Given:
The differential equation is,
y″+xy=0.
Approach:
The point x=x0 is called an ordinary point of the differential equation y″+p(x)y′+q(x)y=0, if p and q are analytic at x=x0.
Suppose their power series expansions are valid for |x−x0|<R then the general solution to the differential equation y″+p(x)y′+q(x)y=0 can be represented as a power series centered at x=x0 with radius of convergence R.
The ratio test is,
limn→∞an+1an=l,
and if l<1 then the series is convergent.
Calculation:
The point x=x0 is called an ordinary point of the differential equation y″+p(x)y′+q(x)y=0, if p and q are analytic at x=x0.
In the given equation y″+xy=0,
p(x)=0 and q(x)=x, both are analytic at x=0. So it is an ordinary point.
Let the general solution of the differential equation is
y(x)=∑n=0∞anxn...(1)
Differentiate equation (1) with respect to x.
y′(x)=∑n=1∞nanxn−1y″(x)=∑n=2∞n(n−1)anxn−2
Substitute ∑n=2∞n(n−1)anxn−2 for y″ and ∑n=0∞anxn for y in the given equation.
∑n=2∞n(n−1)anxn−2+x∑n=0∞anxn=0 and,
∑n=2∞n(n−1)anxn−2+∑n=0∞anxn+1=0
Replace n by k+2 in the first summation and replace n by k-1 in the second summation.
∑k=0∞(k+2)(k+1)ak+2xk+∑k=1∞ak−1xk=0
Separate out the terms corresponding to k=0 and k≥1. 2a2+∑k=1∞[(k+2)(k+1)ak+2+ak−1]xk=0
For k=0 and k≥1, it is obtained that 2a2=0.
Thus, a2=0 and the general relation is,
(k+2)(k+1)ak+2+ak−1=0(k+2)(k+1)ak+2=−ak−1ak+2=−1(k+2)(k+1)ak−1, k=1,2,3,...
Use this relation to determine the appropriate values of the coefficients.
Substitute successively for k into above relation and obtain the coefficient.
When k=1,
a3=−13⋅2a0
When k=2,
a4=−14⋅3a1
When k=3,
a5=−15⋅4a2=0
When k=4,
a6=−16⋅5a1=16⋅5⋅3⋅2a0
When k=5,
a7=−17⋅6a4=17⋅6⋅4⋅3a1
Continue further to obtain a3n.
a3n=(−1)n3n(3n−1)(3n−3)(3n−4)⋅...3⋅2a0...(2)
and
a3n+1=(−1)n(3n+1)(3n)(3n−2)(3n−3)⋅...4⋅3a1...(3)
Use equations (3) and (4) to show that for all values of a0, a1, a solution to the given differential equation is,
y(x)=[a0(1−13⋅2x3+16⋅5⋅3⋅2x6−19⋅8⋅6⋅5⋅3⋅2x9+.......)+a1(x−14⋅3x4+17⋅6⋅4⋅3x7−......)]=[a0[1+∑n=1∞(−1)n3n(3n−1)(3n−3)(3n−4)⋅...3⋅2x3n]+a1[x+(−1)n(3n+1)(3n)(3n−2)(3n−3)⋅...4⋅3x3n+1]]
Substitute 1 for a0 and 0 for a1 in above equation. y1(x)=1+∑n=1∞(−1)n3n(3n−2)(3n−3)(3n−4)⋅...3⋅2x3n=1+∑n=1∞(−1)n(3n−2)(3n−5),......7⋅4⋅13n!x3n
Now, substitute 0 for a0 and 1 for a1.
y2(x)=x+∑n=1∞(−1)n(3n+1)(3n)(3n−2)(3n−3)⋅...4⋅3x3n+1=x+∑n=1∞(−1)n(3n−1)(3n−4),......5⋅2(3n+1)!x3n+1
For checking the solution is valid or not using the ratio test on both foregoing solution
Suppose,
an=∑n=1∞(−1)n(3n−2)(3n−5),......7⋅4⋅13n!x3nan+1=∑n=1∞(−1)n+1(3n+1)(3n−2),......7⋅4⋅13(n+1)!x3n+3
Then by ratio test,
limn→∞an+1an=l
And if l<1 then series is convergent.
So,
limn→∞an+1an=[limn→∞[(∑n=1∞(−1)n+1(3n+1)(3n−2),......7⋅4⋅13(n+1)!x3n+3)*(∑n=1∞3n!(−1)n(3n−2)(3n−5),......7⋅4⋅1x3n)]]=limn→∞(−1)(3n+1)(3n+3)=limn→∞(−1)n(3+1n)n(3+3n)=−13
From above l<1, so series is convergent.
Similarly for y2(x)
an=∑n=1∞(−1)n(3n−1)(3n−4),......5⋅2(3n+1)!x3n+1an+1=∑n=1∞(−1)n+1(3n+2)(3n−1),......5⋅2(3n+4)!x3n+4
Solve the above equations.
limn→∞an+1an=[limn→∞[∑n=1∞(−1)n+1(3n+2)(3n−1),......5⋅2(3n+4)!x3n+4*∑n=1∞(3n+1)((−1)n(3n−1)(3n−4),......5⋅2)x3n+1]]=limn→∞(−1)(3n+4)(3n+3)=0
From above l<1. So, the series is convergent.
Therefore, both series are convergent and the solution are valid on (−∞,∞).
Conclusion:
Hence, the point x=0 is the ordinary point and the two independent solutions are,
y1(x)=1+∑n=1∞(−1)n(3n−2)(3n−5),......7⋅4⋅13n!x3ny2(x)=x+∑n=1∞(−1)n(3n−1)(3n−4),......5⋅2(3n+1)!x3n+1
valid on (−∞,∞).