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

Substitute $y$, $y^{\prime}$ and $y^{\prime\prime}$ into $5x^{2}y^{\prime\prime}-y^{\prime}+4y = 0$. First, note that $x^{2}=[1+(x - 1)]^{2}=1 + 2(x - 1)+(x - 1)^{2}$. Then $5x^{2}y^{\prime\prime}=5\left(1 + 2(x - 1)+(x - 1)^{2}\right)\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 2}=5\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 2}+10\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 1}+5\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n}$. $y^{\prime}=\sum_{n = 1}^{\infty}na_{n}(x - 1)^{n - 1}$ and $4y = 4\sum_{n = 0}^{\infty}a_{n}(x - 1)^{n}$.

Step3: Re - index the series

For the first series in $5x^{2}y^{\prime\prime}$: Let $m=n - 2$, then $\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 2}=\sum_{m = 0}^{\infty}(m + 2)(m+1)a_{m + 2}(x - 1)^{m}$. For the second series in $5x^{2}y^{\prime\prime}$: Let $m=n - 1$, then $\sum_{n = 2}^{\infty}n(n - 1)a_{n}(x - 1)^{n - 1}=\sum_{m = 1}^{\infty}(m + 1)ma_{m+1}(x - 1)^{m}$. The differential equation becomes: [ \begin{align*} 5\sum_{m = 0}^{\infty}(m + 2)(m + 1)a_{m+2}(x - 1)^{m}+10\sum_{m = 1}^{\infty}(m + 1)ma_{m + 1}(x - 1)^{m}+5\sum_{m = 2}^{\infty}m(m - 1)a_{m}(x - 1)^{m}-\sum_{m = 0}^{\infty}(m + 1)a_{m+1}(x - 1)^{m}+4\sum_{m = 0}^{\infty}a_{m}(x - 1)^{m}&=0 \end{align*} ]

Step4: Find the recurrence relation

For $m = 0$: [ \begin{align*} 5\times2\times1a_{2}-a_{1}+4a_{0}&=0\ 10a_{2}&=a_{1}-4a_{0}\ a_{2}&=\frac{a_{1}-4a_{0}}{10} \end{align*} ] For $m = 1$: [ \begin{align*} 5\times3\times2a_{3}+10\times2\times1a_{2}-2a_{2}-a_{2}+4a_{1}&=0\ 30a_{3}+(20 - 2 - 1)a_{2}+4a_{1}&=0\ 30a_{3}+17a_{2}+4a_{1}&=0 \end{align*} ] Substitute $a_{2}=\frac{a_{1}-4a_{0}}{10}$ into the above equation: [ \begin{align*} 30a_{3}+17\times\frac{a_{1}-4a_{0}}{10}+4a_{1}&=0\ 30a_{3}+\frac{17a_{1}-68a_{0}}{10}+4a_{1}&=0\ 30a_{3}&=\frac{68a_{0}-17a_{1}-40a_{1}}{10}\ 30a_{3}&=\frac{68a_{0}-57a_{1}}{10}\ a_{3}&=\frac{68a_{0}-57a_{1}}{300} \end{align*} ]

Step5: Write the power - series solution

[ y(x)=a_{0}+a_{1}(x - 1)+\frac{a_{1}-4a_{0}}{10}(x - 1)^{2}+\frac{68a_{0}-57a_{1}}{300}(x - 1)^{3}+\cdots ]

Answer:

$a_{0}+a_{1}(x - 1)+\frac{a_{1}-4a_{0}}{10}(x - 1)^{2}+\frac{68a_{0}-57a_{1}}{300}(x - 1)^{3}$