Power Series Solution to Linear ODE

[ChemistryNat]

Homework Statement

Let y(x)=$\sum$c_kx^k (k=0 to ∞) be a power series solution of

(x²-1)y''+x³y'+y=2x, y(0)=1, y'(0)=0

Note that x=0 is an ordinary point.

Homework Equations

y(x)=$\sum$c_kx^k (k=0 to ∞)
y'(x)=$\sum$(kc_kx^k-1) (k=1 to ∞)
y''(x)=$\sum$(k(k-1))c_kx^k-2 (k=2 to ∞)

The Attempt at a Solution

(x²-1)$\sum$(k(k-1))c_kx^k-2 (k=2 to ∞) +$\sum$(kc_kx^k+2) (k=1 to ∞)+$\sum$c_kx^k (k=0 to ∞) -2x=0

Making all the series start with the x³ term
2C₂x⁰+$\sum$(k(k-1))c_kx^k (k=3 to ∞)- 2C₂x⁰-6C₃x¹-12C₄x²-$\sum$(k(k-1))c_kx^k-2 (k=5 to ∞) + $\sum$(kc_kx^k+2) (k=1 to ∞)+C₀x⁰+C₁x¹+C₂x²+$\sum$c_kx^k (k=3 to ∞)2C₂x⁰+$\sum$(n(n-1))c_nxⁿ (n=3 to ∞)- 2C₂x⁰-6C₃x¹-12C₄x²-$\sum$(n+2)(n+1))c_n+2xⁿ (n=3 to ∞) + $\sum$((n-2)c_n-2xⁿ) (n=3 to ∞)+C₀x⁰+C₁x¹+C₂x²+$\sum$c_nxⁿ (n=3 to ∞)
2C₂x⁰-2C₂x⁰-6C₃x¹-12C₄x²+C₀x⁰+C₁x¹+C₂x²=0
c₀=0
c₃={c₁-2}/{6}
c₄={c₂}/{12}

c_n+2={n(n-1)c_n+(n-2)c_n-2+c_n}/{(n+2)(n-1)} n=3,4,5...I started with c₅={(7c₃+c₁)}/{10}
but both of my c₁ and c₃ terms are tied up in the same equation, not really sure were to go from here

 

[vela]

Homework Statement

Let y(x)=$\sum$c_kx^k (k=0 to ∞) be a power series solution of

(x²-1)y''+x³y'+y=2x, y(0)=1, y'(0)=0

Note that x=0 is an ordinary point.

Homework Equations

y(x)=$\sum$c_kx^k (k=0 to ∞)
y'(x)=$\sum$(kc_kx^k-1) (k=1 to ∞)
y''(x)=$\sum$(k(k-1))c_kx^k-2 (k=2 to ∞)

The Attempt at a Solution

(x²-1)$\sum$(k(k-1))c_kx^k-2 (k=2 to ∞) +$\sum$(kc_kx^k) (k=1 to ∞)+$\sum$c_kx^k (k=0 to ∞) -2x=0

The y' term should be $x^3 \sum_{k=1}^\infty kc_k x^{k-1} = \sum_{k=1}^\infty kc_k x^{k+2}$.

Making all the series start with the x³ term
2C₂x⁰+$\sum$(k(k-1))c_kx^k (k=3 to ∞)- 2C₂x⁰-6C₃x¹-12C₄x²-$\sum$(k(k-1))c_kx^k-2 (k=5 to ∞) + $\sum$(kc_kx^k+2) (k=1 to ∞)+C₀x⁰+C₁x¹+C₂x²+$\sum$c_kx^k (k=3 to ∞)2C₂x⁰+$\sum$(n(n-1))c_nxⁿ (n=3 to ∞)- 2C₂x⁰-6C₃x¹-12C₄x²-$\sum$(n+2)(n+1))c_n+2xⁿ (n=3 to ∞) + $\sum$((n-2)c_n-2xⁿ) (n=3 to ∞)+C₀x⁰+C₁x¹+C₂x²+$\sum$c_nxⁿ (n=3 to ∞)
2C₂x⁰-2C₂x⁰-6C₃x¹-12C₄x²+C₀x⁰+C₁x¹+C₂x²=0
c₀=0
c₃={c₁-2}/{6}
c₄={c₂}/{12}

c_n+2={n(n-1)c_n+(n-2)c_n-2+c_n}/{(n+2)(n-1)} n=3,4,5...I started with c₅={(7c₃+c₁)}/{10}
but both of my c₁ and c₃ terms are tied up in the same equation, not really sure were to go from here

 

[ChemistryNat]

The y' term should be $x^3 \sum_{k=1}^\infty kc_k x^{k-1} = \sum_{k=1}^\infty kc_k x^{k+2}$.

Sorry, it would seem I made an error on the first line, but after that I think this term has been corrected

 

[vela]

It seems you made some more algebra mistakes. After I fixed all of my mistakes, I had
\begin{align*}
-2c_2 + c_0 &= 0 \\
-6c_3 + c_1 &= 2 \\
2c_2 - 12 c_4 + c_2 &= 0
\end{align*} and
$$[n(n-1)+1]c_n - (n+2)(n+1)c_{n+2} + (n-2)c_{n-2} = 0$$ for the general recurrence relation.

