Question

In Exercises 33-46 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients in each solution. $$ 3 x^{2}\left(2-x^{2}\right) y^{\prime \prime}+x\left(1-11 x^{2}\right) y^{\prime}+\left(1-5 x^{2}\right) y=0 $$

Short answer

In summary, we used the Frobenius method to obtain a fundamental set of solutions for the given differential equation: $$ 3 x^{2}\left(2-x^{2}\right) y^{\prime \prime}+x\left(1-11 x^{2}\right) y^{\prime}+\left(1-5 x^{2}\right) y=0 $$ The fundamental set of Frobenius solutions is given by: $$ y(x) = c_1\left( a_0+\frac{a_0}{2!} x^2 -\frac{5a_0}{3!}x^4+\cdots \right) + c_2\left( xa_1 - \frac{x^3}{3}a_1+\frac{a_1}{5}x^5-\cdots \right) $$ Here, \(c_1\) and \(c_2\) are constants, and \(a_0\) and \(a_1\) are the first coefficients in the respective series.

Step-by-step solution

Rewrite the differential equation in standard form for the Frobenius method

Divide the given differential equation by the coefficient \(3x^2(2-x^2)\): $$ y^{\prime \prime} + \frac{x\left(1-11 x^{2}\right)}{3 x^{2}\left(2-x^{2}\right)}y^{\prime} + \frac{\left(1-5 x^{2}\right)}{3 x^{2}\left(2-x^{2}\right)}y = 0 $$ Now, let \(y(x) = x^r \sum_{n=0}^{+\infty}a_nx^n = x^r a_0 + x^{r+1}a_1 + x^{r+2}a_2+\cdots\). We will plug this into the differential equation and manipulate it to find the indicial equation.

Find the indicial equation

Differentiate y(x) once with respect to x: $$ y'(x) = r x^{r-1}a_0 + (r+1)x^r a_1 + (r+2)x^{r+1}a_2 + \cdots $$ Differentiate y(x) again with respect to x: $$ y''(x) = r(r-1)x^{r-2}a_0 + r(r+1)x^{r-1} a_1 + (r+1)(r+2)x^r a_2 + \cdots $$ Substitute y(x), y'(x), and y''(x) into the rewritten differential equation: $$ \left[r(r-1)x^{r-2}a_0 + r(r+1)x^{r-1} a_1 + (r+1)(r+2)x^r a_2 + \cdots \right] + \frac{x\left(1-11 x^{2}\right)}{3 x^{2}\left(2-x^{2}\right)}\left[r x^{r-1}a_0 + (r+1)x^r a_1 + (r+2)x^{r+1}a_2 + \cdots \right] + \frac{\left(1-5 x^{2}\right)}{3 x^{2}\left(2-x^{2}\right)}\left[x^r a_0 + x^{r+1}a_1 + x^{r+2}a_2+\cdots \right] = 0 $$ Now, expand the terms and collect the coefficients of \(x^r, x^{r+1}, x^{r+2},\cdots\). After comparing, similar powers of x, we identify the indicial equation and the recursion relationships.

Solve the indicial equation and find recursion relationships

Comparing coefficients of like powers of x, we find the indicial equation: $$ r(r-1) = 0 $$ This gives us two possible values for r: \(r_1=0\) and \(r_2=1\). Each value of r will give us a separate recursion relationship from which we will find the coefficients of the Frobenius solution. For \(r_1=0\), the recursion relationship for \(n\geq 1\) is: $$ a_n = -\frac{a_{n-1}}{(n+1)(n(2-n))} $$ For \(r_2=1\), the recursion relationship for \(n\geq 1\) is: $$ a_n = -\frac{(n+1)a_{n-1}}{n(n+2)(2-n)} $$

Determine the Frobenius solutions from the coefficients found

We can now write down the general form of the first few coefficients for the two solutions. We note that, because the indicial equation has a double root, both solutions are linearly independent. For \(r_1=0\): $$ y_1(x) = a_0+\frac{a_0}{2!} x^2 -\frac{5a_0}{3!}x^4+\cdots $$ For \(r_2=1\): $$ y_2(x) = xa_1 - \frac{x^3}{3}a_1+\frac{a_1}{5}x^5-\cdots $$ The fundamental set of Frobenius solutions to our differential equation is: $$ y(x) = c_1\left( a_0+\frac{a_0}{2!} x^2 -\frac{5a_0}{3!}x^4+\cdots \right) + c_2\left( xa_1 - \frac{x^3}{3}a_1+\frac{a_1}{5}x^5-\cdots \right) $$

Key concepts

Indicial Equation

The indicial equation is a pivotal concept in solving differential equations using the Frobenius method. It arises when substituting a power series solution into the differential equation. Essentially, the indicial equation helps determine the potential powers of the leading term in the series solution, known as 'r'. This equation is derived by focusing on the smallest exponent term from the expansion.
In the given differential equation, after substituting the power series into it and collecting terms, we form the indicial equation from the coefficient of the lowest power of x: \[ r(r-1) = 0 \] This specific equation gives two solutions for 'r':

• \( r_1 = 0 \)
• \( r_2 = 1 \)

These values are crucial as they lay the foundation for developing the recursion relationships that determine all other coefficients in the power series solutions.
The presence of a double root in this example indicates that the solutions are linearly independent, offering a complete set of solutions for the differential equation.

Recursion Relationships

Once we determine the values of 'r' from the indicial equation, the next step is to find the recursion relationships. These relationships allow us to calculate all other coefficients in the power series solution, providing a pathway from one coefficient to another.
For each value of 'r', we develop separate recursion formulas. For the first solution \( r_1 = 0 \), the recursion relationship is:

• \[ a_n = -\frac{a_{n-1}}{(n+1)(n(2-n))} \quad \text{for} \quad n \geq 1 \]

This formula specifies how each coefficient \( a_n \) can be found based on its predecessor \( a_{n-1} \). The relationship includes factors involving 'n' that arise from simplifying collected terms in the differential equation.
Similarly, for the second solution \( r_2 = 1 \), the recursion relationship takes the form:

• \[ a_n = -\frac{(n+1)a_{n-1}}{n(n+2)(2-n)} \quad \text{for} \quad n \geq 1 \]

Here, each step builds upon the previous, ensuring a logical and systematic approach to constructing the series solutions.

Differential Equations

Differential equations form the mathematical framework for understanding dynamic systems and changing processes. They are equations that involve derivatives of a function, widely used to describe natural phenomena.
In the context of the Frobenius method, differential equations often cannot be solved through simple techniques. Frobenius provides a powerful method, especially when dealing with equations with variable coefficients or at regular singular points.
For our problem, the differential equation is:\[ 3x^2(2-x^2)y'' + x(1-11x^2)y' + (1-5x^2)y = 0 \]Here, the goal is to find solutions that satisfy this equation for a range of 'x'. By transforming this equation using the series approach, solutions emerge that are expressed in terms of simpler components:

• Power series
• Controlled by recursion
• Dependent on initial conditions given by the indicial equation

This process exemplifies how complex systems described by differential equations can be resolved into more tangible expressions, breaking down complexity into solvable parts.