Question Use an appropriate infinite series method about x = 0 to find two solutions of the given differential equation. (Enter the first four nonzero terms for each linearly Independent solution, if there are fewer than four nonzero terms then enter all terms. Some beginning terms have been provided for you.) y" - xy' - -

Ansy^{\prime \prime}-x y^{\prime}-3 y=0A linear homogerious ODE with constant coefficients\begin{array}{l}p(x) y^{\prime \prime}+q(x) y^{\prime}+\gamma(x) y=0 \\p(x)=1, q(x)=-x, \gamma(x)=-3 .\end{array}plug in x=0 in p(x)=1p(0)=1(0) \neq 0, there fore None.assume a solution form y=\sum_{k=0}^{\infty} a_{k} x^{k}y^{\prime}=\left(\sum_{k=0}^{\infty} a_{k} x^{k}\right)^{1}Rewrite as y^{\prime}=\left(a_{0}+\sum_{k=1}^{\infty} a_{k} x^{k}\right)^{!}(f \pm g)^{\prime}=f^{\prime} \pm g^{\prime} \Rightarrow y^{\prime}=\left(a_{0}\right)^{\prime}+\sum_{k=1}^{\infty} a_{k} k x^{k-1}y^{\prime}=0+\sum_{k=1}^{\infty} a_{k} k x^{k-1}simplify y^{\prime}=\sum_{k=1}^{\infty} a_{k} k x^{k-1}\begin{aligned}y^{\prime \prime} & =\left(\sum_{k=1}^{\infty} a_{k} k x^{k-1}\right)=\left(a_{1}+\sum_{k=2}^{\infty} a_{k} x^{k-1}\right)^{\prime} \\y^{\prime \prime} & =\left(a_{1}\right)^{\prime}+f_{k=2}^{\infty}+g^{\prime} a_{k} a_{k} x^{k-1}(k-1)\end{aligned}a_{1}^{\prime}=0,\left(a_{k} k x^{k-1}\right)^{\prime}=a_{k}^{k x^{k-2}(k-1)}y^{\prime \prime}=0+\sum_{k=2}^{\infty} a k^{k}(k-1) x^{k-2}simplifyy^{\prime \prime}=\sum_{k-2}^{\infty} a_{k} k(k-1) x^{k-2}plug into y^{\prime \prime}-x y^{\prime}-3 y=0\sum_{=2}^{\infty} a_{k} k(k-1) x^{k-2}-x \cdot\left(\sum_{k=1}^{\infty} a_{k}+k^{k-1}\right)-3 \sum_{k=0}^{\infty} a_{k} x^{k}=0iimplify=2 a_{k} k(k-1) x^{k-2}-\sum_{k=1}^{\infty} k a_{k^{x}} x^{k}-\sum_{k=0}^{\infty} 3 a k^{x^{k}}Rewrile starting k=0, and power of x shitting.=\sum_{k=0}^{\infty} a_{k+2}(k+2)(k+1) x^{k}-\sum_{k=0}^{\infty} k a_{k^{2}}-\sum_{k=0}^{\infty} 3 a_{k^{x}}^{k}implify=\sum_{k=0}^{\infty}\left(a_{k+2}(k+2)(k+1)-k a_{k}-3 a_{k}\right) 2^{k}powerseries equat to zero, cocffeccient olso 0\begin{array}{l}a_{k+2}(k+2)(k+1)-1 a_{k}-3 a_{k}=0 \\a_{k+2}(k+2)(k+1)=+\left(+k a_{k}-3 