Differential equation using power series method

[Graham87]

TL;DR

Differential equation using power series method

I am attempting to solve this differential equation with power series

I came with the following solution but I doubt it is correct.

Since x=1 we get:

I doubt its correctness because it looks messy. Also the convergence radian R goes to 0, giving only a solution for x=0 which is not correct, since the beginning condition states y(1)=2.

 

[pasmith]

If you set $y(x) = \sum_{k=0}^\infty c_k(x - 1)^k$ then $x$ cannot appear in the recurrence relation for $c_k$; the $c_k$ are supposed to be independent of $x$. You need to write $$\begin{split}(x + 3)y' &= <br /> (x - 1 + 4)\sum_{k=1}^\infty kc_k (x-1)^{k-1} \\<br /> &= \sum_{k=1}^\infty kc_k (x - 1)^k + 4\sum_{k=1}^\infty kc_k (x- 1)^{k-1} \end{split}$$ and then compare coefficients of $(x-1)^k$.

 

[Graham87]

If you set $y(x) = \sum_{k=0}^\infty c_k(x - 1)^k$ then $x$ cannot appear in the recurrence relation for $c_k$; the $c_k$ are supposed to be independent of $x$. You need to write $$\begin{split}(x + 3)y' &=<br /> (x - 1 + 4)\sum_{k=1}^\infty kc_k (x-1)^{k-1} \\<br /> &= \sum_{k=1}^\infty kc_k (x - 1)^k + 4\sum_{k=1}^\infty kc_k (x- 1)^{k-1} \end{split}$$ and then compare coefficients of $(x-1)^k$.

Oh, I tried that too but did not go through it completely. So I get thefollowing:

However the general solution formula does not seem obvious. Where might I have gone wrong?

 

[pasmith]

You can write the ODE as $$\sum_{k=0}^\infty ( kc_k + 4(k+1)c_{k+1} + 2c_k )(x-1)^k = 0$$ from which it follows that $$c_{k+1} = -\frac{k+2}{4(k+1)}c_k, \quad k \geq 0.$$ This linear first-order recurrence can be easily solved, since the general solution of $$a_{k+1} = f(k)a_k$$ is $$a_k = a_0 \prod_{r=0}^{k-1} f(r).$$ Note that by inspection $y = A(x + 3)^{-2}$ is the general solution of $(x + 3)y' + 2y = 0$ so that you can check your answer against $$\begin{split}<br /> y(x) &= 16y(1)(x + 3)^{-2} \\<br /> &= 16y(1)(x - 1 + 4)^{-2} \\<br /> &= y(1) \left(1 + \frac{x-1}{4}\right)^{-2} \\<br /> &= y(1) \left(1 - 2\frac{x-1}4 + \dots<br /> + \frac{(-2)(-3) \cdots (-n - 1)}{n!}\left(\frac{x-1}4\right)^n + \dots \right)\end{split}$$

 

[Graham87]

You can write the ODE as $$\sum_{k=0}^\infty ( kc_k + 4(k+1)c_{k+1} + 2c_k )(x-1)^k = 0$$ from which it follows that $$c_{k+1} = -\frac{k+2}{4(k+1)}c_k, \quad k \geq 0.$$ This linear first-order recurrence can be easily solved, since the general solution of $$a_{k+1} = f(k)a_k$$ is $$a_k = a_0 \prod_{r=0}^{k-1} f(r).$$ Note that by inspection $y = A(x + 3)^{-2}$ is the general solution of $(x + 3)y' + 2y = 0$ so that you can check your answer against $$\begin{split}<br /> y(x) &= 16y(1)(x + 3)^{-2} \\<br /> &= 16y(1)(x - 1 + 4)^{-2} \\<br /> &= y(1) \left(1 + \frac{x-1}{4}\right)^{-2} \\<br /> &= y(1) \left(1 - 2\frac{x-1}4 + \dots<br /> + \frac{(-2)(-3) \cdots (-n - 1)}{n!}\left(\frac{x-1}4\right)^n + \dots \right)\end{split}$$

I tried that, but I get stuck in the manipulation on the most left sum to k=0?

From your answer it turned to this somehow?

Shouldn't the most left term be

 

[pasmith]

No; $$(x-1)y' = \sum_{k=1}^\infty kc_k(x-1)^k = \sum_{k=0}^\infty kc_k(x-1)^k$$ because $0c_0(x-1)^0 = 0$.

 

[Graham87]

No; $$(x-1)y' = \sum_{k=1}^\infty kc_k(x-1)^k = \sum_{k=0}^\infty kc_k(x-1)^k$$ because $0c_0(x-1)^0 = 0$.

Yeah, I realised that too! Thanks a lot!

 

[Graham87]

No; $$(x-1)y' = \sum_{k=1}^\infty kc_k(x-1)^k = \sum_{k=0}^\infty kc_k(x-1)^k$$ because 0c0(x−1)0=0.

Are you sure about this?
I did like this:

And since x=1 we get the following

And eventually

 

[pasmith]

The fundamental issue here is that $$(x- 1)\sum_{k=1}^\infty kc_k(x-1)^{k-1} = \sum_{k=1}^\infty kc_k(x-1)^{k} = \sum_{k=0}^\infty kc_k(x-1)^k.$$