Answer:
First we write y and its derivatives as power series:
y=∑n=0∞anxn⟹y′=∑n=1∞nanxn−1⟹y′′=∑n=2∞n(n−1)anxn−2
Next, plug into differential equation:
(x+2)y′′+xy′−y=0
(x+2)∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0
x∑n=2∞n(n−1)anxn−2+2∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0
Move constants inside of summations:
∑n=2∞x⋅n(n−1)anxn−2+∑n=2∞2⋅n(n−1)anxn−2+∑n=1∞x⋅nanxn−1−∑n=0∞anxn=0
∑n=2∞n(n−1)anxn−1+∑n=2∞2n(n−1)anxn−2+∑n=1∞nanxn−∑n=0∞anxn=0
Change limits so that the exponents for x are the same in each summation:
∑n=1∞(n+1)nan+1xn+∑n=0∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−∑n=0∞anxn=0
Pull out any terms from sums, so that each sum starts at same lower limit (n=1)
∑n=1∞(n+1)nan+1xn+4a2+∑n=1∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−a0−∑n=1∞anxn=0
Combine all sums into a single sum:
4a2−a0+∑n=1∞(2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an)xn=0
Now we must set each coefficient, including constant term =0 :
4a2−a0=0⟹4a2=a0
2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an=0
We would usually let a0 and a1 be arbitrary constants. Then all other constants can be expressed in terms of these two constants, giving us two linearly independent solutions. However, since a0=4a2 , I’ll choose a1 and a2 as the two arbitrary constants. We can still express all other constants in terms of a1 and/or a2 .
an+2=−(n+1)nan+1+(n−1)an2(n+2)(n+1)
a3=−(2⋅1)a2+0a12(3⋅2)=−16a2=−13!a2
a4=−(3⋅2)a3+1a22(4⋅3)=0=04!a2
a5=−(4⋅3)a4+2a32(5⋅4)=15!a2
a6=−(5⋅4)a5+3a42(6⋅5)=−26!a2
We see a pattern emerging here:
an=(−1)(n+1)n−4n!a2
This can be proven by mathematical induction. In fact, this is true for all n≥0 , except for n=1 , since a1 is an arbitrary constant independent of a0 (and therefore independent of a2 ).
Plugging back into original power series for y , we get:
y=a0+a1x+a2x2+a3x3+a4x4+a5x5+⋯
y=4a2+a1x+a2x2−13!a2x3+04!a2x4+15!a2x5−⋯
y=a1x+a2(4+x2−13!x3+04!x4+15!x5−⋯)
Notice that the expression following constant a2 is =4+ a power series (starting at n=2 ). However, if we had the appropriate x -term, we would have a power series starting at n=0 . Since the other independent solution is simply y1=x, then we can let a1=c1−3c2, a2=c2 , and we get:
y=(c1−3c2)x+c2(4+x2−13!x3+04!x4+15!x5−⋯)
y=c1x+c2(4−3x+x2−13!x3+04!x4+15!x5−⋯)
y=c1x+c2(−0−40!+0−31!x−2−42!x2+3−43!x3−4−44!x4+5−45!x5−⋯)
y=c1x+c2∑n=0∞(−1)n+1n−4n!xn
Learn more about constants here:
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