# Y' '+ y = 0

1. May 5, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
What are the different methods of solving for y?

Using the series method is one way but I seem to be stuck to get to the correct form for y. My problem is shown in post 3.

3. The attempt at a solution
There is guessing that y=Asin(x)+Bcos(x)
Using a series solution and finding sin and cos series solutions

What else?

Last edited: May 5, 2007
2. May 5, 2007

### tim_lou

hmmm... the possibilities are virtually endless.

you can cast the problem into matrix form, x'=Ax, A is a matrix and the solution is x=e^A

you can do Laplace transformation

you can factor the differential operator, and solve for each of them.

you can apply idea from Banach fixed point theorem and try a solution and then iteratively integrating it. (similar to a series solution)

you can keep differentiating the equation and find y, y', y'', y'''... at a particular point then smack it together in a Taylor series fashion.

you can find an integrating factor and integrate.

you can have a system of ODE's, k=-y', y=k', and look at the complex number n=k+iy and solve a first order ODE.

...

Last edited: May 5, 2007
3. May 5, 2007

### pivoxa15

Ok. But if using 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.

Last edited: May 5, 2007
4. May 5, 2007

### Pseudo Statistic

Well, the general method for solving this is to assume y = e^mx, and, once complex roots a +/- bi are obtained for m, assume the solution y = A e^ax cos(bx) + B e^ax cos(bx).
However, as tim_lou stated among others, the Laplace transform may prove helpful in this case-- especially if your initial conditions are given.
I don't see any reason to go ahead and do a series solution.

5. May 5, 2007

### pivoxa15

The series solution is slower but it should still work. I tried it and seemed to be stuck. Look at post 3.

6. May 5, 2007

### George Jones

Staff Emeritus
Write each $c_n$ as

$$\frac{c_1}{a^{\frac{1}{2}}}$$

multiplied by something.

7. May 5, 2007

### pivoxa15

But ideally there should be something that forces me to write it in the way you suggested while only doing the series method. RIght now, the only reasoning I have writing it out that way is because I knew what the solution should be using another method before hand.

Last edited: May 5, 2007
8. May 5, 2007

### HallsofIvy

Staff Emeritus
Is it the "a" that is causing the problem then? Are you able to convert the infinite series to sine and cosine with a= 0?
You recursion relation is $C_{n+2}= (-a/(n+2)(n+1)) a_n$. What do you get as the "closed form" for $C_n$? Did you actually calculate the first three or four terms?

For given $C_0$ and $C_1$ you should get $C_2= (-a/2)C_0$, $C_3= -(a/6)C_1$, $C_4= -(a/12)C_2= (a^2/24)C_0$, $C_5= -(a/20)C_3= -(a^2/120) C_5. You should be able to recognize the denominator of [itex]C_n$ as n! but then you are alternating between $C_0$ and $C_1$. Do you notice that you will always get $C_0$ in the terms with x to an even power and $C_1$ in the terms with x to an odd power? It would make sense to separate the two so that you have one series with even powers only, multiplied by $C_0$ and one series with odd powers only, multiplied by $C_1$, wouldn't it? Now compare the powers of a and x in each of those terms.

9. May 5, 2007

### pivoxa15

I see what you mean. I did separate the series into even and odd powers of x. For both x^(2n) and x^(2n+1), the coefficient is always a^n. With constant C0 and C1 respectively in the two separated series. I guess you could let C1=B/a^.5, B arbitary so that there is an extra a^.5 in the numerator of the odd powers of x hence a^(n+.5)x^(2n+1) = (a^.5 * x)^(2n+1) which is what is recquired to let sin(a^(.5)x) represent the series with odd powers of x. Which is what George Jones suggested.

10. May 5, 2007

### tim_lou

lol...:rofl: I completely misunderstood your question. I thought it was like a game where we try to figure out as many ways as possible to solve that linear ODE...

11. May 5, 2007

### pivoxa15

It was originally meant to be something like that but when I actually tried to solve y''+ay=0 using the series method I got some difficulties. Things didn't turn out to be as smooth as expected. I had to 'shift a constant' or do a change of constants in the coefficients of odd powers of x in order to get it to match a sin function.