Chapter 4: Problem 13
The differential equation corresponding to the family of curves \(y=e^{x}(a x+b)\) is (a) \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-y=0\) (b) \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y=0\) (c) \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=0\) (d) \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}-y=0\)
Short Answer
Step by step solution
Understand the Given Family of Curves
Differentiate to Find the First Derivative
Differentiate Again to Find the Second Derivative
Eliminate Parameters and Form the DE
Select the Correct Option
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-order Differential Equation
For example, acceleration is the second derivative of position with respect to time.
When solving second-order differential equations, the goal is to find the original function, given the equation of its second derivative.
These equations are usually denoted as \( \frac{d^2 y}{dx^2} \) and they may also involve the first derivative \( \frac{dy}{dx} \) and or the function itself \( y \).
The standard form of a second-order linear differential equation can be expressed as:
- \( a(x) \frac{d^2 y}{dx^2} + b(x) \frac{dy}{dx} + c(x)y = f(x) \)
- Where \( a(x), b(x), c(x) \) are coefficient functions and \( f(x) \) is the non-homogeneous term.
Family of Curves
This equation will include one or more parameters. These parameters act as variables within the family, effectively generating multiple curves by varying their values.
In the original exercise, the family of curves is represented by the equation \( y = e^{x}(ax + b) \). Here, \( a \) and \( b \) serve as parameters. Adjusting these parameters will yield different individual curves within the family, all sharing a structural relationship.
Families of curves are significant in calculus and differential equations because they allow for analyzing trends and patterns across multiple solutions derived from varying specific conditions or parameters.
Overall, understanding families of curves aids in crafting a differential equation that accurately models certain phenomena, providing a broader view by considering all possible curves that adhere to the initial equation.
Differentiation Techniques
It is crucial when dealing with differential equations, as it helps in deriving relationships between functions and their rates of change.
In the provided solution, the product rule is utilized to differentiate the function \( y = e^{x}(ax + b) \). The product rule, a key differentiation technique, states:\[\frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v'\] For \( u = ax + b \) and \( v = e^x \), their derivatives are substituted back into the product rule to find the first derivative \( \frac{dy}{dx} \). This process is repeated to obtain the second derivative \( \frac{d^2y}{dx^2} \).
Differentiation techniques, such as the product rule, are instrumental in systematically breaking down complex expressions. Understanding these techniques fully is vital, especially when creating or solving differential equations, as they form the groundwork for more complex mathematical models and analyses.
Elimination of Parameters
The aim is to remove the arbitrary constants or parameters, usually denoted as \( a \) and \( b \) from the equation, so that the resulting differential equation only contains the variable \( x \) and the function \( y \).
In the exercise, parameters are eliminated by utilising the relationships obtained from differentiating the original function. Using equations from each differentiation step:
- Start with \( y = e^x(ax+b) \),
- Use \( \frac{dy}{dx} = e^x(ax+a+b) \),
- And \( \frac{d^2y}{dx^2} = e^x(ax+2a+b) \).