 

[ChemistryNat]

It seems you made some more algebra mistakes. After I fixed all of my mistakes, I had
\begin{align*}
-2c_2 + c_0 &= 0 \\
-6c_3 + c_1 &= 2 \\
2c_2 - 12 c_4 + c_2 &= 0
\end{align*} and
$$[n(n-1)+1]c_n - (n+2)(n+1)c_{n+2} + (n-2)c_{n-2} = 0$$ for the general recurrence relation.

oh wow! so many mistakes! Thank you

So for the recurrence relation, I want it in terms of c_n+2

$$ c_{n+2} = \frac{[n(n-1)+1]c_n + (n-2)c_{n-2}}{(n+2)(n+1)} = 0 $$

and then I solve for n=3 giving me

c₅= $$ \frac{7c_3 c_1}{20} $$ ?

 

[vela]

oh wow! so many mistakes! Thank you

So for the recurrence relation, I want it in terms of c_n+2

$$ c_{n+2} = \frac{[n(n-1)+1]c_n + (n-2)c_{n-2}}{(n+2)(n+1)} = 0 $$

The =0 shouldn't be there on the end.

and then I solve for n=3 giving me

c₅= $$ \frac{7c_3 c_1}{20} $$ ?

Almost — perhaps just a typo. It should be
$$c_5 = \frac{7c_3 + c_1}{20}.$$ Note that you know what $c_3$ is in terms of $c_1$, so you can express $c_5$ in terms of $c_1$.

 

[ChemistryNat]

The =0 shouldn't be there on the end.


Almost — perhaps just a typo. It should be
$$c_5 = \frac{7c_3 + c_1}{20}.$$ Note that you know what $c_3$ is in terms of $c_1$, so you can express $c_5$ in terms of $c_1$.

$$ c_5=\frac{8c_1 - 14}{20} $$
which I think I'm supposed to see some sort of condensed version involving n! of some sort?

and
$$ c_6=\frac{13c_4 + 2c_2}{30} $$
but I have two equations with C₂ and one with both c₂ and c₄?

 

[vela]

$$ c_5=\frac{8c_1 - 14}{20} $$
which I think I'm supposed to see some sort of condensed version involving n! of some sort?

Good luck with that. I don't see anything straightforward, but I didn't look very hard.

and
$$ c_6=\frac{13c_4 + 2c_2}{30} $$
but I have two equations with C₂ and one with both c₂ and c₄?

You can express $c_2$ in terms of $c_0$. Use that to express $c_4$ in terms of $c_0$. Then you can express $c_6$ in terms of $c_0$. See how this is going?

 

[ChemistryNat]

Good luck with that. I don't see anything straightforward, but I didn't look very hard.


You can express $c_2$ in terms of $c_0$. Use that to express $c_4$ in terms of $c_0$. Then you can express $c_6$ in terms of $c_0$. See how this is going?

So all of my odd terms will be expressed in terms of c₁ and all the even terms c₀

which means I need two solutions? I'm not very familiar with power series in general, I struggled with them in first year calc too

 

[vela]

You need two independent solutions because you have a second-order differential equation, and you have two, one proportional to $c_0$ and one proportional to $c_1$. There's also a solution that arises from the 2x term. That's the particular solution. The constants $c_0$ and $c_1$ are the arbitrary constants you generally have when you solve a differential equation with no boundary conditions.

 

[ChemistryNat]

oh wow, the question actually asks for c₁ to c₄ and the radius of converge. which I assume doesn't mean I need to put the solutions into series form.. my apologies.


The radius of convergence is |x-x₀|<R but how do you find R?

 

[vela]

You apply the ratio test to each series.

 

[ChemistryNat]

You apply the ratio test to each series.

oh. so I need to build two series equations, how can I do this? especially since the odd terms contain a weird fraction part?

 

[vela]

Oh, I just saw that you were given initial conditions. You should apply those to get specific values for $c_0$ and $c_1$.

 

[ChemistryNat]

Oh, I just saw that you were given initial conditions. You should apply those to get specific values for $c_0$ and $c_1$.

c₀=1 and c₁= 0? I just took the first term of the power series and it's derivative, unless I actually plug in 0 for x?

 

[vela]

You have to plug in x=0 into the series for y and y' and solve for the constants.

I'm not getting this to work out. I think we may have made a mistake somewhere. Let me look at this again for a bit.

 

[vela]

OK, I was just confusing myself. Your work so far is fine.

I have an idea that might help you find the radius of convergence. I'm not completely sure this method is valid, so you might want to run it by your professor. For the ratio test, you want to calculate the limit
$$\lim_{n \to \infty} \left|\frac{c_{n+2}x^{n+2}}{c_n x^n}\right|.$$ The recurrence relation you have is
$$c_{n+2} = \frac{n(n-1)+1}{(n+1)(n+2)} c_n + \frac{n-2}{(n+1)(n+2)} c_{n-2}.$$ If that second term on the righthand side weren't there, it would be straightforward to apply the ratio test. You could try arguing that in the limit $n \to \infty$, the second term doesn't matter. Do you see why?