To solve this equation, we will express out unknown solution as an infinite series $\displaystyle \sum_{k=0}^{\infty}a_k x^k.$ Then, $$ \begin{array}{lll} &\displaystyle y'-2xy=1 &\mbox{}\\ \implies &\displaystyle \left(\sum_{k=0}^{\infty}a_k x^k\right)'-2x\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=1}^{\infty}a_k k x^{k-1}-2x\sum_{k=0}^{\infty}a_k x^k=0&\mbox{differentiating term by term}\\ \implies &\displaystyle a_1+\sum_{k=2}^{\infty}a_k k x^{k-1}-\sum_{k=0}^{\infty}2 a_k x^{k+1}=0&\mbox{peeling off the first term of the first series}\\ \implies &\displaystyle a_1+\sum_{k=1}^{\infty}a_{k+1} (k+1) x^{k}-\sum_{k=1}^{\infty}2 a_{k-1} x^{k}=0&\mbox{re-indexing both sums}\\ \implies &\displaystyle a_1+\sum_{k=1}^{\infty}\left[a_{k+1} (k+1) x^{k}-2 a_{k-1}\right] x^{k}=0&\mbox{using properties of series}\\ \end{array} $$ The only way the above series can be identically zero for all $x$ is if $a_1=0$ and $(k+1)a_{k+1}-2 a_{k-1}=0$ for all $k \geq 1.$

From these equations, we may determine the coefficients $a_1, a_2, a_3, \ldots$

From the first equation, we know that $a_1=0.$

For the second equation which is a recurrence relation, we rewrite it as $ka_{k} -2 a_{k-2}=0$ for all $k\geq 2,$ which more conveniently becomes $$ a_{k} =\frac{2}{k}a_{k-2} \mbox{ for all } k\geq 2 $$ Then, $$ \begin{array}{lll} \displaystyle a_0 & &\mbox{an unknown constant}\\ a_1 &=\displaystyle 0&\mbox{}\\ a_2 &=\displaystyle \frac{2}{2}a_{0}=a_0&\mbox{}\\ a_3 &=\displaystyle \frac{2}{3}\cdot 0=0&\mbox{}\\ a_4 &=\displaystyle \frac{2}{4} a_{2}=\frac{1}{2}a_2=\frac{1}{2}a_0&\mbox{}\\ a_5 &=\displaystyle \frac{2}{5} a_{3}=0&\mbox{}\\ a_6 &=\displaystyle \frac{2}{6} a_{4}=\frac{1}{3}\cdot \frac{1}{2}a_0&\mbox{}\\ a_7 &=\displaystyle \frac{2}{7} a_{5}=0&\mbox{}\\ a_8 &=\displaystyle \frac{2}{8} a_{6}=\frac{1}{4}\cdot \frac{1}{3}\cdot \frac{1}{2}a_0&\mbox{}\\ a_9 &=\displaystyle \frac{2}{9} a_{7}=0&\mbox{}\\ a_{10} &=\displaystyle \frac{2}{10} a_{8}=\frac{1}{5}\cdot \frac{1}{4}\cdot \frac{1}{3}\cdot \frac{1}{2}a_0&\mbox{}\\ &\vdots& \end{array} $$ From the above we can see the pattern: $$ a_{2k+1}= 0 $$ and $$ a_{2k}=\frac{1}{k}\cdots \frac{1}{5}\cdot \frac{1}{4}\cdot \frac{1}{3}\cdot \frac{1}{2}a_0=\frac{1}{k!}a_0 $$ Our solution is then $$ \begin{array}{ll} y&=\displaystyle \sum_{k=0}^{\infty}a_k x^k\\ &=\displaystyle \sum_{k=0}^{\infty}a_{2k}x^{2k}+\sum_{k=0}^{\infty}a_{2k+1}x^{2k+1}\\ &=\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!}a_0 x^{2k}+\sum_{k=0}^{\infty}0\cdot x^{2k+1}\\ &=\displaystyle a_0\sum_{k=0}^{\infty} \frac{1}{k!} x^{2k}\\ \end{array} $$ We note that $a_0$ is the constant of our general solution. We also recognize the series $\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!} x^{2k}$ as none other than $e^{x^2}.$ Thus, $$ y=a_0e^{x^2} $$

