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Chapter 3: Problem 34

$$ \begin{aligned} &2 x^{2} y^{\prime \prime}+5 x y^{\prime}+y=x^{2}-x \\ &y=c_{1} x^{-1 / 2}+c_{2} x^{-1}+\frac{1}{15} x^{2}-\frac{1}{6} x,(0, \infty) \end{aligned} $$

Short Answer

Expert verified
The given solution fits the differential equation, confirming it is correct.

Step by step solution

01

Verify the given differential equation

The given second order differential equation is \(2x^{2}y'' + 5xy' + y = x^{2} - x\). We will first clarify this equation to confirm it matches the solution form provided.
02

Identify The Solution Structure

The general solution provided is: \(y = c_1 x^{-1/2} + c_2 x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x\). This consists of a homogeneous solution \(c_1 x^{-1/2} + c_2 x^{-1}\) and a particular solution \(\frac{1}{15}x^2 - \frac{1}{6}x\).
03

Verify Homogeneous Solution

Rewrite the homogeneous part \(2x^{2}y''_h + 5xy'_h + y_h = 0\), assume \(y_h = c_1 x^{-1/2} + c_2 x^{-1}\) and compute its derivatives: \(y'_h = -\frac{1}{2}c_1 x^{-3/2} - c_2 x^{-2}\) and \(y''_h = \frac{3}{4}c_1 x^{-5/2} + 2c_2 x^{-3}\). Substitute these into the homogeneous equation to confirm it satisfies \(=0\).
04

Check Particular Solution

Verify the particular solution \(y_p = \frac{1}{15}x^2 - \frac{1}{6}x\) by calculating \(y'_p = \frac{2}{15}x - \frac{1}{6}\) and \(y''_p = \frac{2}{15}\). Substitute \(y_p\), \(y'_p\), and \(y''_p\) into the original equation to ensure the equation \(=x^2-x\) holds.
05

Conclusion of Verification

After substitution, the homogeneous solution cancels out, and the particular solution provides the \(x^2-x\) part of the equation, confirming the solution \(y = c_1 x^{-1/2} + c_2 x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x\) is indeed a valid solution over the interval \((0, \infty)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homogeneous Solution
When we talk about the *homogeneous solution* in the context of differential equations, we refer to the solution of the equation when it is set to zero. In simpler terms, if we have a differential equation like \(2x^{2}y'' + 5xy' + y = 0\), the solution to this is called the homogeneous solution.

The homogeneous solution helps us understand how the system behaves without any external influences or inputs. It's like understanding how a swinging pendulum moves on its own, without any push from outside.

In our specific equation, the homogeneous solution is expressed as \(c_1 x^{-1/2} + c_2 x^{-1}\). Here, \(c_1\) and \(c_2\) are arbitrary constants. This expression comes from solving the reduced form of the differential equation, where we set the right-hand side to zero. By finding the homogeneous solution, we recognize the natural modes of the system. This means we're finding solutions that describe the inherent behavior of the system (i.e., the solution that arises naturally without external forcing).
Particular Solution
In differential equations, the *particular solution* represents a solution to the equation with an external force or input. For instance, if our equation is \(2x^{2}y'' + 5xy' + y = x^2 - x\), the particular solution will satisfy this specific non-homogeneous part.

In this case, the particular solution is given by \(\frac{1}{15}x^2 - \frac{1}{6}x\). This part of the solution is like applying an external influence to a system鈥攕uch as pushing a pendulum to make it swing a certain way.

Finding a particular solution involves assuming a form that is similar to the external force (right-hand side of the equation). We then adjust constants to specifically fit the equation. It's tailored to respond to the specific contextual problem or input described by the equation. This process of determining the particular solution ensures that the equation's unique conditions are met. Thus, it describes how the system responds to specific scenarios or influences.
Second Order Differential Equation
A *second order differential equation* is a type of equation that involves the unknown function and its derivatives up to the second order. It is an essential concept in mathematics because it models many real-world phenomena, such as oscillations and waves.

The general form of a second-order differential equation is \(a(x) y'' + b(x) y' + c(x) y = d(x)\). The equation we have is \(2x^{2}y'' + 5xy' + y = x^2 - x\), which fits this pattern, where:
  • \(a(x) = 2x^2\)
  • \(b(x) = 5x\)
  • \(c(x) = 1\)
  • \(d(x) = x^2 - x\)
Understanding second-order differential equations is crucial as they allow us to describe complex systems where various forces interact. These equations often show up in physics, engineering, finance, and other fields. Learning to solve them by combining homogeneous and particular solutions equips us with tools to model and predict the behavior of diverse systems under different conditions.

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Most popular questions from this chapter

Solve the given system of differential equations by systematic elimination. $$ \begin{aligned} &\frac{d x}{d t}=-x+z \\ &\frac{d y}{d t}=-y+z \\ &\frac{d z}{d t}=-x+y \end{aligned} $$

Find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$ y \quad c_{1}+c_{2} e^{3 x} $$

Consider a pendulum that is released from rest from an initial displacement of \(\theta_{0}\) radians. Solving the linear model (7) subject to the initial conditions \(\theta(0)=\theta_{0}, \theta^{\prime}(0)=0\) gives \(\theta(t)=\theta_{0} \cos \sqrt{g} \| t .\) Theperiod of oscillations predicted by this modelisgivenbythefamiliarformula \(T=2 \pi / \sqrt{g l l}=2 \pi \sqrt{U g}\). The interesting thing about this formula for \(T\) is that it does not depend on the magnitude of the initial displacement \(\theta_{0}\). In other words, the linear model predicts that the time that it would take the pendulum to swing from an initial displacement of, say, \(\theta_{0}=\pi / 2\left(=90^{\circ}\right)\) to \(-\pi / 2\) and back again would be exactly the same time to cycle from, say, \(\theta_{0}=\pi / 360\left(=0.5^{\circ}\right)\) to \(-\pi / 360\). This is intuitively unreasonable; the actual period must depend on \(\theta_{0}\). If we assume that \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) and \(l=32 \mathrm{ft}\), then the period of oscillation of the linear model is \(T=2 \pi \mathrm{s}\). Let us compare this last number with the period predicted by the nonlinear model when \(\theta_{0}=\pi / 4\). Using a numerical solver that is capable of generating hard data, approximate the solution of $$\frac{d^{2} \theta}{d t^{2}}+\sin \theta=0, \quad \theta(0)=\frac{\pi}{4}, \quad \theta^{\prime}(0)=0$$ for \(0 \leq t \leq 2\). As in Problem 24 , if \(t_{1}\) denotes the first time the pendulum reaches the position \(O P\) in Figure 3.11.3, then the period of the nonlinear pendulum is \(4 t_{1} .\) Here is another way of solving the equation \(\theta(t)=0\). Expeniment with small step sizes and advance the time staning at \(t=0\) and ending at \(t=2\). From your hard data, observe the time \(t_{1}\) when \(\theta(t)\) changes, for the first time, from positive to negative. Use the value \(t_{1}\) to determine the true value of the period of the nonlinear pendulum. Compute the percentage relative error in the period estimated by \(T=2 \pi\).

Solve the given system of differential equations by systematic elimination. $$ \begin{aligned} &\frac{d x}{d t}+\frac{d y}{d t}=e^{t} \\ &-\frac{d^{2} x}{d t^{2}}+\frac{d x}{d t}+x+y=0 \end{aligned} $$

Solve the given differential equation by undetermined coefficients. \(y^{\prime \prime}-4 y=\left(x^{2}-3\right) \sin 2 x\)
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