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The given is $$\frac{dy}{dx} = 2xy$$ Then moving everything to one side and changing the terms $$y^{'}-2xy =0 $$ We assume that $y= \sum^\infty_{n=0}C_nx^n$ and taking the derivative is: $$y^{'}=\sum^\infty_{n=1}nC_nx^{n-1}$$ Then plugging back into the given yields: $$\sum^\infty_{n=1}nC_nx^{n-1}-\sum^\infty_{n=0}2nC_nx^{n+1}=0$$ Now we need to reindex: $$C_1+\sum^\infty_{k=0}\left[(k+1)C_{k-1}-2C_{k+1}\right]x^{k}=0$$ Then we can see that $C_1=0$ and our recurrence formula is: $$C_{k+1}=\frac{2C_{k-1}}{k+1}$$ Then after finding the different values of $k$ in the recurrence gives us: $$k=1:C_2=\frac{2C_0}{2}=C_0$$ $$k=2:C_3=\frac{2C_1}{3}$$ $$k=3: C_4=\frac{2C_2}{4}$$ So our final answers are: $$y_1=C_0$$ $$y_2=\sum^\infty_{n=1} \frac{x^{2n}}{n!}$$ Was I correct in my evaluation of the series?

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    $\begingroup$ Yes. Note that you can evaluate the original differential equation by separation of variables which gives you the solution of $e^{x^2}$ and that matches your series. $\endgroup$ Apr 24, 2020 at 1:30
  • $\begingroup$ No, see my detailed answer. For example, clearly $C_0$ is not a solution. $\endgroup$
    – John B
    Apr 24, 2020 at 1:33

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$$\sum^\infty_{n=1}nC_nx^{n-1}-\sum^\infty_{n=0}2C_nx^{n+1}=0$$ the next step is not correct. It should be: $$\sum^\infty_{n=0}(n+1)C_{n+1}x^{n}-\sum^\infty_{n=1}2C_{n-1}x^{n}=0$$ $$C_1+\sum^\infty_{n=1}(n+1)C_{n+1}x^{n}-\sum^\infty_{n=1}2C_{n-1}x^{n}=0$$ $$C_1+\sum^\infty_{n=1}((n+1)C_{n+1}-2C_{n-1})x^{n}=0$$ Hence for $n \ge 1$: $$C_1=0 \\ C_{n+1}=\dfrac {2C_{n-1}}{n+1} $$ The recurrence formula is: $$C_{2n}=\dfrac {C_0}{n!}$$ Therefore the solution is: $$y(x)=C_0\sum_{n=0}^\infty \dfrac {x^{2n}}{n!}=C_0e^{x^2}$$

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Clearly $y_1=C_0$ is not a solution and you forgot to multiply the terms in $y_2$ by $C_0$ that is not a solution also.

Following your own computations, the solutions are $$y(x)=y_1+C_0y_2=C_0\sum^\infty_{n=0} \frac{x^{2n}}{n!}.$$

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If I may suggest, never change the index $$y= \sum^\infty_{n=0}C_n\,x^n \implies y'=\sum^\infty_{n=\color{red}{0}}n\,C_n\,x^{n -1}$$ $$y'-2xy =0\implies \sum^\infty_{n=0}n\,C_n\,x^{n -1}-2\sum^\infty_{n=0}C_n\,x^{n+1}=0$$

So, to have degree $m$ in the first sum, you must make $n-1=m$ that is to say $n=m+1$ and to have degree $m$ in the second sum, you must make $n+1=m$ that is to say $n=m-1$. Then $$(m+1)\, C_{m+1}-2\, C_{m-1}=0\implies C_{m+1}=\frac{2\, C_{m-1}}{m+1}$$

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