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Find solutions of the given equation in the form of power series $$y'' +2xy' = 0$$

My approach:

$$2xy' = 2a_1x + 2\times2a_2x^2 + 2\times3a_3x^3 + ... = \sum 2ia_{i}x^{i}$$ $$y'' = 2a_2 + 3\times2a_3x + 4\times3a_4x^2 + ... = \sum (i+1)(i+2)a_{i+2}x^{i}$$

So to have $y'' +2xy' = 0$ we need to have every coefficient equal to $0$.

So

$x^0$ // $2a_2 + 0 = 0 => a_2=0$

$x^1$ // $3\times2a_3 + 2a_1 = 0 => a_3 = -a_1/3$

$x^2$ // $4\times3a_4 + 2\times2a_2=0 => a_4 = -4a_2/12 = 0$

...

So it was easy to observe that every even coefficient is equal to $0$ and every odd one can be written in terms of $a_0$

Afterwards I wanted to find a recursive formula for odd coefficients and this is a problem that I can't solve.

I tried doing it in that way: $$2n(2n+1)a_{2n+1} + 2(2n-1)a_{2n-1} = 0$$ $$a_{2n+1} = -\frac{2(2n-1)a_{2n-1}}{2n(2n+1)}$$

Now probably it should be somehow easy to write it as a recursive formula in terms of $a_0$, but I can't really see it.

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You already did most of the job.

We continue \begin{align*} \color{blue}{a_{2n+1}}&= -\frac{2(2n-1)}{(2n+1)2n}a_{2n-1}\\ &=+\frac{2(2n-1)2(2n-3)}{(2n+1)2n(2n-1)(2n-2)}a_{2n-3}\\ &=\cdots\\ &=(-1)^n\frac{2^n(2n-1)!!}{(2n+1)!}a_1\tag{1}\\ &=(-1)^n\frac{2^n}{(2n+1)(2n)!!}a_1\tag{2}\\ &\,\,\color{blue}{=\frac{(-1)^n}{(2n+1)n!}a_1}\tag{3} \end{align*}

Comment:

  • In (1) we use the double factorial $(2n-1)!!=(2n-1)(2n-3)\cdots3\cdot 1$.

  • In (2) we use $(2n)!=(2n)!!(2n-1)!!$ and $(2n)!!=(2n)(2n-2)\cdots 4\cdot2$.

  • In (3) we use $(2n)!!= 2^n n!$.

Since we already know that $a_{2n}=0, n\geq 0$, we get from (3) the generating function $A(x)$ \begin{align*} A(x)=\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_{2n+1} x^{2n+1}=a_1\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)n!} x^{2n+1} \end{align*} and the general solution \begin{align*} A(x)+C\color{blue}{=a_1\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)n!} x^{2n+1}+C}\qquad\qquad a_1,C\in\mathbb{R} \end{align*}

Note that $A(x)=\frac{a_1\sqrt{\pi}}{2}\mathrm{erf}(x)$ where $\mathrm{erf}$ is the error-function.

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Let the solution of the given ODE be $y=\sum c_{s}x^{s}$. Then this satisfies the equation $y^{''}+2xy^{'}=0$. Now, $y^{'}=\sum sc_{s}x^{s-1}$ and $y^{''}=\sum s(s-1)c_{s}x^{s-2}$. Substituting these values in the differential equation.


$\sum s(s-1)c_{s}x^{s-2} + 2x \sum sc_{s}x^{s-1}=0$
$=>\sum s(s-1)c_{s}x^{s-2} + 2\sum sc_{s}x^{s}=0$
$=>\sum 0(0-1)c_{0}x^{0} + \sum 1(1-1) c_{1}x^{1} + \sum_{s=2} s(s-1)c_{s}x^{s} + 2\sum 0c_{0}x^{0} + 2\sum_{s=1} sc_{s}x^{s}=0$
$=>\sum_{s=0} (s+1)(s+2)c_{s+2}x^{s} + 2\sum_{s=0} (s+1)c_{s+1}x^{s+1} =0$

Now since the last relation is an identity, therefore

$(s+1)(s+2)c_{s+2}=0$ and $2(s+1)c_{s+1}=0$
$=>(s+1)(s+2)c_{s+2}=2(s+1)c_{s+1}$
$=>c_{s+2}=\frac{2c_{s+1}}{s+2}$
$=>c_{s+1}=\frac{2c_{s}}{s+1}$

Now I think you can see the recurrence relation

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Assume $y = \displaystyle \sum_{j = 0}^{\infty} a_j x^{s+j}$ with $a_0 \ne 0$.

We find $$\sum_{j = 0}^{\infty} a_j (s+j)(s+j-1)x^{s+j-2} + 2 \sum_{j = 0}^{\infty} a_j(s+j) x^{s+j} = 0$$

Substituting for lowest power of $x$ at $j=0$, $$a_0 s (s-1) = 0 \implies s= 0$$ For $j=1$, $$a_1 .s(s+1)+2a_1.(s+1) = 0 \implies a_1 = 0$$

For recurrence relation collect the coefficients of $x^{s+j}$, $$a_{j+2}(s+j+2)(s+j+1) + 2a_{j}(s+j)=0$$ $$a_{j+2} = \frac{-2a_{j}(s+j)}{(s+j+2)(s+j+1)}$$

Since $a_1 = 0$, all the odd coefficients are zero. You will obtain two solutions; one for $s = 0$, other for $s = 1$. Can you proceed from here?

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  • $\begingroup$ I am not really used to the notation with these "s" and "j" letters, so it will take me a while to analyze. Thanks. $\endgroup$ – James Smith Jan 27 at 13:03

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