We will express out unknown solution as an infinite series $\displaystyle \sum_{k=0}^{\infty}a_k x^k.$ Then, $$ \begin{array}{lll} &\displaystyle y''-xy=0 &\mbox{}\\ \implies &\displaystyle \left(\sum_{k=0}^{\infty}a_k x^k\right)''-x\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=2}^{\infty}a_k k(k-1)x^{k-2}-\sum_{k=0}^{\infty}a_k x^{k+1}=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}a_{k+2} (k+2)(k+1)x^{k}-\sum_{k=1}^{\infty}a_{k-1} x^{k}=0&\mbox{re-index sums for common power}\\ \implies &\displaystyle 2a_2+\sum_{k=1}^{\infty}a_{k+2} (k+2)(k+1)x^{k}-\sum_{k=1}^{\infty}a_{k-1} x^{k}=0&\mbox{peel off extra term in first sum}\\ \implies &\displaystyle 2a_2+\sum_{k=1}^{\infty}\left[a_{k+2} (k+2)(k+1)-a_{k-1}\right]x^{k}=0&\mbox{}\\ \end{array} $$ For the above to be identically $0,$ is must be that $a_2=0$ and $a_{k+2} (k+2)(k+1)-a_{k-1}=0$ for all $k\geq 1.$

Then, $$ a_{k+2}=\frac{1}{(k+2)(k+1)}a_{k-1} $$ which becomes $$ a_{k+3}=\frac{1}{(k+3)(k+2)}a_{k} $$ for $k\geq 0.$

We now determine our coefficients noting that $a_0$ and $a_1$ are an undermined constants. $$ \begin{array}{lll} \displaystyle a_0&=\displaystyle a_0 & \mbox{}\\ \displaystyle a_1&=\displaystyle a_1 & \mbox{}\\ \displaystyle a_2&=\displaystyle 0 & \mbox{as explained above}\\ \displaystyle a_3&=\displaystyle \frac{1}{3\cdot 2}a_{0}=\frac{1}{6}a_0 & \mbox{}\\ \displaystyle a_4&=\displaystyle \frac{1}{4\cdot 3}a_{1}=\frac{1}{12}a_1 & \mbox{}\\ \displaystyle a_5&=\displaystyle \frac{1}{5\cdot 4}a_{2}=0 & \mbox{}\\ \displaystyle a_6&=\displaystyle \frac{1}{6\cdot 5}a_{3}=\frac{1}{6\cdot 5}\cdot\frac{1}{6}a_{0}=\frac{1}{180}a_0 & \mbox{}\\ \displaystyle a_7&=\displaystyle \frac{1}{7\cdot 6}a_{4}=\frac{1}{7\cdot 6}\cdot\frac{1}{12}a_1=\frac{1}{504}a_1 & \mbox{}\\ \displaystyle a_8&=\displaystyle \frac{1}{8\cdot 7}a_{5}=0 & \mbox{}\\ \vdots&&\\ \end{array} $$ We write our general solution as $$ \begin{array}{lll} \displaystyle y&\displaystyle=\sum_{k=0}^{\infty}a_k x^k &\mbox{}\\ &\displaystyle= a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5+a_6 x^6+a_7 x^7+a_8 x^8+\cdots &\mbox{}\\ &\displaystyle= a_0+a_1 x+0\cdot x^2+\frac{1}{6}a_0 x^3+\frac{1}{12}a_1 x^4+0\cdot x^5+\frac{1}{180}a_0 x^6+\frac{1}{504}a_1 x^7+0\cdot x^8+\cdots &\mbox{}\\ &\displaystyle= a_0+a_1 x+\frac{1}{6}a_0 x^3+\frac{1}{12}a_1 x^4+\frac{1}{180}a_0 x^6+\frac{1}{504}a_1 x^7+\cdots &\mbox{}\\ &\displaystyle= a_0+\frac{1}{6}a_0 x^3+\frac{1}{180}a_0 x^6+\cdots+a_1 x+\frac{1}{12}a_1 x^4+\frac{1}{504}a_1 x^7+0\cdot x^8+\cdots &\mbox{}\\ \displaystyle &\displaystyle=a_0\left(1+\frac{1}{6}x^3+\frac{1}{180}x^6+\cdots\right)+a_1\left(x+\frac{1}{12}x^4+\frac{1}{504}x^7+\cdots\right) &\mbox{}\\ \end{array} $$

