Differential equation using power series method

[Graham87]

TL;DR Summary

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

 

[Graham87]

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

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

 

[Graham87]

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

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; [tex]
(x-1)y' = \sum_{k=1}^\infty kc_k(x-1)^k = \sum_{k=0}^\infty kc_k(x-1)^k[/tex] because [itex]0c_0(x-1)^0 = 0[/itex].

 

[Graham87]

No; [tex]
(x-1)y' = \sum_{k=1}^\infty kc_k(x-1)^k = \sum_{k=0}^\infty kc_k(x-1)^k[/tex] because [itex]0c_0(x-1)^0 = 0[/itex].

Yeah, I realised that too! Thanks alot!

 

[Graham87]

No; [tex]
(x-1)y' = \sum_{k=1}^\infty kc_k(x-1)^k = \sum_{k=0}^\infty kc_k(x-1)^k[/tex] 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 [tex](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.[/tex]