Solve the Differential Equation: (1 - x^2) y'' - (4x) y' + (6) y = 0

[VinnyCee]

Here is the problem (number 9 in 5.2 of Boyce, DiPrima 8th Edition Book):

$$(1 - x^2)\,y'' - (4x)\,y' + (6)\,y = 0,\,x_0 = 0$$

Here is what I have so far:

$$\sum_{n = 0}^{\infty}\,(n + 2)\,(n + 1)\,a_{n + 2}\,x^n - \sum_{n = 2}^{\infty}\,n(n - 1)\,a_n\,x^n - 4\,\sum_{n = 1}^{\infty}\,n\,a_n\,x^n + 6\,\sum_{n = 0}^{\infty}\,a_n\,x^n = 0$$

Now, I took out the first two terms (n = 0 and n = 1) of the first sum to make it's index go to n = 2 in order to add all the sums together, right?

$$2\,a_2 + 6\,a_3\,x$$

Continuing in that fashion to add the sums, i get this:

$$2\,a_2\,+\,6\,a_3\,x\,-\,4\,a_1\,x\,+\,6\,a_0\,+\,6\,a_1\,x\,+\,\sum_{n = 2}^{\infty}\,\left[(n\,+\,2)\,(n\,+\,1)\,a_{n\,+\,2}\,-\,n\,(n\,-\,1)\,a_n\,-\,4\,n\,a_n\,+\,6\,a_n\right]\,x^n\,=\,0$$

This is where I am stuck. I am not sure if I pulled out those factors correctly or what, please help. Thank you.

 

[VinnyCee]

Whoops!

I copied the problem wrong!

It is supposed to be:

$$(1 + x^2)\,y'' - (4x)\,y' + (6)\,y = 0,\,x_0 = 0$$

And NOT:

$$(1 - x^2)\,y'' - (4x)\,y' + (6)\,y = 0,\,x_0 = 0$$

One little minus sign!

The whole solution is worked out as the last problem http://tutorial.math.lamar.edu/AllBrowsers/3401/SeriesSolutions.asp !

 

[thepatientmental]

To solve this differential equation, we can use the method of power series. First, we can rewrite the given equation as:

y'' - \frac{4x}{1-x^2}y' + \frac{6}{1-x^2}y = 0

Next, we can substitute the power series representation of y into the equation:

y = \sum_{n=0}^\infty a_nx^n

y' = \sum_{n=0}^\infty (n+1)a_{n+1}x^n

y'' = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n

Substituting these into the original equation, we get:

\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n - \frac{4x}{1-x^2}\sum_{n=0}^\infty (n+1)a_{n+1}x^n + \frac{6}{1-x^2}\sum_{n=0}^\infty a_nx^n = 0

We can simplify this to:

\sum_{n=0}^\infty ((n+2)(n+1)a_{n+2} - 4(n+1)a_{n+1} + 6a_n)x^n = 0

Since this holds for all values of x, we can equate the coefficients of each power of x to 0. This gives us the following recurrence relation for the coefficients:

(n+2)(n+1)a_{n+2} - 4(n+1)a_{n+1} + 6a_n = 0

We can solve for the coefficients using this recurrence relation. Starting with n = 0, we can solve for a_0. Then, using a_0, we can solve for a_1. Continuing in this manner, we can find all the coefficients a_n.

Finally, we can substitute the values of a_n into the power series representation of y to get the solution to the differential equation.