We will express out unknown solution as an infinite series $\displaystyle \sum_{k=0}^{\infty}a_k x^k.$ Then, $$ \begin{array}{lll} &\displaystyle y''+(x-1)y'+y=0 &\mbox{}\\ \implies &\displaystyle \left(\sum_{k=0}^{\infty}a_k x^k\right)''+(x-1)\left(\sum_{k=0}^{\infty}a_k x^k\right)'+\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=2}^{\infty}a_k k(k-1)x^{k-2}+(x-1)\sum_{k=1}^{\infty}a_k k x^{k-1}+\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}a_{k+2} (k+2)(k+1)x^{k}+\color{magenta}{(x-1)\sum_{k=0}^{\infty} a_{k+1} (k+1) x^{k}}+\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}a_{k+2} (k+2)(k+1)x^{k}+\color{magenta}{x\sum_{k=0}^{\infty} a_{k+1} (k+1) x^{k}-\sum_{k=0}^{\infty} a_{k+1} (k+1) x^{k}}+\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}a_{k+2} (k+2)(k+1)x^{k}+\sum_{k=0}^{\infty} a_{k+1} (k+1) x^{k+1}-\sum_{k=0}^{\infty} a_{k+1} (k+1) x^{k}+\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}a_{k+2} (k+2)(k+1)x^{k}+\color{blue}{\sum_{k=1}^{\infty} a_{k} k x^{k}}-\sum_{k=0}^{\infty} a_{k+1} (k+1) x^{k}+\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}a_{k+2} (k+2)(k+1)x^{k}+\color{blue}{\sum_{k=0}^{\infty} a_{k} k x^{k}}-\sum_{k=0}^{\infty} a_{k+1} (k+1) x^{k}+\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ %\implies &\displaystyle \sum_{k=0}^{\infty}a_{k+2} (k+2)(k+1)x^{k}+\sum_{k=0}^{\infty} a_{k} k x^{k}-\sum_{k=0}^{\infty} a_{k+1} (k+1) x^{k}+a_0+\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}\left[a_{k+2} (k+2)(k+1)+ a_{k} k - a_{k+1} (k+1) +a_k \right]x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}\left[a_{k+2} (k+2)(k+1)+ a_{k}(k+1) - a_{k+1} (k+1)\right]x^k=0&\mbox{}\\ \end{array} $$ For the above to be identically $0$ for all $x,$ it must be that $$ a_{k+2} (k+2)(k+1)x^{k}+ a_{k}(k+1) - a_{k+1} (k+1)=0 $$ that is, $$ a_{k+2}=\frac{a_{k+1} (k+1)-a_{k} (k+1)}{(k+2)(k+1)}=\frac{a_{k+1}-a_k}{k+2} $$ for all $k \geq 0.$

Treating $a_0$ and $a_1$ as undetermined constants, we may determine now determine $a_2$ and $a_3.$ $$ \begin{array}{lll} \displaystyle a_2&\displaystyle=\frac{a_1-a_0}{2}=\frac{1}{2}(a_1-a_0) &\mbox{}\\ \displaystyle a_3&\displaystyle=\frac{a_2-a_1}{3}=\frac{\frac{1}{2}(a_1-a_0)-a_1}{3}=\frac{-\frac{1}{2}a_1-\frac{1}{2}a_0}{3}=-\frac{1}{6}(a_1+a_0) &\mbox{}\\ \end{array} $$ So, $$ \begin{array}{lll} \displaystyle y&\displaystyle=\sum_{k=0}^{\infty}a_k x^k &\mbox{}\\ \displaystyle &\displaystyle=a_0+a_1x+a_2x^2+a_3x^3+\cdots &\mbox{}\\ \displaystyle &\displaystyle=a_0+a_1x+\frac{1}{2}(a_1-a_0)x^2-\frac{1}{6}(a_1+a_0)x^3+\cdots &\mbox{}\\ \displaystyle &\displaystyle=a_0\left(1-\frac{1}{2}x^2-\frac{1}{6}x^3+\cdots\right)+a_1\left(x+\frac{1}{2}x^2-\frac{1}{6}x^3+\cdots\right) &\mbox{}\\ \end{array} $$

