Find the series solution of the IVP $y''+3xy'+3y=0, y(0)=2, y'(0)=3$

Question

I'm using the method that we were taught in class for a previous problem.

$x=0$ is an ordinary point of the given ODE.

Differentiating the equation repeatedly, we get: $y^m(0)=-3xy^{m-1}(0)-3y^{m-2}(0)=-3y^{m-2}(0)$ since I have $x=0$. This is valid for $m \geq 2$

This gives me $y''(0)=-6, y'''(0)=-9, y^{IV}(0)=18, y^V(0)=27$

So my solution becomes $y=2\left[ 1- \dfrac{3x^2}{2!}+\dfrac{3^2x^4}{4!}+ \cdots\right]+3\left[x-\dfrac{3x^3}{3!}+\dfrac{3^2x^5}{5!}+\cdots\right]$

However, the answer given to us is $y=2\left[1+\sum\limits_{k=1}^{\infty} \dfrac{(-3)^k}{2^k k!} x^{2k}\right]+3\left[x+\sum\limits_{k=1}^{\infty}\dfrac{(-6)^kk!}{(2k+1)!}x^{2k+1}\right]$

Clearly, the answers don't match. What am I doing wrong here?

Answer 1

HINT

Here is an alternative approach

\begin{align*} y^{\prime\prime} + 3xy^{\prime} + 3y = 0 \Longleftrightarrow y^{\prime\prime} + (3xy)^{\prime} = 0 \Longleftrightarrow y^{\prime} + 3xy = k \end{align*}

Then you can apply the series method to the last ODE.

Answer 2

Setting $y=\sum a_kx^k$ you get the coefficient relation $$ (n+2)(n+1)a_{n+2}+3na_n+3a_n=0\iff a_{n+2}=-\frac{3}{n+2}a_n. $$ This means that $$ a_{2k}=\frac{(-3)^k}{2^kk!}a_0\text{ and }a_{2k+1}=\frac{(-3)^k}{(2k+1)(2k-1)\cdots 3\cdot 1}a_1=\frac{(-6)^kk!}{(2k+1)!}. $$