Chapter 9: Problem 7
Solve the differential equation. $$ y^{\prime}=x-y $$
Short Answer
Expert verified
Solution: \( y = x - 1 + Ce^{-x} \).
Step by step solution
01
Identify the type of differential equation
The given differential equation is \( y' = x - y \). This is a first-order linear differential equation.
02
Rearrange the equation to the standard form
Rearrange the equation into the standard form \( y'+py=q \), where \( p = -1 \) and \( q = x \). The equation becomes \( y' + y = x \).
03
Find the integrating factor
The integrating factor \( \mu(x) \) is given by \( e^{\int p \, dx} \). Here \( p = 1 \), so \( \mu(x) = e^{\int 1 \, dx} = e^x \).
04
Multiply through by the integrating factor
Multiply the whole differential equation by the integrating factor \( e^x \): \( e^x y' + e^x y = e^x x \).
05
Recognize the left-hand side as a derivative
Notice that \( e^x y' + e^x y = \frac{d}{dx}(e^x y) \). This means we can rewrite the equation as \( \frac{d}{dx}(e^x y) = e^x x \).
06
Integrate both sides with respect to \( x \)
Integrate both sides: \( \int \frac{d}{dx}(e^x y) \, dx = \int e^x x \, dx \). This gives \( e^x y = \int e^x x \, dx \).
07
Solve the right-side integral
Use integration by parts for \( \int e^x x \, dx \). Let \( u = x \) and \( dv = e^x \, dx \). Then \( du = dx \) and \( v = e^x \). The integral becomes \( e^x x - \int e^x \, dx = e^x x - e^x + C \).
08
Divide by the integrating factor
Solve for \( y \) by dividing through by \( e^x \): \( y = x - 1 + Ce^{-x} \).
09
State the general solution
The general solution to the differential equation \( y' = x - y \) is \( y = x - 1 + Ce^{-x} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equation
A first-order linear differential equation is a relationship between a function and its first derivative. It takes the form \(y' + p(x)y = q(x)\), where \(p(x)\) and \(q(x)\) are functions of \(x\).
In these equations, the highest order derivative is the first derivative, hence the term 'first-order'. The equation we have, \(y' = x - y\), fits this form once rearranged to \(y' + y = x\).
This type of differential equation is called linear because the function \(y\) and its derivative appear to the power of one, and purely in a linear form. They are important in mathematical modeling of systems and phenomena because they can describe a wide range of physical processes.
In these equations, the highest order derivative is the first derivative, hence the term 'first-order'. The equation we have, \(y' = x - y\), fits this form once rearranged to \(y' + y = x\).
This type of differential equation is called linear because the function \(y\) and its derivative appear to the power of one, and purely in a linear form. They are important in mathematical modeling of systems and phenomena because they can describe a wide range of physical processes.
Integrating Factor
An integrating factor is a crucial method for solving linear first-order differential equations. It is typically denoted as \(\mu(x)\) and is used to simplify the equation.
To find the integrating factor, we compute \(\mu(x) = e^{\int p(x) \, dx}\), where \(p(x)\) is the coefficient of \(y\) in the standard form.
For the equation \(y' + y = x\), we identify \(p(x) = 1\). Thus, the integrating factor is \(e^x\).
By multiplying the entire differential equation by this factor, the left-hand side becomes an exact derivative, allowing straightforward integration. This simplifies the approach to finding a solution.
To find the integrating factor, we compute \(\mu(x) = e^{\int p(x) \, dx}\), where \(p(x)\) is the coefficient of \(y\) in the standard form.
For the equation \(y' + y = x\), we identify \(p(x) = 1\). Thus, the integrating factor is \(e^x\).
By multiplying the entire differential equation by this factor, the left-hand side becomes an exact derivative, allowing straightforward integration. This simplifies the approach to finding a solution.
Integration by Parts
Integration by parts is a technique to integrate products of functions. It is based on the formula \[\int u \, dv = uv - \int v \, du\]. This method is particularly handy when dealing with an integral like \(\int e^x x \, dx\).
In the process for solving such integrals, we choose parts: \(u = x\) and \(dv = e^x \, dx\). Then, differentiate and integrate respectively to find \(du = dx\) and \(v = e^x\).
Substituting these into the formula, we split the original integral into simpler parts and compute: \(e^x x - \int e^x \, dx\). Solving this leads to \(e^x x - e^x + C\), which is key in solving the differential equation.
In the process for solving such integrals, we choose parts: \(u = x\) and \(dv = e^x \, dx\). Then, differentiate and integrate respectively to find \(du = dx\) and \(v = e^x\).
Substituting these into the formula, we split the original integral into simpler parts and compute: \(e^x x - \int e^x \, dx\). Solving this leads to \(e^x x - e^x + C\), which is key in solving the differential equation.
General Solution
The general solution of a differential equation is an expression that encompasses all possible solutions. For first-order linear equations, the general solution includes a constant \(C\), representing any particular solution dependent on initial conditions.
In our problem, after solving the differential equation \(y' = x - y\) using integrating factors and integration by parts, we arrive at \(y = x - 1 + Ce^{-x}\).
Here, \(x - 1\) is the particular solution to the non-homogeneous part, and \(Ce^{-x}\) is the solution to the homogeneous part (when \(x = 0\)).
The constant \(C\) allows for adjustment to specific initial conditions, ensuring the solution fits given criteria in a real-world context.
In our problem, after solving the differential equation \(y' = x - y\) using integrating factors and integration by parts, we arrive at \(y = x - 1 + Ce^{-x}\).
Here, \(x - 1\) is the particular solution to the non-homogeneous part, and \(Ce^{-x}\) is the solution to the homogeneous part (when \(x = 0\)).
The constant \(C\) allows for adjustment to specific initial conditions, ensuring the solution fits given criteria in a real-world context.