a_{k}\right)\end{array}Divide by bothsides (k+2)(k+1)\begin{array}{l} \frac{a_{k+2}(k+2)(k+1)}{(k+2)(k+1)} \cdot \frac{k a_{k}-3 a_{k}}{(k+2)(k+1)} \\a_{k+2}=\frac{+a_{k}(k+3)}{(k+2)(k+1)} \\a_{k}=\frac{a_{k-2}(k+1)}{(k-1) k}\end{array}solve\begin{aligned}a_{2} n & =\frac{a_{2 n}-2(2 n+1)}{(2 n-1) \cdot 2 n} \\a_{2} & =\frac{a_{0}(3)}{(1)(2)} \\a_{4} & =\frac{a_{2}(5)}{(3)(4)}=\frac{a_{0} 5}{(1)(2)(4)} \\& \vdots \\a_{2 n} & =\frac{a_{0}(5)(2 n+1)}{(1)(2)(4) \ldots(2 n-1)(2 n)}\end{aligned}Therefore the solution isy_{1}=a_{0}\left(1+\frac{x^{2}(3)}{(1)(1)}+\frac{x^{4}(5)}{(1)(2)(4)}+\cdots+\frac{x^{2 n}(5)}{(1)(2)(4) \ldots}+\cdots\right.simplifyy_{1}=a_{0}\left(1+\frac{3 x^{2}}{2}+\frac{5 x y}{8}+\cdots+\frac{x^{2 n}(5)(2 n+1)}{(1)(2)(4)-(2 n-1)}\right)ao replaced with c_{1}y_{1}=c_{1}\left(1+\frac{3 x^{2}}{2}+\frac{5 x^{4}}{8}+\cdots-\frac{x^{2 n(5) \cdot(2 n+1)}+\cdots}{(1)(2)(4) \ldots(2 n-1)}\right)solve for k=2 n+1\begin{array}{l}a_{2 n+1}=\frac{a_{2 n-1}(n+1)}{n(2 n+1)} \\a_{3}=\frac{a_{1}(4)}{(2)(3)} \\a_{5}=\frac{a_{3}(6)}{(4)(5)}=\frac{a_{1}(6)}{(2)(5)(5)} \\1 \\a_{2} n+1=\frac{a_{1}(6)(n+1)}{(2)(3)(5) \ldots n(2 n+1)}\end{array}y_{2}=a_{1}\left(x+\frac{x^{3}(4)}{(2)(3)}+\frac{x^{5}(6)}{(2)(3)(5)}+\cdots+\frac{x^{2 n+1}(6)(n+1)}{(2)(3)(5)-n(2 n+1)}\right.Simplifyy_{2}=a_{1}\left(x+\frac{2 x^{3}}{3}+\frac{x^{5}}{5}+\cdots+\frac{x^{2 n+1}(6)(n+1)}{(2)(3)(5) \ldots n(2 n+1)}+\cdots\right)a_{1} replaced with c_{2}\begin{array}{l} y_{2}=c_{2}\left(x+\frac{2 x^{3}}{3}+\frac{x^{5}}{5}+\cdots+\frac{x^{2 n+1}(6)(n+1)}{(2)(3)(5) \cdots n(2 n+1)}\right)^{+\cdots} \\y=c_{1}\left(1+\frac{3 x^{2}}{2}+\frac{5 x^{4}}{8}+\cdots+\frac{x^{2 n}(5)(2 n+1)}{(1)(2)(4) \ldots(2 n-1)(2 n)}+\cdots\right. \\+c_{2}\left(x+\frac{2 x^{3}}{3}+\frac{x^{5}}{5}+\cdots+\frac{x^{2 n+1}(6)(n+1)}{(2)(3)(5) \cdots n(2 n+1)}+\cdots\right. \\y_{1}=\left(1+\frac{3}{2} x^{2}+\frac{5}{8} x^{4}+\cdots+\frac{x^{2 n}(5)(2 n+1)}{(1)(2)(4) \cdots(2 n-1)(n)}\right. \\y_{2}=\left(x+\frac{2}{3} x^{3}+\frac{x^{5}}{5}+\cdots+\frac{x^{2 n+1}(6)(n+1)}{}+\cdots\right)(3)(5) \cdots n(2 n+1)\end{array}Thank yowplease give me upvote ...