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a)Find the Unique Power series solution of, $y''+xy'+y=0$ given $y(0)=1$ and $y'(0)=0$

I have done the work and find the unique solution by using initial condition where the constant $a_0=1$ and $a_1=0$. After calculating the unique Power Series solution as follow:

Solution : $y(x)= 1-x^2/2+x^4/(2×4)-x^6/(2×4×6)+x^8/(2×4×6×8)-...$

b)However I can't express the unique power series solution I have found in term of elementary function. *Express the unique power series solution that have found in term of elementary function. (e.x:exponential, logarithmic, Sine Cosine )*Please help me to write this solution in term of elementary function. Please describe how it can be done.thank You :)

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I am not sure what you have done, but the solution is $y(x) = e^{-x^2/2}$. – Amzoti Dec 1 '13 at 3:36
Also, your series is correct, you just need to figure out that it is exponential I mention above. You can just do a Taylor Series expansion of that exponential and show it is your series. Regards – Amzoti Dec 1 '13 at 3:49
Thank You Sir . – shownTiger Dec 1 '13 at 3:50
You are very welcome - great work! – Amzoti Dec 1 '13 at 3:51

Note that we have $xy'+y = (xy)'$. Hence, $$y'' + xy' + y = y''+(xy)' = 0 \implies y'+xy = c \implies e^{x^2/2}y' + xe^{x^2/2}y = c e^{x^2/2}$$ This gives us $$\left(y e^{x^2/2}\right)' = c e^{x^2/2} \implies y(x) = ce^{-x^2/2} \int_0^xe^{t^2/2} dt + ke^{-x^2/2}$$ Now at $x=0$, we have $y(0)=1$ and $y'(0) = 0$. Hence, $c=0$ and $k=1$. This gives us $$y(x) = e^{-x^2/2}$$ The power series you have obtained is nothing but the power series of $e^{-x^2/2}$, which you should be able to check with relative ease.

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$$ \cos x = \sum \frac{(-1)^nx^{2n}}{(2n)!} \; \; \; (proof?) $$

and that is not complex analysis.

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YES it would be great if you can verify the process. Thank You – shownTiger Dec 1 '13 at 3:37

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