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Mathematics LibreTexts

Another Example

Solve

\[ (x^2-1)y''+xy'-y=0\]

Solution

Let

\[ y = \sum_{n=0}^{\infty} a_n\,x^n\]

then

\[ y' = \sum_{n=1}^{\infty} n\,a_n\,x^{n-1}\]

and

\[ y'' = \sum_{n=2}^{\infty} n(n-1)\, a_n\,x^{n-2}\]

We can write the original differential equation as

\[ x^2 y'' - y'' + xy' -y =0 \]

Substituting back into this differential equation

\[ (x^2-1)\sum_{n=2}^{\infty} n(n-1)\, a_n\,x^{n-2}+x\sum_{n=1}^{\infty} n\,a_n\,x^{n-1}-\sum_{n=0}^{\infty} a_n\,x^n=0\]

and multiplying the \(x^2\) through gives

\[ \sum_{n=2}^{\infty} n(n-1)\, a_n\,x^{n} - \sum_{n=2}^{\infty} n(n-1)\, a_n\,x^{n-2} + \sum_{n=1}^{\infty} n\,a_n\,x^{n}-\sum_{n=0}^{\infty} a_n\,x^n=0\]

We next need to make the second term has the nth power of \(x\) instead of \(n-2\). For this term, we let

           \( u  =  n – 2\) and  \(n  =  u + 2\)

The second term becomes

\[ \sum_{u=0}^{\infty} (u+2)(u+1) a_{u+2}x^u\]

now changing this back to \(n\) and placing the term back into the differential equation gives

\[ \sum_{n=2}^{\infty} n(n-1)\, a_n\,x^{n} - \sum_{n=0}^{\infty} (n+2)(n+1)\, a_{n+2}\,x^{n} + \sum_{n=1}^{\infty} n\,a_n\,x^{n}-\sum_{n=0}^{\infty} a_n\,x^n=0\]

Since they sums do not all start at the same number, we pull out the \(n = 0\) and \(n = 1\) terms to get

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

or

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

We can now combine the series to get

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

We are looking for two linearly independent solutions, so we let the first one be such that

\( y(0)  =  0 \)  and  \(y’(0)  =  1\)

this implies that

\( a_0  =  0 \) and \( a_1  =  1\)

from our last equation we have

\[ 0  =  -2a_2  – a_0  =  -2a_2\]

or

\[ a_2  =  0 \]

We also have

\[ 0  =   - 6a_3 \]

or

\[ a_3  =  0\]

The terms from the series must all be zero, since that is what it means for a polynomial to be zero. Hence

\[ n(n-1)\,a_n -(n+2)(n+1)\,a_{n+2} + n\,a_n - a_n = 0\]

\[ (n+2)(n+1)\,a_{n+2} = \left[ n(n-1)+n-1 \right] a_n = (n-1)(n+1)a_n\]

\[ a_{n+2} = \dfrac{n-1}{n+1} a_n\]

Notice that since

\[   a_2  =  a_3  =  0\]

All of the rest of the coefficients must be zero, since they are each a multiple of the coefficient two before them. Therefore the first linear independent solution is

\[  y_1  =  x \]

For the second linearly independent solution, we let

\( y(0)  =  1 \) and \(y’(0)  =  0\)

this implies that

\( a_0  =  1 \)  and \(a_1  =  0\)

from our last equation we have

\[  1  =  -2a_2 – a_0  =  -2a_2 - 1 \]

or

\[  a_2  =  -1 \]

We also have

\[       0  =  -6a_3   \]

or

\[ a_3  =  0 \]

we still have

\[ a_{n+2} =\dfrac{n-1}{n+1}a_n\]

so notice that the odd terms are all zero. For the even terms, we have

\[ a_4 = -\dfrac{1}{3}\]

\[a_6 = \dfrac{3}{5}\dfrac{1}{3}= \dfrac{1}{5}\]

\[a_8 = -\dfrac{5}{7}\dfrac{1}{5} = -\dfrac{1}{7}\]

\[a_{2n}= \dfrac{(-1)^{n+1}}{2n-1}\]

This one has the series representation

\[ y_2 = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{2n-1} x^{2n} \]

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