to model and predict behaviors more effectively. To handle IVPs, solve the differential equation using a chosen method (e.g., integrating factor), and then substitute the initial condition to find any unknown constants. This step ensures that the solution faithfully represents the specific situation defined by the initial conditions. Understanding IVPs is essential because they often reflect real-world processes where the state (or situation) is known at a starting time or position.
