Question

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

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

Identify the differential equation and its standard form

The given differential equation is $x(x+1)y^{″}-(x-1)y^{\prime }+y=0$. Rewrite in standard form: $$y^{″}+\frac{-(x-1)}{x(x+1)}{y}^{\prime }+\frac{1}{x(x+1)}y=0$$.

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.

Find the indicial equation

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

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 $$s=1$$.

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.

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.

Key concepts

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:

$$x(x+1)y^{″}-(x-1)y^{\prime }+y=0$$

we first rewrite it in its standard form:

$$y^{″}+\frac{-(x-1)}{x(x+1)}{y}^{\prime }+\frac{1}{x(x+1)}y=0$$

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

$$y^{″}+\frac{-(x-1)}{x(x+1)}{y}^{\prime }+\frac{1}{x(x+1)}y=0$$

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:

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

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.