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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 ...