y' - 3y = BRACKETS{ 3sin(x)/x when x != 0, 3 x = 0
y(0) = 1
Using series, how would I sold this problem? My answer should be in the form _ + _x +____x^2 + ....
I know how to solve Homogenous equations, but how do I do it with the multiple functions on the right side?
Thanks
You don't have "multiple functions on the right side", you have one function (say, $f$) which is defined piece-wise. And the only reason it's defined piecewise is that $3\sin x/x$ is not meaningful at $x=0$, so $3$ (its value at $x=0$) is used for $f(0)$. Begin by expanding $f$ into Taylor series:
$$f(x) = \sum_{k=0}^\infty 3\frac{(-1)^k}{(2k+1)!} x^{2k}$$
You are looking for $y(x)= \sum_{n=1}^\infty c_n x^n$ such that $y'-3y=f$, subject to initial condition $c_0=1$. (This is a first order equation, not second-order). Since $$y'(x)-3y(x) = \sum_{n=0}^\infty ((n+1)c_{n+1}-3c_n)x^n$$ it follows that $$\begin{split} (n+1)c_{n+1}-3c_n&=0\quad\text{when $n$ is odd}\\ (n+1)c_{n+1}-3c_n&= 3\frac{(-1)^k}{(2k+1)!} \quad \text{when $n=2k$}\end{split}$$ If I understand the question correctly, you are asked to find $c_0,c_1,c_2$ only: this I would do directly from the above relation. Writing down a general formula for $c_n$ takes a bit more work (but it can be done by induction, focusing on the odd-numbered coefficients first).