Chapter 11: Problem 8
Find the general solution of the given first-order linear differential equation. State an interval over which the general solution is valid. $$ y^{\prime}=\frac{2 y}{x}+x^{3} e^{x}-1 $$
Short Answer
Expert verified
The solution is \( y = x^2(x - 1)e^x + x + Cx^2 \), valid for \( x \neq 0 \).
Step by step solution
01
Identify and Rewrite Differential Equation
The given first-order linear differential equation is \( y' = \frac{2y}{x} + x^3 e^x - 1 \). Rewriting it in standard linear form, it becomes \( y' - \frac{2y}{x} = x^3 e^x - 1 \). The equation is \( y' + P(x)y = Q(x) \), where \( P(x) = -\frac{2}{x} \) and \( Q(x) = x^3 e^x - 1 \).
02
Find Integrating Factor
To solve the linear differential equation, we need to find the integrating factor \( \mu(x) \). The integrating factor is given by \( \mu(x) = e^{\int P(x) \ dx} = e^{\int -\frac{2}{x} \ dx} = e^{-2\ln|x|} = |x|^{-2} = \frac{1}{x^2} \).
03
Multiply Through by Integrating Factor
Multiply the entire differential equation by the integrating factor \( \frac{1}{x^2} \):\[\frac{1}{x^2}y' - \frac{2}{x^3}y = \frac{x e^x}{x^2} - \frac{1}{x^2}.\] This gives:\[\frac{d}{dx} \left( \frac{y}{x^2} \right) = x e^x - \frac{1}{x^2}.\]
04
Integrate Both Sides
Integrate both sides with respect to \( x \):\[\int \frac{d}{dx} \left( \frac{y}{x^2} \right) \ dx = \int \left( x e^x - \frac{1}{x^2} \right) \ dx\]\[\frac{y}{x^2} = \int x e^x \ dx - \int \frac{1}{x^2} \ dx + C,\]where \( C \) is the constant of integration.
05
Solve the Integrals
The integral \( \int x e^x \ dx \) can be solved using integration by parts, resulting in \(( x - 1)e^x\). The integral \( \int \frac{1}{x^2} \ dx \) is \( -\frac{1}{x} \). Thus, \[\frac{y}{x^2} = (x - 1)e^x + \frac{1}{x} + C.\]
06
Solve for y
Solve for \( y \):\[y = x^2 \left((x - 1)e^x + \frac{1}{x} + C\right).\]This simplifies to:\[y = x^2(x - 1)e^x + x + Cx^2.\]
07
Determine Interval of Validity
The interval of validity is determined by the restrictions on \( x \). Since the differential equation contains \( P(x) = -\frac{2}{x} \), \( x \) cannot be zero. Thus, the solution is valid for \( x \in (-\infty, 0) \cup (0, \infty) \).
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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
First-order linear differential equations are types of equations that involve the rate of change of a function with respect to one variable. These equations are written in a form where the highest derivative is first-order. Typically, they appear as \( y' + P(x)y = Q(x) \). The goal is to find the function \( y(x) \) that satisfies this relationship.
To simplify the identification and solution processes, these equations:
To simplify the identification and solution processes, these equations:
- Contain a single derivative \( y' \)
- Have coefficients \( P(x) \) and \( Q(x) \) which are functions of the independent variable \( x \)
integrating factor
The integrating factor is a clever mathematical tool used to solve first-order linear differential equations. Once a differential equation is rewritten in its standard form \( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is introduced to transform the equation into a solvable state.
The integrating factor is calculated as:
The integrating factor is calculated as:
- \( \mu(x) = e^{\int P(x) \ dx} \)
- This involves computing the integral of the function \( P(x) \)
- The resulting function is then used to multiply through the entire equation
interval of validity
The interval of validity refers to the range of \( x \) values over which the solution to a differential equation is valid. This is crucial because certain values of \( x \) might make terms in the differential equation undefined.
Common factors affecting the interval include:
Common factors affecting the interval include:
- Points where denominators in the equation become zero, leading to undefined mathematical operations
- Discontinuities in \( P(x) \) and \( Q(x) \)
- Initial conditions provided with a specific problem, often determining a starting point