Find a fundamental set of solutions for $t^{2}y^{(4)}+ty^{(3)}+y^{(2)}-4y=0$. 
Find a fundamental set of solutions for $t^{2}y^{(4)}+ty^{(3)}+y^{(2)}-4y=0$.

I haven't learned any techniques on how to solve an $n$th order DE with nonconstant coefficients besides the Euler-Cauchy equation. Any ideas will be helpful!
 A: This can be tackled with the Frobenius method, which seeks to find a power series solution for the differential equation. Let $\displaystyle y=\sum_{n=0}^\infty a_nt^{n+r}$ and see that we then have
$$r(r-1)[(r-2)^2+1]a_0t^{r-2}+(r+1)r[(r-1)^2+1]a_1t^{r-1}+\\ \sum_{n=0}^\infty[(n+r+2)(n+r+1)[(n+r)^2+1]a_{n+2}-4a_n]t^{n+r}=0$$
Equating coefficients gives $r=0$ and
$$a_{n+2}=\frac4{(n+2)(n+1)(n^2+1)}a_n$$
This gives us two solutions:
$$y_1={}_0F_3\left(;\frac12,-\frac i2,\frac i2;\frac14t^2\right)=1+2t^2+\frac2{15}t^4+\mathcal O(t^6)$$
$$y_2=t\cdot{}_0F_3\left(;1,\frac{1-i}2,\frac{1+i}2;\frac14t^2\right)=t+\frac13t^3+\frac1{150}t^5+\mathcal O(t^7)$$
When $a_1=0$ we also get two more solutions from $r=2\pm i$ and
$$a_{n+2}=\frac4{(n+4\pm i)(n+3\pm i)[(n+2\pm i)^2+1]}a_n$$
which resolve to
$$y_3=t^{2-i}\cdot{}_0F_3\left(;\frac{4-i}2,\frac{3-i}2,1-i;\frac14t^2\right)=t^{2-i}+\frac{2+9i}{170}t^{4-i}+\mathcal O(t^{6-i})$$
$$y_4=t^{2+i}\cdot{}_0F_3\left(;\frac{4+i}2,\frac{3+i}2,1+i;\frac14t^2\right)=t^{2+i}+\frac{2-9i}{170}t^{4+i}+\mathcal O(t^{6+i})$$
Since these are conjugates of each other, we can add subtract them from each other to form their real counterparts:
$$y_+=\frac{y_3+y_4}2$$
$$y_-=\frac{y_3-y_4}{2i}$$
Notes:


*

*The case of $a_1=0$ and $r=1$ is just a special case of the original with $r=0$. The same is true for $a_0=0$ and $r=-1,1\pm i$. This is simply because they are integer shifts of $r$.

*${}_pF_q$ is the generalized hypergeometric function.
