Solving y' + y = 0 with series and other methods

[pivoxa15]

Homework Statement

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.

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?

 

[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.

...

 

[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.

 

[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.

 

[pivoxa15]

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

 

[George Jones]

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.

Write each $c_n$ as

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

multiplied by something.

 

[pivoxa15]

Write each $c_n$ as

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

multiplied by something.

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.

 

[HallsofIvy]

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 $C_2= (-a/2)C_0$, $C_3= -(a/6)C_1$, $C_4= -(a/12)C_2= (a^2/24)C_0$, [itex]C_5= -(a/20)C_3= -(a^2/120) C_5.<br /> <br /> You should be able to recognize the denominator of $C_n$ 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.[/itex]

 

[pivoxa15]

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 $C_2= (-a/2)C_0$, $C_3= -(a/6)C_1$, $C_4= -(a/12)C_2= (a^2/24)C_0$, [itex]C_5= -(a/20)C_3= -(a^2/120) C_5.<br /> <br /> You should be able to recognize the denominator of $C_n$ 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.[/itex]

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

 

[tim_lou]

lol... 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...

 

[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.