We will express out unknown solution as an infinite series $\displaystyle \sum_{k=0}^{\infty}a_k x^k.$ Then, $$ \begin{array}{lll} &\displaystyle (4+x)y''+(1-3x)y'+(-1-4x)y=0&\mbox{}\\ \implies &\displaystyle (4+x)\left(\sum_{k=0}^{\infty}a_k x^k\right)''+(1-3x)\left(\sum_{k=0}^{\infty}a_k x^k\right)'+(-1-4x)\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle (4+x)\sum_{k=2}^{\infty}k(k-1)a_k x^{k-2}+(1-3x)\sum_{k=1}^{\infty}ka_k x^{k-1}+(-1-4x)\sum_{k=0}^{\infty}a_k x^{k}=0&\mbox{}\\ \implies &\displaystyle 4\sum_{k=2}^{\infty}k(k-1)a_k x^{k-2}+x\sum_{k=2}^{\infty}k(k-1)a_k x^{k-2}+\sum_{k=1}^{\infty}ka_k x^{k-1}-3x\sum_{k=1}^{\infty}ka_k x^{k-1}-\sum_{k=0}^{\infty}a_k x^{k}-4x\sum_{k=0}^{\infty}a_k x^{k}=0&\mbox{}\\ \implies &\displaystyle \sum_{k=2}^{\infty}4k(k-1)a_k x^{k-2}+\sum_{k=2}^{\infty}k(k-1)a_k x^{k-1}+\sum_{k=1}^{\infty}ka_k x^{k-1}-\sum_{k=1}^{\infty}3ka_k x^{k}-\sum_{k=0}^{\infty}a_k x^{k}-\sum_{k=0}^{\infty}4a_k x^{k+1}=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}4(k+2)(k+1)a_{k+2} x^{k}+\sum_{k=1}^{\infty}(k+1)ka_{k+1} x^{k}+\sum_{k=0}^{\infty}(k+1)a_{k+1} x^{k}-\sum_{k=1}^{\infty}3ka_k x^{k}-\sum_{k=0}^{\infty}a_k x^{k}-\sum_{k=1}^{\infty}4a_{k-1} x^{k}=0&\mbox{}\\ \implies &\displaystyle \color{magenta}{8a_2}+\sum_{\color{magenta}{k=1}}^{\infty}4(k+2)(k+1)a_{k+2} x^{k}+\sum_{k=1}^{\infty}(k+1)ka_{k+1} x^{k}+\color{magenta}{a_1}+\sum_{\color{magenta}{k=1}}^{\infty}(k+1)a_{k+1} x^{k}-\sum_{k=1}^{\infty}3ka_k x^{k}-\color{magenta}{a_0}-\sum_{\color{magenta}{k=1}}^{\infty}a_k x^{k}-\sum_{k=1}^{\infty}4a_{k-1} x^{k}=0&\mbox{}\\ \implies &\displaystyle 8a_2+a_1-a_0+\sum_{k=1}^{\infty}[4(k+2)(k+1)a_{k+2} +(k+1)ka_{k+1}+(k+1)a_{k+1} -3ka_k-a_k -4a_{k-1}] x^{k}=0&\mbox{}\\ \implies &\displaystyle 8a_2+a_1-a_0+\sum_{k=1}^{\infty}[4(k+2)(k+1)a_{k+2} +((k+1)k+(k+1))a_{k+1} -(3k+1)a_k -4a_{k-1}] x^{k}=0&\mbox{}\\ \implies &\displaystyle 8a_2+a_1-a_0+\sum_{k=1}^{\infty}[4(k+2)(k+1)a_{k+2} +(k^2+2k+1)a_{k+1} -(3k+1)a_k -4a_{k-1}] x^{k}=0&\mbox{}\\ \end{array} $$ For the above to be identically $0,$ is must be that $$ 8a_2+a_1-a_0=0 $$ or $$ a_2=\frac{a_0-a_1}{8} $$ and $$ 4(k+2)(k+1)a_{k+2} +(k^2+2k+1)a_{k+1} -(3k+1)a_k -4a_{k-1}=0 $$ or $$ a_{k+2}=\frac{-(k^2+2k+1)a_{k+1} +(3k+1)a_k +4a_{k-1}}{4(k+2)(k+1)} $$ where $k \geq 1.$

We now determine the values of $a_0$ and $a_1$ by satisfying the initial conditions.

Expressing $y(x)=\displaystyle \sum_{k=0}^{\infty}a_k x^k$ as $a_0+a_1 x+a_2 x^2+a_3 x^3+\cdots,$ satisfying the initial conditions is relatively straightforward. $$ \begin{array}{lll} &\displaystyle y(0)=2&\mbox{}\\ \implies &\displaystyle a_0+a_1\cdot 0+a_2\cdot 0^2+a_3\cdot 0^3+\cdots=2&\mbox{}\\ \implies &\displaystyle a_0=2&\mbox{}\\ \end{array} $$ and $$ \begin{array}{lll} &\displaystyle y'(0)=1&\mbox{}\\ \implies &\displaystyle a_1+2a_2\cdot 0+3a_3\cdot 0^2+\cdots=1&\mbox{}\\ \implies &\displaystyle a_1=1&\mbox{}\\ \end{array} $$ We may now have everything we need generate as many terms of the solution as we like.

