Unfortunately for us there is nothing from the first example that can be reused here. Changing to \({x_0} = - 2\) completely changes the problem. In this case our solution will be,

\[y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} \]

The derivatives of the solution are,

\[y'\left( x \right) = \sum\limits_{n = 1}^\infty {n{a_n}{{\left( {x + 2} \right)}^{n - 1}}} \hspace{0.25in}y''\left( x \right) = \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} \]

Plug these into the differential equation.

\[\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} - x\sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} = 0\]

We now run into our first real difference between this example and the previous example. In this case we can’t just multiply the \(x\) into the second series since in order to combine with the series it must be \(x+2\). Therefore, we will first need to modify the coefficient of the second series before multiplying it into the series.

\[\begin{align*}\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} - \left( {x + 2 - 2} \right)\sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} & = 0\\ \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} - \left( {x + 2} \right)\sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} + 2\sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} & = 0\\ \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{{\left( {x + 2} \right)}^{n - 2}}} - \sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^{n + 1}}} + \sum\limits_{n = 0}^\infty {2{a_n}{{\left( {x + 2} \right)}^n}} & = 0\end{align*}\]

We now have three series to work with. This will often occur in these kinds of problems. Now we will need to shift the first series down by 2 and the second series up by 1 the get common exponents in all the series.

\[\sum\limits_{n = 0}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{{\left( {x + 2} \right)}^n}} - \sum\limits_{n = 1}^\infty {{a_{n - 1}}{{\left( {x + 2} \right)}^n}} + \sum\limits_{n = 0}^\infty {2{a_n}{{\left( {x + 2} \right)}^n}} = 0\]

In order to combine the series we will need to strip out the \(n=0\) terms from both the first and third series.

\[\begin{align*}2{a_2} + 2{a_0} + \sum\limits_{n = 1}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{{\left( {x + 2} \right)}^n}} - \sum\limits_{n = 1}^\infty {{a_{n - 1}}{{\left( {x + 2} \right)}^n}} + \sum\limits_{n = 1}^\infty {2{a_n}{{\left( {x + 2} \right)}^n}} & = 0\\ 2{a_2} + 2{a_0} + \sum\limits_{n = 1}^\infty {\left[ {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} - {a_{n - 1}} + 2{a_n}} \right]{{\left( {x + 2} \right)}^n}} & = 0\end{align*}\]

Setting coefficients equal to zero gives,

\[\begin{align*} & n = 0 & 2{a_2} + 2{a_0} & = 0\\ & n = 1,2,3, \ldots & \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} - {a_{n - 1}} + 2{a_n} & = 0\end{align*}\]

We now need to solve both of these. In the first case there are two options, we can solve for \(a_{2}\) or we can solve for \(a_{0}\). Out of habit I’ll solve for \(a_{0}\). In the recurrence relation we’ll solve for the term with the largest subscript as in previous examples.

\[\begin{align*} & n = 0 & {a_2} & = - {a_0}\\ & n = 1,2,3, \ldots & {a_{n + 2}} & = \frac{{{a_{n - 1}} - 2{a_n}}}{{\left( {n + 2} \right)\left( {n + 1} \right)}}\end{align*}\]

Notice that in this example we won’t be having every third term drop out as we did in the previous example.

At this point we’ll also acknowledge that the instructions for this problem are different as well. We aren’t going to get a general formula for the \(a_{n}\)’s this time so we’ll have to be satisfied with just getting the first couple of terms for each portion of the solution. This is often the case for series solutions. Getting general formulas for the \(a_{n}\)’s is the exception rather than the rule in these kinds of problems.

To get the first four terms we’ll just start plugging in terms until we’ve got the required number of terms. Note that we will already be starting with an \(a_{0}\) and an \(a_{1}\) from the first two terms of the solution so all we will need are three more terms with an \(a_{0}\) in them and three more terms with an \(a_{1}\) in them.

\[n = 0\hspace{0.25in}{a_2} = - {a_0}\]

We’ve got two \(a_{0}\)’s and one \(a_{1}\).

\[n = 1\hspace{0.25in}{a_3} = \frac{{{a_0} - 2{a_1}}}{{\left( 3 \right)\left( 2 \right)}} = \frac{{{a_0}}}{6} - \frac{{{a_1}}}{3}\]

We’ve got three \(a_{0}\)’s and two \(a_{1}\)’s.

\[n = 2\hspace{0.25in}{a_4} = \frac{{{a_1} - 2{a_2}}}{{\left( 4 \right)\left( 3 \right)}} = \frac{{{a_1} - 2\left( { - {a_0}} \right)}}{{\left( 4 \right)\left( 3 \right)}} = \frac{{{a_0}}}{6} + \frac{{{a_1}}}{{12}}\]

We’ve got four \(a_{0}\)’s and three \(a_{1}\)’s. We’ve got all the \(a_{0}\)’s that we need, but we still need one more \(a_{1}\). So, we’ll need to do one more term it looks like.

\[n = 3\hspace{0.25in}{a_5} = \frac{{{a_2} - 2{a_3}}}{{\left( 5 \right)\left( 4 \right)}} = - \frac{{{a_0}}}{{20}} - \frac{1}{{10}}\left( {\frac{{{a_0}}}{6} - \frac{{{a_1}}}{3}} \right) = - \frac{{{a_0}}}{{15}} + \frac{{{a_1}}}{{30}}\]

We’ve got five \(a_{0}\)’s and four \(a_{1}\)’s. We’ve got all the terms that we need.

Now, all that we need to do is plug into our solution.

\[\begin{align*}y\left( x \right) & = \sum\limits_{n = 0}^\infty {{a_n}{{\left( {x + 2} \right)}^n}} \\ & = {a_0} + {a_1}\left( {x + 2} \right) + {a_2}{\left( {x + 2} \right)^2} + {a_3}{\left( {x + 2} \right)^3} + {a_4}{\left( {x + 2} \right)^4} + {a_5}{\left( {x + 2} \right)^5} + \cdots \\ & = {a_0} + {a_1}\left( {x + 2} \right) - {a_0}{\left( {x + 2} \right)^2} + \left( {\frac{{{a_0}}}{6} - \frac{{{a_1}}}{3}} \right){\left( {x + 2} \right)^3} + \\ & \hspace{0.25in}\left( {\frac{{{a_0}}}{6} + \frac{{{a_1}}}{{12}}} \right){\left( {x + 2} \right)^4} + \left( { - \frac{{{a_0}}}{{15}} + \frac{{{a_1}}}{{30}}} \right){\left( {x + 2} \right)^5} + \cdots \end{align*}\]

Finally collect all the terms up with the same coefficient and factor out the coefficient to get,

\[\begin{align*}y\left( x \right) & = {a_0}\left\{ {1 - {{\left( {x + 2} \right)}^2} + \frac{1}{6}{{\left( {x + 2} \right)}^3} + \frac{1}{6}{{\left( {x + 2} \right)}^4} - \frac{1}{{15}}{{\left( {x + 2} \right)}^5} + \cdots } \right\} + \\ & \,\,\,{a_1}\left\{ {\left( {x + 2} \right) - \frac{1}{3}{{\left( {x + 2} \right)}^3} + \frac{1}{{12}}{{\left( {x + 2} \right)}^4} + \frac{1}{{30}}{{\left( {x + 2} \right)}^5} + \cdots } \right\}\end{align*}\]

That’s the solution for this problem as far as we’re concerned. Notice that this solution looks nothing like the solution to the previous example. It’s the same differential equation but changing \(x_{0}\) completely changed the solution.