Question

The amount A in a fixed account at any time t is known to satisfy the the differential equation

d²A/dt²+tdA/dt+(t²-4)A=0. Express A as a series in powers of t.

Answer 1

$\text{The given differential equation can be written as} \\A''+tA'+(t^2-4)A=0 \\\text{The entire set of real numbers are ordinary points, therefore we find the} \\\text{series solution around the point 0} \\\text{Let A = $\sum_{n=0}^{\infty}a_nt^n$, $A' = \sum_{n=1}^{\infty}na_nt^{n-1}$, $A''=\sum^{\infty}_{n=2}n(n-1)a_nt^{n-2}$} \\\sum^{\infty} _{n=2}n(n-1)a_nt^{n-2} + t\sum^\infty_{n=1}na_nt^{n-1} + \sum^\infty_{n=0}a_nt^n+\sum^\infty_{n=0}a_{n-2}t^n=0 \\ \text{Shifting index of summation, we have } \\\sum^\infty_{n=2}[(n+2)(n+1)a_{n+2}-4a_n+a_n+a_{n-2}]x^n=0 \\\implies a_{n+2}=\frac{3a_n-a_{n-2}}{(n+2)(n+1)} \\\text{Next we compute $a_{n+2}$ for n=0,1,2,3,$\cdots$} \\a_2 = \frac{a_0}{2}, a_3 = \frac{a_1}{2} \\\implies a_{n+2}= \frac{3a_n-a_{n-2}}{(n+2)(n+1)}, n = 2,3,4,\cdots \\a_4 = \frac{a_0}{24}, a_5 = \frac{a_1}{40} \\a_6= -\frac{a_0}{80} \\=a_0+a_1t+\frac{a_0}{2}t^2+\frac{a_1}{2}t^3+\frac{a_0}{24}t^4+\frac{a_1}{40}t^5$