From the above, $$ a_2=\frac{a_0-a_1}{8}=\frac{2-1}{8}=\frac{1}{8} $$ Using our recursion formula for $\color{magenta}{k=1},$ $$ \begin{array}{lll} \displaystyle a_3&\displaystyle=a_{\color{magenta}{1}+2} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-(\color{magenta}{1}^2+2(\color{magenta}{1})+1)a_{\color{magenta}{1}+1} +(3(\color{magenta}{1})+1)a_{\color{magenta}{1}} +4a_{\color{magenta}{1}-1}}{4(\color{magenta}{1}+2)(\color{magenta}{1}+1)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-4a_{2} +4a_1 +4a_{0}}{24} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-4\cdot \frac{1}{8}+4(1)+4(2)}{24} &\mbox{}\\ \displaystyle &\displaystyle=\frac{23}{48} &\mbox{}\\ \end{array} $$ Using our recursion formula for $\color{magenta}{k=2},$ $$ \begin{array}{lll} \displaystyle a_4&\displaystyle=a_{\color{magenta}{2}+2} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-((\color{magenta}{2})^2+2(\color{magenta}{2})+1)a_{\color{magenta}{2}+1} +(3(\color{magenta}{2})+1)a_{\color{magenta}{2}} +4a_{\color{magenta}{2}-1}}{4(\color{magenta}{2}+2)(\color{magenta}{2}+1)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-9a_{3} +7a_2 +4a_{1}}{48} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-9\cdot \frac{23}{48} +7\cdot \frac{1}{8} +4(1)}{48} &\mbox{}\\ \displaystyle &\displaystyle=\frac{3}{256} &\mbox{}\\ \end{array} $$ Thus, the solution to our initial value problem showing the first $5$ terms is $$ y(x)=2+x+\frac{1}{8}x^2+\frac{23}{48}x^3+\frac{3}{256}x^4+\cdots $$

Expressing the unknown solution as an infinite series, $\displaystyle \sum_{k=0}^{\infty}a_k x^k,$ we have $$ \begin{array}{lll} &\displaystyle x^2y''-y=0&\mbox{}\\ \implies &\displaystyle x^2\left(\sum_{k=0}^{\infty}a_k x^k\right)''-\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle x^2\sum_{k=2}^{\infty}k(k-1)a_k x^{k-2}-\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=2}^{\infty}k(k-1)a_k x^{k}-\sum_{k=0}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=2}^{\infty}k(k-1)a_k x^{k}-a_0-a_1x-\sum_{k=2}^{\infty}a_k x^k=0&\mbox{}\\ \implies &\displaystyle -a_0-a_1x+\sum_{k=2}^{\infty}[k(k-1)a_k -a_k] x^k=0&\mbox{}\\ \implies &\displaystyle -a_0-a_1x+\sum_{k=2}^{\infty}[k(k-1) -1]a_k x^k=0&\mbox{}\\ \implies &\displaystyle -a_0-a_1x+\sum_{k=2}^{\infty}(k^2-k-1)a_k x^k=0&\mbox{}\\ \end{array} $$ In order for the above to be identically $0$ for all $x,$ it must be that $$ -a_0-a_1x=0 $$ for all $x$ and $$ (k^2-k-1)a_k=0 $$ for all $k\geq 2.$

The only way $-a_0-a_1x$ could be $0$ for all $x$ is if $a_1=0,$ and that would force $a_0$ to also be $0.$

Also, since $k^2-k-1\neq 0$ for any $k,$ the only way $(k^2-k-1)a_k$ could be zero for $k\geq 2$ is if $a_k=0.$

Consequently, $y(x)\equiv 0$ is the only solution that we can find using our current method.

However, there are more solutions.

What happened??????????????????????

We express the unknown solution as an infinite series about $x_0=1,$ that is $\displaystyle \sum_{k=0}^{\infty}a_k (x-1)^k.$ $$ \begin{array}{lll} &\displaystyle x^2y''-y=0&\mbox{}\\ \implies &\displaystyle x^2\left(\sum_{k=0}^{\infty}a_k (x-1)^k\right)''-\sum_{k=0}^{\infty}a_k (x-1)^k=0&\mbox{}\\ \implies &\displaystyle x^2\sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k-2}-\sum_{k=0}^{\infty}a_k (x-1)^k=0&\mbox{}\\ \end{array} $$ Now, the coefficient $x^2$ does not play nice with the powers of the form $(x-1)^k.$

So, we express $x^2$ as a sum of powers of $x-1$ by performing a Taylor expansion around $x_0=1.$ $$ \begin{array}{lll} \displaystyle p(x)&\displaystyle=x^2 &\mbox{}\\ \displaystyle p'(x)&\displaystyle=2x &\mbox{}\\ \displaystyle p''(x)&\displaystyle=2 &\mbox{}\\ \displaystyle p^{(k)}(x)&\displaystyle=0 &\mbox{for $k\geq 3$}\\ \end{array} $$ Then $$ \begin{array}{lll} \displaystyle \frac{p(1)}{0!}&\displaystyle=\frac{1}{1}=1 &\mbox{}\\ \displaystyle \frac{p'(1)}{1!}&\displaystyle=\frac{2}{1}=2 &\mbox{}\\ \displaystyle \frac{p''(1)}{2!}&\displaystyle=\frac{2}{2}=1 &\mbox{}\\ \displaystyle \frac{p^{(k)}(1)}{k!}&\displaystyle=\frac{0}{k!}=0 &\mbox{for $k\geq 3$}\\ \end{array} $$ Thus, $p(x)=x^2=1+2(x-1)+(x-1)^2.$

