Problem 70
Question
Second-Order Potpourri For each of the following second-order differential equations, find at least one particular solution. You will need to call on past experience with functions you have differentiated. For a significantly greater challenge, find the general solution (which will involve two unknown constants) (a)\(y^{\prime \prime}=x\) (b)\(y^{\prime \prime}=-x\) (c)\(y^{\prime \prime}=-\sin x\) (d)\(y^{n}=y\) (e)\(y^{\prime \prime}=-y\)
Step-by-Step Solution
Verified Answer
The solutions for the exercises are: (a) \( y = \frac{1}{6}x^3 + Cx + D \), (b) \( y = -\frac{1}{6}x^3 + Cx + D \), (c) \( y = -\sin x + Cx + D \), (d) \( y = Ax^n + Bx + C \), and (e) \( y = A\sin(x) + B\cos(x) \)
1Step 1: Exercise (a)
Consider \( y^{\prime \prime}=x \). If one chooses a parabolic function such as \( y = \frac{1}{6}x^3 + Cx + D \), they will find that the second derivative \( y^{\prime \prime} = x \), thus solving the equation. Here, C and D are arbitrary constants.
2Step 2: Exercise (b)
For \( y^{\prime \prime}=-x \), a function like \( y = -\frac{1}{6}x^3 + Cx + D \) is needed as the second derivative \( y^{\prime \prime} = -x \). Again C and D are arbitrary constants.
3Step 3: Exercise (c)
In the case of \( y^{\prime \prime}=-\sin x \), using a function of the form \( y = -\sin x + Cx + D \) will give \( y^{\prime \prime} = -\sin x \). Here, C and D are arbitrary constants.
4Step 4: Exercise (d)
Exercise (d) is a simple power rule dimensional analysis. Here, any constant scaled power rule dimension such as \( y = Ax^n + Bx + C \), where A and B are scaling constants and C is an offset constant.
5Step 5: Exercise (e)
In \( y^{\prime \prime}=-y \), the solution is derived from the equation of harmonic oscillator. The general solution of such differential equation is \( y = A\sin(x) + B\cos(x) \), where A and B are arbitrary constants.
Key Concepts
Particular SolutionGeneral SolutionHarmonic Oscillator
Particular Solution
In differential equations, a particular solution is one that satisfies the differential equation as well as initial conditions or specific values within the context. For instance, when working with the second-order differential equation like those given in Exercises (a), (b), and (c), the particular solution is a function that, when substituted into the equation, will directly yield the expression on the right-hand side of the equation. It's like finding a key that fits perfectly into a lock.
In the examples provided, a particular solution is obtained by selecting functions whose second derivatives match the non-homogeneous term of the differential equation. A simple trick here is to think about what the original function might look like, given that you know its second derivative. This often requires recalling past experience with differentiation. For equations dealing with polynomials, sines, and cosines, the process can be intuitive, with practice. As proposed in Exercise (a) with the solution being a polynomial function, we see that after differentiating twice, the resulting expression directly matches the non-homogeneous term, confirming that we have indeed found a particular solution.
In the examples provided, a particular solution is obtained by selecting functions whose second derivatives match the non-homogeneous term of the differential equation. A simple trick here is to think about what the original function might look like, given that you know its second derivative. This often requires recalling past experience with differentiation. For equations dealing with polynomials, sines, and cosines, the process can be intuitive, with practice. As proposed in Exercise (a) with the solution being a polynomial function, we see that after differentiating twice, the resulting expression directly matches the non-homogeneous term, confirming that we have indeed found a particular solution.
General Solution
A general solution, in contrast to a particular solution, encompasses a family of solutions with arbitrary constants representing an infinite set of possible particular solutions. The general solution of a second-order differential equation includes two arbitrary constants because it is a second-order equation, implying two initial conditions are needed to find a unique solution.
For instance, in Exercises (a) and (b), the general solution included two arbitrary constants C and D alongside the unique terms derived from the particular solution. These constants are placeholders for any real number and can be determined if additional information is given – such as initial values or boundary conditions. The general solution contains all possible trajectories consistent with the differential equation's behavior, like a map of all paths that could be taken to satisfy the equation's rules. In Exercise (e), the general solution to the harmonic oscillator equation is given, where A and B are these essential constants, without which we couldn't capture the full richness of the motion described by the equation.
For instance, in Exercises (a) and (b), the general solution included two arbitrary constants C and D alongside the unique terms derived from the particular solution. These constants are placeholders for any real number and can be determined if additional information is given – such as initial values or boundary conditions. The general solution contains all possible trajectories consistent with the differential equation's behavior, like a map of all paths that could be taken to satisfy the equation's rules. In Exercise (e), the general solution to the harmonic oscillator equation is given, where A and B are these essential constants, without which we couldn't capture the full richness of the motion described by the equation.
Harmonic Oscillator
The harmonic oscillator is a fundamental concept ubiquitous in physics, representing systems where the force is proportional to the displacement but in the opposite direction. The beauty of the harmonic oscillator lies in its simplicity and its applicability to a myriad of physical situations, from pendulums to molecular vibrations.
As illustrated in Exercise (e), the general solution to the harmonic oscillator's second-order differential equation, where the acceleration (the second derivative of the displacement) is proportional and opposite to the position, yields the sine and cosine functions. The equation is of the form
As illustrated in Exercise (e), the general solution to the harmonic oscillator's second-order differential equation, where the acceleration (the second derivative of the displacement) is proportional and opposite to the position, yields the sine and cosine functions. The equation is of the form
y'' = -ky, where k is a positive constant, describing the system's natural frequency. The solution involves both sine and cosine because these functions represent oscillatory motion through their periodic nature. This allows us to cover a full range of oscillatory behaviors depending on the choice of initial conditions, which are captured within the constants A (amplitude of the sine term) and B (amplitude of the cosine term). It's like configuring a swing to match exactly how we want it to move through the air, with A and B dictating the starting position and velocity of the swing.Other exercises in this chapter
Problem 68
In Exercises 67 and \(68,\) make a substitution \(u=\cdots(\) an expression in \(x), \quad d u=\cdots .\) Then (a) integrate with respect to \(u\) from \(u(a)\)
View solution Problem 69
Differential Equation Potpourri For each of the following differential equations, find at least one particular solution. You will need to call on past experienc
View solution Problem 71
True or False By \(u\) -substitution, \(\int_{0}^{\pi / 4} \tan ^{3} x \sec ^{2} x d x=\) \(\int_{0}^{\pi / 4} u^{3} d u .\) Justify your answer.
View solution Problem 72
True or False If \(f\) is positive and differentiable on \([a, b],\) then $$\int_{a}^{b} \frac{f^{\prime}(x) d x}{f(x)}=\ln \left(\frac{f(b)}{f(a)}\right) .$$ J
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