find indicial equation for (x-7)y" - 2x^2y' + 9y=0

To find the indicial equation for the given differential equation, we assume a power series solution of the form: \[y(x) = \sum_{n=0}^{\infty} a_nx^{n+r},\] where \(r\) is the unknown exponent to be determined and \(a_n\) are the coefficients to be found. We differentiate this series twice to find the derivatives needed to substitute into the differential equation. First, we find the first derivative: \[y'(x) = \sum_{n=0}^{\infty} a_n(n+r)x^{n+r-1}.\] Next, we find the second derivative: \[y''(x) = \sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2}.\] Now we substitute the series solution and its derivatives back into the original differential equation: \[(x-7)\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2} - 2x^2\sum_{n=0}^{\infty} a_n(n+r)x^{n+r-1} + 9\sum_{n=0}^{\infty} a_nx^{n+r} = 0.\] Rearranging the terms and combining the series, we obtain: \[\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r} - 7\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2} - 2\sum_{n=0}^{\infty} a_n(n+r)x^{n+r+2} + 9\sum_{n=0}^{\infty} a_nx^{n+r} = 0.\] Now we shift the index of summation in the second term by replacing \(n\) with \(n-2\): \[\sum_{n=2}^{\infty} a_{n+2}(n+r)(n+r-1)x^{n+r} - 7\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2} - 2\sum_{n=0}^{\infty} a_n(n+r)x^{n+r+2} + 9\sum_{n=0}^{\infty} a_nx^{n+r} = 0.\] Next, we combine all the series into a single series and set the coefficient of each power of \(x\) to zero: \[\sum_{n=0}^{\infty} \left[(n+r)(n+r-1)a_n + a_{n+2}(n+r)(n+r+1) - 7(n+r)(n+r-1)a_n - 2(n+r)a_n - 2a_n(n+r)(n+r+1) + 9a_n\right] x^{n+r} = 0.\] Since this equation holds for all values of \(x\), the coefficient of each power of \(x\) must be zero. Therefore, we can write: \[(n+r)(n+r-1)a_n + a_{n+2}(n+r)(n+r+1) - 7(n+r)(n+r-1)a_n - 2(n+r)a_n - 2a_n(n+r)(n+r+1) + 9a_n = 0.\] Simplifying this equation gives: \[(n+r)( \[(n+r)(n+r-1)a_n + a_{n+2}(n+r)(n+r+1) - 7(n+r)(n+r-1)a_n - 2(n+r)a_n - 2a_n(n+r)(n+r+1) + 9a_n = 0.\] Simplifying further, we have: \[(n+r)(n+r-1)a_n + a_{n+2}(n+r)(n+r+1) - 7(n+r)(n+r-1)a_n - 2(n+r)a_n - 2a_n(n+r)(n+r+1) + 9a_n = 0.\] Grouping the terms with common coefficients of \(a_n\) together, we obtain: \[(n+r)(n+r-1 - 7(n+r)(n+r-1) - 2(n+r) - 2(n+r)(n+r+1) + 9)a_n + a_{n+2}(n+r)(n+r+1) = 0.\] Now, to satisfy this equation for all values of \(n\), we equate the coefficient of each power of \(n\) to zero. Therefore, we have: \[(n+r)(n+r-1 - 7(n+r)(n+r-1) - 2(n+r) - 2(n+r)(n+r+1) + 9) = 0.\] This is the indicial equation for the given differential equation.