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Chapter 5: Problem 5

Apply the power series method. Do this by band, not by a \(\mathrm{CAS},\) so that you get a feel for the method, e.g., why a series may terminate, or has even powers only, or has no constant or Linear terms, etc. Show the details of your work. $$(2+x) y^{\prime}=y$$

Short Answer

Expert verified
The solution is a constant function, specifically \( y = a_0 \).

Step by step solution

01

Set up the power series

Assume that the solution can be represented as a power series: \[ y = \sum_{n=0}^{\infty} a_n x^n \]Differentiate to find \(y'\): \[ y' = \sum_{n=1}^{\infty} na_n x^{n-1} \]Substitute these into the differential equation \((2+x)y' = y\).
02

Substitute the power series into the equation

Substitute the power series into the equation:\[(2 + x)\sum_{n=1}^{\infty} na_n x^{n-1} = \sum_{n=0}^{\infty} a_n x^n\]Multiply out the left side:\[2\sum_{n=1}^{\infty} na_n x^{n-1} + x\sum_{n=1}^{\infty} na_n x^{n-1} = \sum_{n=0}^{\infty} a_n x^n\]This becomes: \[2\sum_{n=1}^{\infty} na_n x^{n-1} + \sum_{n=1}^{\infty} na_n x^{n} = \sum_{n=0}^{\infty} a_n x^n\].
03

Align the series powers

Re-index the first term on the left so all series are compatible. For the first series:\[2\sum_{n=1}^{\infty} na_n x^{n-1} = 2\sum_{m=0}^{\infty} (m+1)a_{m+1} x^{m} \] For the second term, there is no need for a re-index:\[\sum_{n=1}^{\infty} na_n x^n = \sum_{m=1}^{\infty} ma_m x^m \] However, extend it down to start at \(m=0\) because the coefficient of \(x^0\) must be zero:\[\sum_{m=0}^{\infty} ma_m x^m \] (with \(m_0 = 0\)).Make sure the series on the right also start at \(m=0\): \[\sum_{m=0}^{\infty} a_m x^m\].
04

Combine and equate coefficients

Combine the series and equate coefficients for each power of \(x\) to find expressions for \(a_n\):\[2(m+1)a_{m+1} + ma_{m} = a_{m}\]Simplifying this equality, we find:\[2(m+1)a_{m+1} = 0\]Thus:\[a_{m+1} = 0\] implies \(a_{1}, a_{2}, \ldots \text{ all } 0\) meaning the series terminates.
05

Determine constant term and solution

The equating implies all higher \(a_n\) coefficients are zero beyond the first one. Thus:\[y = a_0 \]This indicates the solution is simply a constant, and further computation shows:\[y(x) = a_0\cdot 1\].

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Key Concepts

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

Differential Equations
Differential equations form a fundamental part of calculus and mathematical modeling. They are equations that involve a function and its derivatives. Essentially, they help us describe how things change. For example, they can model how populations grow, how heat transfers, or how objects move.
  • In our example, we have the differential equation \((2+x) y^{\prime}=y\). This is a first-order differential equation, which means it involves only the first derivative of \(y\).
  • The goal is to find a function \(y(x)\) that satisfies this relationship for all \(x\).
This specific equation signifies that the change in \(y\), represented by \(y'\), is proportional to \(y\) itself, modified by the factor \((2+x)\). Such equations can often be challenging to solve with direct methods, which is why we explore techniques like the power series solution.
Power Series Solution
The power series method is a powerful technique used to solve differential equations, especially when a neat, straightforward algebraic solution is elusive. In essence, it represents functions as an infinite sum of terms, each term being a multiple of a power of \(x\).
  • In our exercise, we assume the solution \(y\) can be expressed as a power series: \( y = \sum_{n=0}^{\infty} a_n x^n \).
  • This lets us handle more complex functions by breaking them down into simpler polynomial terms.
We then differentiate this series to find \(y'\), and substitute both \(y\) and \(y'\) back into the original differential equation. This substitution lets us convert a differential equation into an algebraic equation involving power series. This simplification is crucial, as it enables us to solve the equation term-by-term rather than dealing with the whole function head-on.
Re-indexing Series
Re-indexing a series is a process of adjusting the starting point and index of a series. It ensures that all related series have a compatible form, which simplifies comparing and combining them.
  • In our solution, we re-index the series to have common starting indices, so every term for each power of \(x\) can match easily.
  • For example, the term \(2\sum_{n=1}^{\infty} na_n x^{n-1}\) is re-indexed as \(2\sum_{m=0}^{\infty} (m+1)a_{m+1} x^{m}\).
Why do this? Because when you rearrange terms to align, you can easily equate coefficients from series on both sides of an equation. This is a key step in solving power series solutions for differential equations, as it allows you to find the coefficients \(a_n\) systematically, moving one power of \(x\) at a time.
Terminating Series
In the context of power series, a terminating series is one where almost all the coefficients vanish鈥攑ractically, it behaves like a polynomial of finite degree.
  • In our example, after re-indexing and equating powers, we found that \(a_{m+1} = 0\) whenever \(m \geq 1\).
  • This tells us that the series does not continue indefinitely; it stops after the first constant term, making it a terminating series.
The important insight here is that in certain equations, conditions on coefficients cause the series to "terminate," or be cut short, due to zero values for \(a_n\). This result simplifies our solution drastically, as in this case, where \(y(x) = a_0\cdot 1\), a constant function, is the solution. Terminating series often appear in boundary value problems and indicate solutions that have a finite expression, even when initially expressed as an infinite series.

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Most popular questions from this chapter

Using the indicated substitutions, find a general solution in terms of \(J_{r}\) and \(J_{-r}\) or indicate when this is not possible. (This is just a sample of various ODEs reducible to Bessel's equation. Some more follow in the next problem set Show the details of your work.) $$x^{2} y^{\prime \prime}+x y^{\prime}+\left(\lambda^{2} x^{2}-y^{2}\right) y=0 \quad(\lambda x=z)$$

Find a power series solution in powers of \(x\). (Show the details of your work.) $$y^{\prime \prime}+x y=0$$

Showing the details of your calculations, develop: $$x^{2}-x^{2}+x-1$$

Determine the radius of convergence (Show the details.) $$\sum_{m=1}^{\infty} \frac{1}{\pi^{m}}\left(x-\frac{1}{2} \pi\right)^{m}$$

Find a power series solution in powers of \(x\). (Show the details of your work.) $$y^{\prime \prime}-4 x y^{\prime}+\left(4 x^{2}-2\right) y=0$$
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