Chapter 6: Problem 7
Solving a First-Order Linear Differential Equation In Exercises \(5-14,\) solve the first-order linear differential equation. $$ y^{\prime}-y=16 $$
Short Answer
Expert verified
The solution for the given first-order linear differential equation is \( y(x) = -16 + ke^{x} \)
Step by step solution
01
Rearrange the equation
Let's start by moving all the terms involving \( y \) to one side of the equation. This gives us: \( y' - y = 16 \)
02
Solve the homogeneous equation
Next, we solve the homogeneous equation obtained by setting the right hand side of the equation equal to zero, \( y' - y = 0 \). This is a separable differential equation, and the solution is given by \( y = Ce^x \), for a constant \( C \).
03
Apply the integrating factor method
We then apply the integrating factor method to solve the inhomogeneous equation. To do this, we multiply both sides of the original equation by the integrating factor, which is \( e^{-x} \). This gives us \( e^{-x}y' - e^{-x}y = 16e^{-x} \).
04
Apply properties of derivatives
The left side of this equation is now the derivative of \( y \times \exp(-x) \). Thus, we can express the equation in differential form as \( (ye^{-x})' = 16e^{-x} \). Then integrate both sides of the equation to get \( ye^{-x} = -16e^{-x} + k \), with \( k \) as an integration constant.
05
Solve for y(x)
Finally, solve the equation for \( y(x) \) by multiplying by \( e^{x} \) on both sides. This gives the solution: \( y(x) = -16 + ke^{x} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor Method
In the realm of solving first-order linear differential equations, the Integrating Factor Method is a powerful tool. The idea is to simplify an equation into a form that is easier to integrate. For a linear equation of the form \( y' + P(x)y = Q(x) \), an integrating factor \( \mu(x) \) is chosen such that it makes the left side a perfect derivative.
This integrating factor is typically \( e^{\int P(x) \, dx} \). It's used to multiply through the equation, converting it into:
This integrating factor is typically \( e^{\int P(x) \, dx} \). It's used to multiply through the equation, converting it into:
- \( \mu(x)(y' + P(x)y) = \mu(x)Q(x) \)
- \( (\mu(x)y)' = \mu(x)Q(x) \)
Homogeneous Equations
Homogeneous equations are those where every term is a direct function of \( y \) and potentially other variables. For example, the equation \( y' - y = 0 \) is homogeneous because both terms are functions of \( y \). Solving them usually involves recognizing their separable nature or exponential solutions.
In these equations, if you set \( y' \) equal to zero, you effectively acknowledge the presence of terms interacting with each other linearly. Solving this leads us to solutions like \( y = Ce^{x} \), where \( C \) is a constant derived from initial conditions or further evaluation. Homogeneous equations give insights into equilibrium states where changes cancel each other out perfectly, a useful concept in both mathematics and applied sciences.
In these equations, if you set \( y' \) equal to zero, you effectively acknowledge the presence of terms interacting with each other linearly. Solving this leads us to solutions like \( y = Ce^{x} \), where \( C \) is a constant derived from initial conditions or further evaluation. Homogeneous equations give insights into equilibrium states where changes cancel each other out perfectly, a useful concept in both mathematics and applied sciences.
Separable Differential Equations
Separable differential equations are those that can be manipulated algebraically to separate variables into two different sides of the equation. For example, in homogenized forms like \( y' - y = 0 \), teeny rearrangements expose the separable nature.Reformulating yields:
- \( \frac{dy}{dx} = y \)
- \( \frac{dy}{y} = dx \)
- \( \ln|y| = x + C \)