We may now proceed with our first calculation. $$ \begin{array}{lll} &\displaystyle x^2y''-y=0&\mbox{}\\ \implies &\displaystyle x^2\left(\sum_{k=0}^{\infty}a_k (x-1)^k\right)''-\sum_{k=0}^{\infty}a_k (x-1)^k=0&\mbox{}\\ \implies &\displaystyle x^2\sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k-2}-\sum_{k=0}^{\infty}a_k (x-1)^k=0&\mbox{}\\ \implies &\displaystyle \left[1+2(x-1)+(x-1)^2\right]\sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k-2}-\sum_{k=0}^{\infty}a_k (x-1)^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k-2}+2(x-1)\sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k-2}+(x-1)^2\sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k-2}-\sum_{k=0}^{\infty}a_k (x-1)^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k-2}+\sum_{k=2}^{\infty}2k(k-1)a_k (x-1)^{k-1}+\sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k}-\sum_{k=0}^{\infty}a_k (x-1)^k=0&\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}(k+2)(k+1)a_{k+2} (x-1)^{k}+\sum_{k=1}^{\infty}2(k+1)ka_{k+1} (x-1)^{k}+\sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k}-\sum_{k=0}^{\infty}a_k (x-1)^k=0&\mbox{}\\ \implies &\displaystyle \color{magenta}{2a_2+6a_3(x-1)}+\sum_{\color{magenta}{k=2}}^{\infty}(k+2)(k+1)a_{k+2} (x-1)^{k}+\color{magenta}{4a_2(x-1)}+\sum_{\color{magenta}{k=2}}^{\infty}2(k+1)ka_{k+1} (x-1)^{k}+\sum_{k=2}^{\infty}k(k-1)a_k (x-1)^{k}\color{magenta}{-a_0-a_1(x-1)}-\sum_{\color{magenta}{k=2}}^{\infty}a_k (x-1)^k=0&\mbox{}\\ \implies &\displaystyle 2a_2+6a_3(x-1)+4a_2(x-1)-a_0-a_1(x-1)+\sum_{k=2}^{\infty}\left[(k+2)(k+1)a_{k+2}+2(k+1)ka_{k+1}+k(k-1)a_k-a_k\right] (x-1)^k=0&\mbox{}\\ \implies &\displaystyle 2a_2-a_0+(6a_3+4a_2-a_1)(x-1)+\sum_{k=2}^{\infty}\left[(k+2)(k+1)a_{k+2}+2(k+1)ka_{k+1}+k(k-1)a_k-a_k\right] (x-1)^k=0&\mbox{}\\ \end{array} $$ For the above to be identically $0,$ we must have $$ 2a_2-a_0=0 $$ $$ 6a_3+4a_2-a_1=0 $$ and $$ (k+2)(k+1)a_{k+2}+2(k+1)ka_{k+1}+k(k-1)a_k-a_k=0 $$ or $$ a_{k+2}=\frac{-2(k+1)ka_{k+1}-k(k-1)a_k+a_k}{(k+2)(k+1)}=\frac{-2(k+1)ka_{k+1}-(k^2-k-1)a_k}{(k+2)(k+1)} $$ for $k \geq 2.$

Letting $a_0$ and $a_1$ be undetermined constants, $$ a_2=\frac{1}{2}a_0 $$ and $$ \begin{array}{lll} \displaystyle a_3 &\displaystyle=\frac{-4a_2+a_1}{6} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-4\left(\frac{1}{2}a_0\right)+a_1}{6} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-2a_0+a_1}{6} &\mbox{}\\ \end{array} $$ From here we may generate as many terms as we like. $$ \begin{array}{lll} \displaystyle a_4&\displaystyle=a_{\color{magenta}{2}+2} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-2(\color{magenta}{2}+1)(\color{magenta}{2})a_{\color{magenta}{2}+1}-(\color{magenta}{2}^2-\color{magenta}{2}-1)a_{\color{magenta}{2}}}{(\color{magenta}{2}+2)(\color{magenta}{2}+1)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-12a_3-a_2}{12} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-12\left(\frac{-2a_0+a_1}{6}\right)-\frac{1}{2}a_0}{12} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-2\left(-2a_0+a_1\right)-\frac{1}{2}a_0}{12} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-4\left(-2a_0+a_1\right)-a_0}{24} &\mbox{}\\ \displaystyle &\displaystyle=\frac{8a_0-4a_1-a_0}{24} &\mbox{}\\ \displaystyle &\displaystyle=\frac{7a_0-4a_1}{24} &\mbox{}\\ \end{array} $$ Our series expansion, showing the first $5$ terms is $$ y(x)=a_0+a_1(x-1)+\frac{a_0}{2}(x-1)^2+\frac{-2a_0+a_1}{6}(x-1)^3+\frac{7a_0-4a_1}{24}(x-1)^4+\cdots $$ and we may choose $a_0$ and $a_1$ to satisfy initial conditions if necessary.

