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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 $

$42a_7+ (p + \frac{1}{2})a_5 - \frac{1}{4}a_3=0$

esmeraldabrown
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  • You've mispositioned $n+2=m$ and it is glued to subsequent expression. It looks like your part a) is correct. To solve second part you can consider two basis solutions. $y_1(x)$ such that $a_0=1$, $a_1=0$ and $y_2(x)$ such that $a_0=0$, $a_1=1$. Then you can write expressions for $y_1(x)$ and $y_2(x)$ (at least first seven terms). Then general solution is $y=C_1 y_1+C_2 y_2$, where $C_1=y(0)$ and $C_2=y'(0)$. – Ivan Kaznacheyeu Apr 20 '23 at 15:47
  • Can you explain me step by step this second part? How is it applied to this exercise to arrive at the general solution? I'm just getting started on the subject and I really don't understand – esmeraldabrown Apr 20 '23 at 16:30
  • Differential equation is linear. Then if $y_1$ and $y_2$ are solutions, then $C_1y_1+C_2y_2$ is also solution. Equation is of second order. Then solution is unique for given $y(0)$ and $y'(0)$. Let consider two solutions: $y_1(x)$ such that $y_1(0)=1$, $y_1'(0)=0$ and $y_2(x)$ such that $y_2(0)=0$, $y_2'(0)=1$. Then solution $y(x)=C_1 y_1(x)+C_2 y_2(x)$ has $y(0)=C_1$ and $y'(0)=C_2$. Then all solutions are covered by this formula. $y_1(0)=1$, $y_1'(0)=0$ is equivalent to $a_0=1$, $a_1=0$ in $y_1=a_0+a_1x+a_2x^2+\ldots$ – Ivan Kaznacheyeu Apr 21 '23 at 14:50
  • Coefficient at $a_3$ for $m=1$ is wrong. – Ivan Kaznacheyeu Apr 21 '23 at 14:54
  • One can write general solution in terms of $a_0$ and $a_1$ instead of $C_1$ and $C_2$. It looks too long. $$y=\frac{1}{483840}(a_0(483840-20160(1+2p)x^2+336(13+4p+4p^2)x^4-3(43+90p+12p^2+8p^3)x^6)+a_1x(483840-12096(1+2p)x^2+144(21+4p+4p^2)x^4-(63+130p+12p^2+8p^3)x^6))+\ldots$$ – Ivan Kaznacheyeu Apr 21 '23 at 15:04
  • How do you find the coefficients until you reach a_7 I did them this way but I don't know if they are correct $2a_2 + (p+ \frac{1}{2})a_0=0$ – esmeraldabrown Apr 22 '23 at 13:37
  • $a_2= - \frac{1}{2}(p + \frac{1}{2})a_0=0$. For $m=1$ $3a_3 + ( p+ \frac{1}{2} a_1=0$ $a_3= -\frac{1}{3} (p+ \frac{1}{2})a_1$ – esmeraldabrown Apr 22 '23 at 13:45
  • How do you find the coefficients until you reach a_7 I did them this way but I don't know if they are correct – esmeraldabrown Apr 22 '23 at 18:15
  • I have added it to my exercise in part b – esmeraldabrown Apr 22 '23 at 18:17
  • I believe for $m=1$ you should have $6a_3+(p+\frac12)a_1=0$ – Ivan Kaznacheyeu Apr 29 '23 at 15:36

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