The Differential Equation
$y'' + ( p + \frac{1}{2} - \frac{1}{4}x^2)y=0$
where p is a constant, it has a series solution of the form
$y =\displaystyle\sum {a_n} x^n$ prove that:
a) the coefficients $a_n$ are related by the three-term recurrence formula
$(n+1)(n+2)a_{n+2}+( p + \frac{1}{2})$$a_n - \frac{1}{4}$$a_{n-2}=0 $
b) find its general solution by series $y_1(x)$ and $y_2(x)$ at least the first seven terms.
a)If we derive the series we have:
$y'= \sum na_nx^{n-1}$
$y''= \sum n( n-1)a_n x^{n-2} $
replacing in the D.E. $ \sum n( n-1)a_n x^{n-2} + (p+ \frac{1}{2} - \frac{1}{4}x^2) \sum a_n x^n = 0$ $\sum n(n-1)a_n x^{n-2}+ (p + \frac{1}{2}) \sum a_n x^n - \frac{1}{4} \sum a_n x^{n+2}=0$ looking at the coefficients of the power $x^m$ of degree $m$ in the first summation we take: $n-2= m$ and in the last and remains: $n+2=m$ $ (m+2)(m+1)a_{m+2}+ ( p+ \frac{1}{2})a_m - \frac{1}{4}a_{m-2}=0$ for $m≥2$
For $m=1$
$3a_3 + (p+ \frac{1}{2})a_1=0$
for $m=0$
$2a_2+ (p+ \frac{1}{2})a_0=0$
is this part a) correct? how to solve part b) can you develop it step by step? for $m=0$:
$2a_2+\left(p+\dfrac{1}{2}\right)a_0=0$ $a_2=-\dfrac{1}{2}\left(p+\dfrac{1}{2}\right)a_0$
dor $m=1$:
$3a_3+\left(p+\dfrac{1}{2}\right)a_1=0$
$a_3=-\dfrac{1}{3}\left(p+\dfrac{1}{2}\right)a_1$
for $m=2$:
$12a_{4}+\left(p+\dfrac{1}{2}\right)a_2-\dfrac{1}{4}a_{0}=0$
$a_4= \frac{1}{48}a_0+ \frac{1}{24} ( p+ \frac{1}{2})^2 a_0$
for $m=3$
$20a_5+(p+ \frac{1}{2})a_3- \frac{1}{4}a_1 =0$
$a_5= \frac{1}{80}a_1 + \frac{1}{120}(p+ \frac{1}{2})^2a_1$
Para $m=4$
$30a_6 + (p+ \frac{1}{2})a_4 - \frac{1}{4}a_2=0$
$a_6=- \frac {7}{240}(p + \frac{1}{2} - \frac{1}{24}(p+ \frac{1}{2})^2a_0$
for $m=5 $
$42a_7+ (p + \frac{1}{2})a_5 - \frac{1}{4}a_3=0$
How is the term $a_7$? Are the calculations I have done for the other terms correct? How do you end the exercise? what is the general solution?