To confirm our calculations, let's choose initial conditions $y(1)=1$ and $y'(1)=1$ (that is, $a_0=1$ and $a_1=1$) and compare the graph of the exact solution and the $4$th-degree Taylor polynomial approximation $\displaystyle \color{#3e7bb7}{P_4(x)=1+(x-1)+\frac{1}{2}(x-1)^2-\frac{1}{6}(x-1)^3+\frac{1}{8}(x-1)^4}.$

To solve this equation, we will express out unknown solution as an infinite series $\displaystyle \sum_{k=0}^{\infty}a_k x^k.$ Then, $$ \begin{array}{lll} &\displaystyle y'-2xy=1 &\mbox{}\\ \implies &\displaystyle \left(\sum_{k=0}^{\infty}a_k x^k\right)'-2x\sum_{k=0}^{\infty}a_k x^k=1&\mbox{}\\ \implies &\displaystyle \sum_{k=1}^{\infty}a_k k x^{k-1}-2x\sum_{k=0}^{\infty}a_k x^k=1&\mbox{differentiating term by term}\\ \implies &\displaystyle a_1+\sum_{k=2}^{\infty}a_k k x^{k-1}-\sum_{k=0}^{\infty}2 a_k x^{k+1}=1&\mbox{peeling off the first term of the first series}\\ \implies &\displaystyle a_1+\sum_{k=1}^{\infty}a_{k+1} (k+1) x^{k}-\sum_{k=1}^{\infty}2 a_{k-1} x^{k}=1&\mbox{re-indexing both sums}\\ \implies &\displaystyle a_1+\sum_{k=1}^{\infty}\left[a_{k+1} (k+1) x^{k}-2 a_{k-1}\right] x^{k}=1&\mbox{using properties of series}\\ \implies &\displaystyle (a_1-1)+\sum_{k=1}^{\infty}\left((k+1)a_{k+1} -2 a_{k-1}\right) x^{k}=0&\mbox{}\\ \end{array} $$ The only way the above series can be zero for all $x$ is if $a_1-1=0$ and $(k+1)a_{k+1}-2 a_{k-1}=0$ for all $k \geq 1.$

From these equations, we may determine the coefficients $a_1, a_2, a_3, \ldots$

From the first equation, we know that $a_1=1.$

For the second equation which is a recurrence relation, we rewrite it as $ka_{k} -2 a_{k-2}=0$ for all $k\geq 2,$ which more conveniently becomes $$ a_{k} =\frac{2}{k}a_{k-2} \mbox{ for all } k\geq 2 $$ Then, $$ \begin{array}{lll} \displaystyle a_0 & &\mbox{an unknown constant}\\ a_1 &=\displaystyle 1&\mbox{}\\ a_2 &=\displaystyle \frac{2}{2}a_{0}=a_0&\mbox{}\\ a_3 &=\displaystyle \frac{2}{3}a_{1}=\frac{2}{3}&\mbox{}\\ a_4 &=\displaystyle \frac{2}{4} a_{2}=\frac{1}{2}a_2=\frac{1}{2}a_0&\mbox{}\\ a_5 &=\displaystyle \frac{2}{5} a_{3}=\frac{2}{5}\cdot \frac{2}{3}&\mbox{}\\ a_6 &=\displaystyle \frac{2}{6} a_{4}=\frac{1}{3}\cdot \frac{1}{2}a_0&\mbox{}\\ a_7 &=\displaystyle \frac{2}{7} a_{5}=\frac{2}{7}\cdot \frac{2}{5}\cdot \frac{2}{3}&\mbox{}\\ a_8 &=\displaystyle \frac{2}{8} a_{6}=\frac{1}{4}\cdot \frac{1}{3}\cdot \frac{1}{2}a_0&\mbox{}\\ a_9 &=\displaystyle \frac{2}{9} a_{7}=\frac{2}{9}\cdot \frac{2}{7}\cdot \frac{2}{5}\cdot \frac{2}{3}&\mbox{}\\ a_{10} &=\displaystyle \frac{2}{10} a_{8}=\frac{1}{5}\cdot \frac{1}{4}\cdot \frac{1}{3}\cdot \frac{1}{2}a_0&\mbox{}\\ &\vdots& \end{array} $$ From the above we can see the pattern: $$ a_{2k+1}= \frac{2}{(2k+1)}\cdots \frac{2}{9}\cdot \frac{2}{7}\cdot \frac{2}{5}\cdot \frac{2}{3}=\frac{2^{k}}{3\cdot 5 \cdot 7 \cdots (2k+1)} $$ and $$ a_{2k}=\frac{1}{k}\cdots \frac{1}{5}\cdot \frac{1}{4}\cdot \frac{1}{3}\cdot \frac{1}{2}a_0=\frac{1}{k!}a_0 $$ Our solution is then $$ \begin{array}{ll} y&=\displaystyle \sum_{k=0}^{\infty}a_k x^k\\ &=\displaystyle \sum_{k=0}^{\infty}a_{2k}x^{2k}+\sum_{k=0}^{\infty}a_{2k+1}x^{2k+1}\\ &=\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!}a_0 x^{2k}+\sum_{k=0}^{\infty}\frac{2^{k}}{3\cdot 5 \cdot 7 \cdots (2k+1)}x^{2k+1}\\ &=\displaystyle a_0\sum_{k=0}^{\infty} \frac{1}{k!} x^{2k}+\sum_{k=0}^{\infty}\frac{2^{k}}{3\cdot 5 \cdot 7 \cdots (2k+1)}x^{2k+1}\\ \end{array} $$ We note that $a_0$ is the constant of our general solution. We also recognize the series $\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!} x^{2k}$ as none other than $e^{x^2}.$ Thus, $$ y=a_0e^{x^2}+\sum_{k=0}^{\infty}\frac{2^{k}}{3\cdot 5 \cdot 7 \cdots (2k+1)}x^{2k+1} $$ We see that $y_h=a_0e^{x^2}$ solves the homogeneous equation (our first example) and $\displaystyle y_p=\sum_{k=0}^{\infty}\frac{2^{k}}{3\cdot 5 \cdot 7 \cdots (2k+1)}x^{2k+1}$ is the particular solution.

