Solve the differential equations by the Frobenius method; observe that you get only one solution. (Note, also, that the two values of \(s\) are equal or differ by an integer, and in the latter case the larger \(s\) gives the one solution.) Show that the conditions of Fuchs's theorem are satisfied. Knowing that the second solution is \(\ln x\) times the solution you have, plus another Frobenius series, find the second solution. $$x(x+1) y^{\prime \prime}-(x-1) y^{\prime}+y=0$$

Short Answer

Expert verified
Solve the differential equation by verifying Fuchs's theorem, find the indicial equation roots, and construct one solution using Frobenius method with the larger root. The second solution is constructed by multiplying the primary solution by \[ \text{ln}(x) \] plus another Frobenius series.

Step by step solution

01

Identify the differential equation and its standard form

The given differential equation is \( x(x+1)y'' - (x-1)y' + y = 0 \). Rewrite in standard form: \[ y'' + \frac{-(x-1)}{x(x+1)}y' + \frac{1}{x(x+1)}y = 0 \].
02

Verify the conditions of Fuchs's theorem

Identify the coefficients and ensure they are analytic near \( x = 0 \). We see that \[ P(x) = \frac{-(x-1)}{x(x+1)}, \ Q(x) = \frac{1}{x(x+1)} \] are both analytic near \( x = 0 \). Thus, the conditions of Fuchs's theorem are satisfied.
03

Find the indicial equation

Assume a solution of the form \[ y = x^s \with \ P(x) = \frac{-(x-1)}{x(x+1)}, \ Q(x) = \frac{1}{x(x+1)} \]. Substitute and simplify to form the indicial equation: \[ s(s-1) + \frac{-(x-1)}{(x+1)}s + \frac{1}{(x+1)} = 0 \].
04

Solve the indicial equation

Solve for \( s \) in the simplified polynomial. Solve using algebraic methods to obtain the possible values of \( s \). The values are found to be \[ s = 0 \] and \[ s = 1 \].
05

Construct the Frobenius series solution

Use the larger \( s \) value since it gives the non-trivial solution. Assume a solution of the form \[ y = x^1 \times \text{series} \]. Evaluate and find the leading coefficients. Since only one solution arises, identify it as the primary solution.
06

Find the second solution using the known form

Since the second solution is known to be of form \[ y_2 = y_1 \times \text{ln}(x) + \text{Frobenius series} \], construct \[ y_2 \] using the given information.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Fuchs's Theorem
Understanding Fuchs's Theorem is essential in solving differential equations using the Frobenius method. This theorem provides conditions under which a point is classified as a regular singular point of a differential equation. For a differential equation of the form:

\[x(x+1)y'' - (x-1)y' + y = 0\]

we first rewrite it in its standard form:

\[ y'' + \frac{-(x-1)}{x(x+1)}y' + \frac{1}{x(x+1)}y = 0\]

Here, the functions \[ P(x) = \frac{-(x-1)}{x(x+1)} \text{and}\ Q(x) = \frac{1}{x(x+1)} \] should be analytic near the point of interest, which in our problem is \ x = 0 \. We can observe that both \ P(x) \text{and}\ Q(x) \ do not have any essential singularities near \ x = 0 \. Hence, \ x = 0 \ is a regular singular point and the conditions of Fuchs's theorem are satisfied. This validation allows us to use the Frobenius method for solving the differential equation.
Indicial Equation
The indicial equation plays a crucial role in the Frobenius method. It helps identify the roots that determine the coefficients of our series solution. We start by assuming a solution of the form \ y = x^s \. Substituting this assumed solution and its derivatives into the differential equation

\[ y'' + \frac{-(x-1)}{x(x+1)} y' + \frac{1}{x(x+1)} y = 0 \]

gives us an equation that can be simplified to find the values of \ s \. This results in the indicial equation:

\[ s(s-1) + \frac{-(x-1)}{x(x+1)}s + \frac{1}{x(x+1)} = 0 \]

Solving this polynomial, we find two potential values for \ s \: \ s = 0 \ and \ s = 1 \. The indicial equation tells us where to start our series solution and informs us whether we will have one or two linearly independent solutions.
Series Solution
Once the indicial equation gives us the values of \ s \, we can construct the series solution for the differential equation. We use the larger value of \ s \, which in this case is \ s = 1 \, to find the primary solution. We assume the series solution in the form of:

\[ y = x^s \times \text{(series)} \]

By plugging this into the original equation and evaluating, we find the leading coefficients of our series.

