find the first four nonzero terms in a power series expansion about x0 for a general solution to the given…

Explanation:

Step1: Assume power - series solution

Let (y(x)=\sum_{n = 0}^{\infty}a_{n}(x - 1)^{n}), then (y^{\prime}(x)=\sum_{n = 1}^{\infty}na_{n}(x - 1)^{n - 1}) and (y^{\prime\prime}(x)=\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 2}).

Step2: Substitute into the differential equation

First, rewrite (5x^{2}=5((x - 1)+1)^{2}=5((x - 1)^{2}+2(x - 1)+1)). The differential equation (5x^{2}y^{\prime\prime}-y^{\prime}+4y = 0) becomes: [5((x - 1)^{2}+2(x - 1)+1)\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 2}-\sum_{n = 1}^{\infty}na_{n}(x - 1)^{n - 1}+4\sum_{n = 0}^{\infty}a_{n}(x - 1)^{n}=0] [5\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n}+10\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 1}+5\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 2}-\sum_{n = 1}^{\infty}na_{n}(x - 1)^{n - 1}+4\sum_{n = 0}^{\infty}a_{n}(x - 1)^{n}=0]

Step3: Adjust the indices of the sums

For the first sum: (\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n}) For the second sum: Let (m=n - 1), then (10\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 1}=10\sum_{m = 1}^{\infty}(m + 1)ma_{m+1}(x - 1)^{m}) For the third sum: Let (k=n - 2), then (5\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 2}=5\sum_{k = 0}^{\infty}(k + 2)(k + 1)a_{k+2}(x - 1)^{k}) For the fourth sum: (\sum_{n = 1}^{\infty}na_{n}(x - 1)^{n - 1}) For the fifth sum: (\sum_{n = 0}^{\infty}a_{n}(x - 1)^{n}) The equation is: [5\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n}+10\sum_{n = 1}^{\infty}(n + 1)na_{n+1}(x - 1)^{n}+5\sum_{n = 0}^{\infty}(n + 2)(n + 1)a_{n+2}(x - 1)^{n}-\sum_{n = 1}^{\infty}na_{n}(x - 1)^{n - 1}+4\sum_{n = 0}^{\infty}a_{n}(x - 1)^{n}=0]

Step4: Find the recurrence relation

For (n = 0): [5\times2\times1a_{2}-a_{1}+4a_{0}=0\Rightarrow a_{2}=\frac{a_{1}-4a_{0}}{10}] For (n = 1): [10\times2\times1a_{3}+5\times2\times1a_{2}-a_{1}+4a_{1}=0] Substitute (a_{2}=\frac{a_{1}-4a_{0}}{10}) into the above - equation: [20a_{3}+10\times\frac{a_{1}-4a_{0}}{10}+3a_{1}=0] [20a_{3}+a_{1}-4a_{0}+3a_{1}=0\Rightarrow a_{3}=\frac{4a_{0}-4a_{1}}{20}=\frac{a_{0}-a_{1}}{5}]

Step5: Write the power - series solution

[y(x)=a_{0}+a_{1}(x - 1)+\frac{a_{1}-4a_{0}}{10}(x - 1)^{2}+\frac{a_{0}-a_{1}}{5}(x - 1)^{3}+\cdots]

Answer:

(a_{0}+a_{1}(x - 1)+\frac{a_{1}-4a_{0}}{10}(x - 1)^{2}+\frac{a_{0}-a_{1}}{5}(x - 1)^{3})