We have the ODE, $y''+3xy'+3y=0$. We have that $x_0=0$ is an ordinary point as the coefficient of $y''$ is 1. Then, I let, $$y(x)=\sum_{n=0}^{\infty}\,a_nx^n,$$ such that, $$y'(x)=\sum_{n=1}^{\infty}\,na_nx^{n-1},$$ and, $$y''(x)=\sum_{n=2}^{\infty}\,n(n-1)a_nx^{n-2}.$$ We can hence rewrite the differential equation as, $$\sum_{n=2}^{\infty}\,n(n-1)a_nx^{n-2}+3x\sum_{n=1}^{\infty}\,na_nx^{n-1}+3\sum_{n=0}^{\infty}\,a_nx^n=0.$$ We pull the coefficients under the sum to get, $$\sum_{n=2}^{\infty}\,n(n-1)a_nx^{n-2}+\sum_{n=1}^{\infty}\,3na_nx^{n}+\sum_{n=0}^{\infty}\,3a_nx^n=0.$$ Re-indexing the first term gives us, $$\sum_{n=0}^{\infty}\,(n+2)(n+1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}\,3na_nx^n+\sum_{n=0}^{\infty}\,3a_nx^n=0.$$ The only way to start the first and last sums at $n=1$ is to pull out the $n=0$ term from both. Then we have, $$2a_2+3a_0+\sum_{n=1}^{\infty}\,((n+2)(n+1)a_{n+2}+3na_n+3a_n)x^n=0.$$ I'm having trouble figuring out what the coefficients are supposed to be. Are $2a_2=0$, and $3a_0=0$? Or is $2a_2+3a_0=0$. I know the sum inside the summation is $0$. Wasn't sure about the external terms however.

Additionally, I am given boundary conditions, $y_1(0)=1$, and $y_1'(0)=0$. I'm not really sure what $y_1$ is referring to here. Any help would be appreciated.


We want the polynomial on the LHS to be zero for $\forall x$, therefore every coefficient (including the constant term) must be zero. Thus we have

$$ 2a_2 + 3a_0 = 0 $$ $$ (n+2)a_{n+2} + 3a_n = 0 $$

We can disregard the first equation, since it's effectively the second with $n=0$. So now we have the following recurrence relation

$$ a_0 = y(0) = 1 $$ $$ a_1 = y'(0) = 0 $$ $$ a_{n} = \frac{3a_{n-2}}{n} $$

Since $a_1 = 0$, all the following odd-powered coefficients must also be $0$.

For even-powered coefficients, we can prove from observation $$ a_n = \frac{3^{n/2}}{n(n-2)(n-4)\cdots 6\cdot 4\cdot 2}a_0 $$

Setting $n=2k$ gives a "nicer" form $$ a_{2k} = \left(\frac{3}{2}\right)^k \frac{1}{k!} $$


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