Chapter 9: Problem 12
Solve the following initial value problems. $$y^{\prime}(x)=-y+2, y(0)=-2$$
Short Answer
Expert verified
Question: Find the solution for the given initial value problem: \(y^{\prime}(x) = -y + 2 \) with the initial condition \(y(0) = -2\).
Answer: \(y(x) = -2\)
Step by step solution
01
Identify Differential Equation
We are given the following first order linear differential equation:
$$y^{\prime}(x) = -y + 2$$
with the initial condition \(y(0) = -2\).
02
Find the Integrating Factor (IF)
In order to solve this differential equation using the Integrating Factor method, we first need to find the integrating factor. For a first-order linear differential equation in the form \(y^{\prime}(x) + P(x)y = Q(x)\), the integrating factor is given by:
$$IF(x) = e^{\int P(x)dx}$$
In our case, \(P(x) = -1\), so we have:
$$IF(x) = e^{\int (-1)dx} = e^{-x}$$
03
Multiply the Differential Equation by the Integrating Factor
Now, we multiply our equation by the integrating factor that we found in the previous step:
$$e^{-x}y^{\prime}(x) - e^{-x}y = 2e^{-x}$$
Notice that the left side of the equation is now the derivative of the product of the unknown function y(x) and the integrating factor (IF):
$$\frac{d}{dx}(ye^{-x}) = 2e^{-x}$$
04
Integrate Both Sides of the Equation
Now, we are ready to integrate both sides of the equation with respect to x:
$$\int \frac{d}{dx}(ye^{-x})dx = \int 2e^{-x}dx$$
Integration yields:
$$ye^{-x} = -2e^{-x} + C$$
where C is the constant of integration.
05
Solve for the Unknown Function y(x)
To find y(x), we multiply both sides by \(e^{x}\) to eliminate the exponential term:
$$y(x) = -2 + Ce^{x}$$
06
Apply Initial Condition to Find Constant of Integration
We're given the initial condition \(y(0) = -2\). Plugging this into our expression for y(x), we get:
$$-2 = -2 + Ce^{0} \Rightarrow C = 0$$
07
Write the Final Solution
With \(C=0\), our final solution for the initial value problem is:
$$y(x) = -2$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Value Problems
Initial value problems (IVPs) are a special type of differential equation problem where the solution is sought based on initial conditions provided at a specific point. These problems involve both a differential equation and a specified value for the function at a certain point, known as the initial condition.
In our example, the differential equation is given by \(y'(x) = -y + 2\), and the initial condition is \(y(0) = -2\).
In our example, the differential equation is given by \(y'(x) = -y + 2\), and the initial condition is \(y(0) = -2\).
- The initial condition helps to find the constant of integration when solving the differential equation.
- This ensures a unique solution that fits both the equation and the initial value.
Integrating Factor Method
The integrating factor method is a powerful technique used to solve first-order linear differential equations. This method is particularly useful because it transforms a non-exact differential equation into an exact equation that can be easily integrated.
To apply the integrating factor method, consider a first-order linear differential equation in the form \(y'(x) + P(x)y = Q(x)\). The integrating factor, denoted as \(IF(x)\), is calculated as follows:
This simplification allows us to integrate both sides with respect to \(x\), resulting in a straightforward expression from which we can solve for \(y(x)\).
To apply the integrating factor method, consider a first-order linear differential equation in the form \(y'(x) + P(x)y = Q(x)\). The integrating factor, denoted as \(IF(x)\), is calculated as follows:
- Calculate \(IF(x) = e^{\int P(x)dx}\).
- In our problem, \(P(x) = -1\), giving us \(IF(x) = e^{-x}\).
This simplification allows us to integrate both sides with respect to \(x\), resulting in a straightforward expression from which we can solve for \(y(x)\).
First Order Linear Differential Equations
First order linear differential equations are a fundamental class of differential equations, characterized by being linear and involving only the first derivative of the unknown function. They are generally expressed in the form \(y'(x) + P(x)y = Q(x)\).
These equations are pivotal in mathematical modeling as they describe a broad range of phenomena where the rate of change of a quantity is proportional to the quantity itself or to some external influence.
These equations are pivotal in mathematical modeling as they describe a broad range of phenomena where the rate of change of a quantity is proportional to the quantity itself or to some external influence.
- The linearity of these equations means the unknown function and its derivative appear to the first power.
- They are easier to handle than nonlinear equations because they follow the principle of superposition.