Answer to Question #160235 in Differential Equations for Sohana Behera

Question

2xy′′ + (x+1)y ′ +3y = 0; Find the series solution of the following equations near x = 0:

Answer 1

We will substitute solution of the form $y=\sum_{k=0}^{+\infty}A_kx^k$ in the equation. We receive: $y'=\sum_{k=1}^{+\infty}kA_kx^{k-1}$ and $y''=\sum_{k=2}^{+\infty}k(k-1)A_kx^{k-2}$. After subsitution we receive: $2x\sum_{k=2}^{+\infty}k(k-1)A_kx^{k-2}+(x+1)\sum_{k=1}^{+\infty}kA_kx^{k-1}+3\sum_{k=0}^{+\infty}A_kx^k=0$.

We simplify and receive:

$2\sum_{k=2}^{+\infty}k(k-1)A_kx^{k-1}+\sum_{k=1}^{+\infty}kA_kx^{k}+\sum_{k=1}^{+\infty}kA_kx^{k-1}+3\sum_{k=0}^{+\infty}A_kx^k=0$.

We change summation indices and get:

$2\sum_{j=1}^{+\infty}(j+1)jA_{j+1}x^{j}+\sum_{k=1}^{+\infty}kA_kx^{k}+\sum_{j=0}^{+\infty}(j+1)A_{j+1}x^{j}+3\sum_{k=0}^{+\infty}A_kx^k=0$.

Finally, we get: $A_1+3A_0+2\sum_{k=1}^{+\infty}(k+1)kA_{k+1}x^{k}+\sum_{k=1}^{+\infty}kA_kx^{k}+\sum_{k=1}^{+\infty}(k+1)A_{k+1}x^{k}+3\sum_{k=1}^{+\infty}A_kx^k=0$

We obtain:

$A_1+3A_0+\sum_{k=1}^{+\infty}((k+1)(2k+1)A_{k+1}+(k+3)A_k)x^k=0$

Thus, we can choose an arbitrary $A_0$ and then set $A_1=-3A_0$, $6A_2+4A_1=0$, $15A_3+5A_2=0$

As a result, we received the recurrence formula for the coefficients given by: $A_1=-3A_0,$ $A_2=-\frac23A_1$, $A_3=-\frac13A_2,$... In general, for $k\geq1$ we get: $(k+1)(2k+1)A_{k+1}+(k+3)A_k=0$.