Given that \ s = 0 \ and \ s = 1 \ yield the same solution, only one solution emerges naturally. To find the second solution, which is generally of the form:

\[ y_2 = y_1 \time \text{ln}(x) + \text{(Frobenius series)} \]

we use the known primary solution. This logarithmic term suggests that the second solution requires handling the repeated root scenario in the Frobenius method, where the second solution includes a logarithmic component multiplied by the first solution.

Hence, the final complete solution intertwines the primary and secondary solutions effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove the least squares approximation property of Legendre polynomials [see (9.5) and (9.6)] as follows. Let \(f(x)\) be the given function to be approximated. Let the functions \(p_{l}(x)\) be the normalized Legendre polynomials, that is, $$p_{l}(x)=\sqrt{\frac{2 l+1}{2}} P_{l}(x) \quad \text { so that } \quad \int_{-1}^{1}\left[p_{l}(x)\right]^{2} d x=1$$ Show that the Legendre series for \(f(x)\) as far as the \(p_{2}(x)\) term is $$f(x)=c_{0} p_{0}(x)+c_{1} p_{1}(x)+c_{2} p_{2}(x) \quad \text { with } \quad c_{l}=\int_{-1}^{1} f(x) p_{l}(x) d x$$ Write the quadratic polynomial satisfying the least squares condition as \(b_{0} p_{0}(x)+\) \(b_{1} p_{1}(x)+b_{2} p_{2}(x)\) (by Problem 5.14 any quadratic polynomial can be written in this form). The problem is to find \(b_{0}, b_{1}, b_{2}\) so that $$I=\int_{-1}^{1}\left[f(x)-\left(b_{0} p_{0}(x)+b_{1} p_{1}(x)+b_{2} p_{2}(x)\right)\right]^{2} d x$$ is a minimum. Square the bracket and write \(I\) as a sum of integrals of the individual terms. Show that some of the integrals are zero by orthogonality, some are 1 because the \(p_{t}\) 's are normalized, and others are equal to the coefficients \(c_{l}\). Add and subtract \(c_{0}^{2}+c_{1}^{2}+c_{2}^{2}\) and show that $$I=\int_{-1}^{1}\left[f^{2}(x)+\left(b_{0}-c_{0}\right)^{2}+\left(b_{1}-c_{1}\right)^{2}+\left(b_{2}-c_{2}\right)^{2}-c_{0}^{2}-c_{1}^{2}-c_{2}^{2}\right] d x$$ Now determine the values of the \(b\) 's to make \(I\) as small as possible. (Hint: The smallest value the square of a real number can have is zero.) Generalize the proof to polynomials of degree \(n\).

Show that the set of functions \(\sin n x\) is not a complete set on \((-\pi, \pi)\) by trying to expand the function \(f(x)=1\) on \((-\pi, \pi)\) in terms of them.

By Leibniz' rule, write the formula for \(\left(d^{n} / d x^{n}\right)(u v)\).

Prove that the functions \(L_{n}(x)\) are orthogonal on \((0, \infty)\) with respect to the weight function \(e^{-x} .\) Hint: Write the differential equation (22.21) as $$e^{x} \frac{d}{d x}\left(x e^{-x} y^{\prime}\right)+n y=0,$$ and see Sections 7 and \(19 .\)

Solve the following differential equations by series and also by an elementary method and verify that your solutions agree. Note that the goal of these problems is not to get the answer (that's easy by computer or by hand) but to become familiar with the method of series solutions which we will be using later. Check your results by computer. $$x y^{\prime}=y$$

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free