Quick clarification on series solutions to differential equations. Watch

Preeka
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The question reads,

 4x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - y = 0 .
Using the series solution method, find two linearly independent solutions of this equation. Express the solution in closed form and write down the general solution.

I have  y_{1}(x) = a_{0}( 1 + \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{(2n)!}) and  y_{2}(x) = a_{0}\sqrt x( 1 + \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{(2n + 1)!})

Just want to make sure the solutions itself are correct. It's just been a while since I got back to revising this topic. My main questions are more based on the technical terms involved. What exactly is meant by "expressing the solution in closed form" and "general solution". I'm assuming general solutions means putting  y_{1}(x) and  y_{2}(x) together and sticking constants next to them but what about closed form?

And lastly, this might be a really daft question but when changing the summation from n = 0 to n = 1, I end up getting 1 plus whatever summation. I'm a bit confused as to where the one comes from. I thought this was from subbing in n = 0 into the summation and obtaining that value and putting it outside. This is wrong but I can't see what the other justification is so I'd appreciate some clarification on this because right now I just stick a one outside as a rule of thumb but I'd like to know why it's done that way. Thank you.
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steve10
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Closed form means using elementary functions. So if your solution was, for example. y=\sum_0^\infty{\frac{x^n}{n!}}, then you could write it as y=e^x.

About the 1, if I'm reading you correct, you will get from the summation (starting with n = 0), \frac{x^0}{0!}=\frac{1}{1}=1
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Preeka
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(Original post by steve10)
Closed form means using elementary functions. So if your solution was, for example. y=\sum_0^\infty{\frac{x^n}{n!}}, then you could write it as y=e^x.

About the 1, if I'm reading you correct, you will get from the summation (starting with n = 0), \frac{x^0}{0!}=\frac{1}{1}=1
Thanks.

I thought so too about subbing one in but there are cases where I either end up dividing by zero or not getting one when a worked example states one is what I should get.
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