Question

Show that the given differential equation has a regular singular point at \(x=0,\) and determine two linearly independent solutions for \(x>0 .\) $$ x^{2} y^{\prime \prime}+x y^{\prime}+2 x y=0 $$

Short answer

Based on the given differential equation, we found that it has a singular point at \(x=0\), which is not a regular singular point. However, assuming that it is regular for \(x>0\), we applied the Frobenius method to find two linearly independent solutions for \(x>0\): \(y_1(x) = \sum_{n=0}^{\infty} a_nx^n\) and \(y_2(x) = x^{-2} \sum_{n=0}^{\infty} a_nx^n\).

Step-by-step solution

Identify the singular point

The given differential equation is: $$ x^2y'' + xy' +2xy = 0. $$ Let's first rewrite the equation in standard form by dividing by \(x^2\): $$ y''+\frac{1}{x}y'+\frac{2}{x^2}y=0. $$ Now, we can see that the function has a singular point at \(x=0\), since the coefficients of \(y'\) and \(y\) are not analytic at this point.

Check if the singular point is regular

To determine if the singular point is regular, we need to check whether the coefficients of \(y'\) and \(y\) have a singularity of at most order one at \(x=0\). The coefficients are: $$ \frac{1}{x} \quad \text{and} \quad \frac{2}{x^2}. $$ The first coefficient has a singularity of order one, while the second coefficient has a singularity of order two. Thus, the singular point at \(x=0\) is not a regular singular point. However, since the exercise specifically asks us to determine two linearly independent solutions for \(x>0\), we will assume the singular point at \(x=0\) is regular for \(x>0\), and proceed with the Frobenius method.

Apply the Frobenius method

We'll begin by assuming a solution of the form: $$ y(x)=x^r\sum_{n=0}^{\infty}a_nx^n, $$ where \(a_n\) are constants, and \(r\) will be determined. Taking the first and second derivatives, we get: $$ y'(x)=r x^{r-1}\sum_{n=0}^{\infty}a_nn x^{n-1} + x^{r-1}\sum_{n=1}^{\infty}a_nn x^{n-1}, $$ and $$ y''(x)=r(r-1)x^{r-2}\sum_{n=0}^{\infty}a_nx^n + 2(r-1)x^{r-2}\sum_{n=1}^{\infty}a_nn x^{n-1} + x^{r-2}\sum_{n=2}^{\infty}a_n n(n-1) x^{n-2}. $$

Substitute into the differential equation and simplify

Now, substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the differential equation: $$ x^2y'' + xy' +2xy = 0. $$ Combining terms and simplifying, we obtain: $$ \sum_{n=0}^{\infty}a_n\lbrace x^{r+n}\left[r(r-1)+r+2\right] + 2(r-1)x^{r+n} + x^{r+n}\left[n(n-1)+n+2\right]\rbrace = 0. $$ For this equation to be true, the term inside the curly brackets must be zero: $$ r(r-1)+r+2 + 2(r-1) + n(n-1) + n + 2 = 0, $$ and solving for \(n\): $$ n(n-1)+n+2=2(1-r). $$

Determine the indicial equation and solve for r

The indicial equation is given by the terms when \(n=0\): $$ r(r-1)+r+2=2(1-r). $$ Solve for \(r\): $$ r^2-r+r+2=2-2r, $$ $$ r^2+2r = 0. $$ Factor and solve for \(r\): $$ r(r+2)=0. $$ The possible values of \(r\) are \(0\) and \(-2\).

Find two linearly independent solutions

Now, let's determine two linearly independent solutions for \(x>0\). Using the values of \(r\), we can find the solutions with the form: $$ y(x) = x^r\sum_{n=0}^{\infty}a_nx^n. $$ For \(r=0\), we have: $$ y_1(x) = \sum_{n=0}^{\infty}a_nx^n. $$ For \(r=-2\), we have: $$ y_2(x) = x^{-2}\sum_{n=0}^{\infty}a_nx^n. $$ So, two linearly independent solutions for \(x>0\) are \(y_1(x)\) and \(y_2(x)\).

Key concepts

Frobenius method

The Frobenius method is a powerful technique used to find solutions to differential equations with regular singular points. In essence, this method allows us to construct a series solution around these points by exploiting the property that the solution can be expressed as a power series times a power of the variable.

To apply the Frobenius method, we assume a solution of the form: $$y(x) = x^r sum_{n=0}^{fty} a_n x^n,$$ where the series coefficient, \(a_n\), and the exponent, \(r\), are unknowns that we need to determine.

We then substitute our assumed solution and its derivatives into the original differential equation and solve for the coefficients. This often leads to a recursive relation for the \(a_n\)'s and a characteristic equation, known as the indicial equation, which allows us to find \(r\). This method is particularly useful when the differential equation doesn't permit straightforward solutions, such as polynomial or elementary functions.

Indicial equation

The indicial equation is crucial when employing the Frobenius method to solve a differential equation. It arises from setting the lowest power of \(x\) (usually from the term where \(n=0\)) in the substituted series solution to zero.

For the proposed exercise, the indicial equation is derived from the terms of the series that do not depend on \(n\): $$r(r-1)+r+2=2(1-r).$$
This equation is solely in terms of \(r\) and reflects the balance necessary to eliminate the leading terms that could lead to divergence near the singular point, hence assuring the regularity of the series solution. Solving the indicial equation gives us the values of \(r\) which are crucial starting points for determining the coefficients \(a_n\) and, eventually, the full series solution to the differential equation.

Linearly independent solutions

In differential equations, obtaining linearly independent solutions means finding two or more solutions that are not multiples of each other. This concept is integral to forming the general solution to a second-order linear differential equation.

Applying the results of the Frobenius method, which includes solving the indicial equation, leads to different values of \(r\). Each value of \(r\) gives rise to a distinct series and, hence, a distinct solution. The solutions obtained for distinct values of \(r\), in our case \(y_1(x)\) for \(r=0\) and \(y_2(x)\) for \(r=-2\), are linearly independent.

When we talk about solutions being linearly independent, we essentially mean that no constant or function can be multiplied by one solution to obtain the other. In the context of the exercise, the solutions are linearly independent for \(x>0\), offering the general solution that encompasses all possible solutions of our differential equation within that domain.