Chapter 19: Problem 5
Solve the differential equation. \(x y^{\prime}+y+x=e^{x}\)
Short Answer
Expert verified
The solution to the differential equation is \(y = \frac{e^x}{x} - \frac{x}{2} + \frac{C}{x}\).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \(x y' + y + x = e^x\). This can be rearranged into the standard form of a linear first-order differential equation: \(y' + P(x)y = Q(x)\). First, divide the equation by \(x\) to make it easier to identify: \(y' + \frac{1}{x}y = \frac{e^x}{x} - 1\). It is now clear that this is a linear first-order differential equation.
02
Find the Integrating Factor
For a linear differential equation of the form \(y' + P(x)y = Q(x)\), the integrating factor is given by \(\mu(x) = e^{\int P(x) \, dx}\). Here, \(P(x) = \frac{1}{x}\), so the integrating factor \(\mu(x) = e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = |x|\). Since \(x\) is assumed to be positive (as we are dividing by \(x\)), \(\mu(x) = x\).
03
Multiply Through by the Integrating Factor
Multiply every term in the equation \(y' + \frac{1}{x}y = \frac{e^x}{x} - 1\) by the integrating factor \(x\): \[ x \cdot y' + y = e^x - x. \] This simplification shows that the left side is the derivative of \(xy\).
04
Express as Derivative of a Product
Recognize that the left-hand side \(x y' + y\) is the derivative of the product \(xy\). Thus, we can write: \[ \frac{d}{dx}(xy) = e^x - x. \]
05
Integrate Both Sides
Integrate both sides with respect to \(x\): \[ \int \frac{d}{dx}(xy) \, dx = \int (e^x - x) \, dx. \] This yields: \[ xy = \int e^x \, dx - \int x \, dx = e^x - \frac{x^2}{2} + C, \] where \(C\) is the constant of integration.
06
Solve for y(x)
Solve for \(y\) by dividing both sides of the equation by \(x\): \[ y = \frac{e^x}{x} - \frac{x}{2} + \frac{C}{x}. \] This is the general solution to the differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear First-Order Differential Equations
A linear first-order differential equation is a type of differential equation that can be written in the form: \( y' + P(x)y = Q(x) \). This form reveals two components,\( P(x) \) and \( Q(x) \), which are both functions of \( x \). These equations are called 'first-order' because the highest derivative is the first derivative \( y' \), and 'linear' because \( y \) and its derivatives appear to the power of one.
To solve these equations:
To solve these equations:
- First, rearrange or transform the equation into the standard linear form, if necessary.
- Identify \( P(x) \), the coefficient of \( y \).
- Identify \( Q(x) \), the standalone term after rearranging.
Integrating Factor
The integrating factor is a powerful tool for solving linear first-order differential equations. It transforms the equation into a form where it can be easily integrated. The formula for the integrating factor \( \mu(x) \) is given by:\[ \mu(x) = e^{\int P(x) \, dx}.\]This factor is designed specifically to make the left side of the equation a derivative of some product, simplifying the process greatly.
In our example:
In our example:
- We identified \( P(x) = \frac{1}{x} \).
- The integral of \( P(x) \) is \( \int \frac{1}{x} \, dx = \ln|x| \).
- Thus, the integrating factor \( \mu(x) \) becomes \( e^{\ln|x|} = |x| \).
Product Rule
The product rule is a fundamental concept from calculus used to compute the derivative of the product of two functions. The rule states:\[ \frac{d}{dx}(uv) = u'v + uv',\]where \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives. Understanding this rule is essential for recognizing the structure of certain differential equations that can be written as a derivative of a product.
In the context of our example,
In the context of our example,
- The left side of the equation becomes \( (xy)' \) after multiplying by the integrating factor \( x \).
- This means that \( xy' + y = \frac{d}{dx}(xy) \), a key realization for integrating both sides directly.