Solution:

The point x=0 is an ordinary point and the two linearly 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 (−∞,∞).

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 (−∞,∞).