I have been asked to assume a power series solution of the form

$$y(x)= \sum_{n=0}^\infty a_n x^n $$

is a solution of the homogeneous linear ODE

$$y^\prime-(1+2x)y=0, $$

Obtain a recursion for the coefficients and use these recursions to determine the series representation of the solution.

I know the solution to the ODE is $y(x)=e^{x(x+1)}$ from a previous problem.

So far I have managed to simply the equation down to

$$ a_1-a_0+\sum_{m=1}^\infty [a_{m+1}(m+1)-a_m-2a_{m-1}]x^m = 0$$

I'm not sure where to go from here to complete the problem.


The function you have written on the left can only be equal to zero if the coefficient of $x^m$ is equal to zero for each $m$ (otherwise, the $m$th derivative at zero will not be zero, right?). Therefore you have the recurrence relation $$ a_{m+1} = \frac{a_m+2a_{m-1}}{m+1} $$ for every $m \geq 1$. The constant term gives you $a_0=a_1$ in the same way. It's not obvious what the solution to this recurrence is, so stick in the first few and let's see what happens: set $a_0=A$, and we find $$ a_0=A,\quad a_1 = A, \quad a_2 = \frac{a_1+2a_0}{1+1} = \frac{3}{2}A, \quad a_3 = \frac{3/2+2}{3}A = \frac{7}{6}A $$

Unfortunately there is, as far as I know, no closed form for this. (Knowing the solution, you can find a representation using the Cauchy product of the series for $e^{x^2}$ and $e^x$, but that's a lot harder to produce from the recurrence.)


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