Letting our solution be an infinite series $\displaystyle \sum_{k=0}^{\infty}a_k x^k,$ we have $$ \begin{array}{lll} &\displaystyle y''-xy=\frac{1}{1-x} &\mbox{}\\ \implies &\displaystyle \left(\sum_{k=0}^{\infty}a_k x^k\right)''-x\sum_{k=0}^{\infty}a_k x^k=\sum_{k=0}^{\infty} x^k &\mbox{}\\ \implies &\displaystyle \sum_{k=2}^{\infty}k(k-1)a_k x^{k-2}-\sum_{k=0}^{\infty}a_k x^{k+1}-\sum_{k=0}^{\infty} x^k=0 &\mbox{}\\ \implies &\displaystyle \sum_{k=0}^{\infty}(k+2)(k+1)a_{k+2} x^{k}-\sum_{k=1}^{\infty}a_{k-1} x^{k}-\sum_{k=0}^{\infty} x^k=0 &\mbox{}\\ \implies &\displaystyle 2a_2+\sum_{k=1}^{\infty}(k+2)(k+1)a_{k+2} x^{k}-\sum_{k=1}^{\infty}a_{k-1} x^{k}-1-\sum_{k=1}^{\infty} x^k=0 &\mbox{}\\ \implies &\displaystyle 2a_2-1+\sum_{k=1}^{\infty}[(k+2)(k+1)a_{k+2}-a_{k-1}-1] x^k=0 &\mbox{}\\ \end{array} $$ For the above to be identically $0$ for all $x,$ it must be that $$ 2a_2-1=0 $$ and $$ (k+2)(k+1)a_{k+2} -a_{k-1}-1=0 $$ that is, $$ a_{k+2}=\frac{a_{k-1}+1}{(k+2)(k+1)} $$ for all $k\geq 1.$

The initial conditions give us that $a_0=0$ and $a_1=0,$ and the above gives that $\displaystyle a_2=\frac{1}{2}.$

Using our recursion formula, we may now generate as many terms as we like. $$ \begin{array}{lll} \displaystyle a_3&\displaystyle=\frac{a_{0}+1}{(1+2)(1+1)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{0+1}{(3)(2)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{6} &\mbox{}\\ \end{array} $$ $$ \begin{array}{lll} \displaystyle a_4&\displaystyle=\frac{a_{1}+1}{(2+2)(2+1)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{0+1}{(4)(3)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{12} &\mbox{}\\ \end{array} $$ $$ \begin{array}{lll} \displaystyle a_5&\displaystyle=\frac{a_{2}+1}{(3+2)(3+1)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\frac{1}{2}+1}{(5)(4)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{3}{40} &\mbox{}\\ \end{array} $$ Thus, our series solution, showing the first $4$ non-zero terms is $$ y(x)=\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{12}x^{4}+\frac{3}{40}x^{5}+\cdots $$