# Math Help - y''+ay=0

1. ## y''+ay=0

I like to use the series method, I am having problems constructing the solution to y''+ay=0, a>0

The solution should be y=Asin(a^.5x)+Bcos(a^.5x)

In the recursion relation I get -aCn/(n+2)(n+1) = Cn+2

However to construct sin, I need 'a' to have fractional power which I cannot seem to get.

i.e. for odd n, Cn should have 'a' with a fractional power like n+.5 or a^(n+.5) but I can only see a^n in Cn for all n. Ideally I like to deduce the correct form of the power of 'a' in the series method instead of writing it because I know the correct solution before hand using another method.

2. Originally Posted by theoretician
I like to use the series method, I am having problems constructing the solution to y''+ay=0, a>0

The solution should be y=Asin(a^.5x)+Bcos(a^.5x)

In the recursion relation I get -aCn/(n+2)(n+1) = Cn+2

However to construct sin, I need 'a' to have fractional power which I cannot seem to get.

i.e. for odd n, Cn should have 'a' with a fractional power like n+.5 or a^(n+.5) but I can only see a^n in Cn for all n. Ideally I like to deduce the correct form of the power of 'a' in the series method instead of writing it because I know the correct solution before hand using another method.
after typing up the solution i realized at the end i ran into the same problem you did, lol, let's wait for some real mathematicians to handle this one. i came up with an arithmetic series to handle the powers of a in the sine function as you described, but it doesn't account for the first term, only the ones after

3. Here.

4. Originally Posted by ThePerfectHacker
Here.
yup, i got that as well. it's the odd ones that are difficult. maybe he has a typo in the question...or maybe not

5. Originally Posted by Jhevon
yup, i got that as well. it's the odd ones that are difficult. maybe he has a typo in the question...or maybe not
The odd ones are just slightly more difficult.
Look at attachment.