Using $$y(t) := \sum^{ \infty} _{k = 0} a_kt^k$$ to solve $$y'' - t^2y' = \ln(1+3t)$$ with y(0) = 1 and y'(0) = 0
I first expand $ln(1+3t)$ using Taylor series at t = 0: $$\sum_{k=0}^{\infty}(-1)^{k-1} \frac{3^{k}x^{k}}{{k}}$$
Then differentiate it twice and plug into the DE and get the following:
$$\sum_{k=2}^{\infty}a_k(k)(k-1)t^{k-2} - t^2\sum_{k=1}^{\infty}a_k(k)t^{k-1} =\sum_{k=0}^{\infty}(-1)^{k-1} \frac{3^{k}x^{k}}{{k}} $$
Then we get: $$\sum_{k=0}^{\infty}a_{k+2}(k+2)(k+1)t^k - \sum_{k=0}^{\infty} a_{k+1} (k+1) t^k$$
By expanding the terms and compare the coefficients, we get:
$$x^0: 0 = 2a_2 - a_1$$ $$x^1: 3 = 6a_3 - 2a_2$$ $$x^2: \frac{-9}{2} = 12a_4-3a_3$$
The question required to find the first four non - zero terms. How can I find that?