Theorem used:
Let r1 and r2 be the roots of the indicial equation F(r)=r(r−1)+p0r+q0 with r1≥r2.
If r1 and r2 are real, then in either the interval −ρ<x<0 or the interval 0<x<ρ, there exists a solution of the form y1(x)=|x|r1[1+∑n=1∞an(r1)xn] where an(r1) are the given by recurrence relation with a0=1 and r=r1.
If r1−r2 is not zero or a positive integer, then in either the interval −ρ<x<0 or the interval 0<x<ρ, there exists a second solution of the form y2(x)=|x|r2[1+∑n=1∞an(r2)xn] where an(r2) are also determined by the recurrence relation with a0=1 and r=r2.
Further the power series converges at least for |x|<ρ.
If r1=r2, then the second solution is y2(x)=y1(x)ln|x|+|x|r1[1+∑n=1∞bn(r1)xn].
If r1−r2=N, positive integer, then y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn].
Calculation:
The given differential equation is given as x2y"+2xy'+xy=0.
Compare the equation x2y"+2xy'+xy=0 with P(x)y"+Q(x)y'+R(x)y=0 and note that P(x)=x2, Q(x)=2x and R(x)=x.
Note that the singular points occur when P(x0)=0.
Thus, the singular point of the given equation is x0=0.
If x0=0 is a regular singular point, check whether the following limits are finite.
p0=limx→0xQ(x)P(x) and q0=limx→0x2R(x)P(x)
For x0=0, obtain the value of p0 as follows.
p0=limx→0x2xx2=2
Similarly, for x0=0, obtain the value of q0 as follows.
q0=limx→0x2x2x2=0.
Since both are finite, x0=0 is a regular singular point of the given differential equation.
Since p0=2 and q0=0, the corresponding indicial equation is as follows:
F(r)=r(r−1)+p0r+q0
Substitute the value of p0=2 and q0=0 in F(r)=r(r−1)+p0r+q0=0.
F(r)=r(r−1)+2r+0=r2−r+2r=r2+r
Let F(r)=0 in the above equation it becomes as follows:
r2+r=0r(r+1)=0
Therefore, the roots of the equation r2−r+2r is r=0,r=−1.
Hence, x0 is a regular singular point.
From the above Theorem, the first solution is given as y1(x)=|x|r1[1+∑n=1∞an(r1)xn].
Substitute the value of r=0 in y1(x)=|x|r1[1+∑n=1∞an(r1)xn].
y1(x)=1+∑n=1∞an(0)xn
Differentiate the equation y1(x)=1+∑n=1∞an(0)xn.
y1'(x)=∑n=1∞(n+1)an+1(0)xn
The coefficients an(0), apart from a0(0)=1 are determined by the recurrence relation F(r+n)an+∑k=1n−1ak[(r+k)pn−k+qn−k]=0 in case of r=r1.
n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0, where ∑n=0∞pnxn=xQ(x)P(x)=2 and ∑n=0∞qnxn=x2R(x)P(x)=x.
Hence, the value of p0=2,pn=0,∀n∈N and q1=1,qn=0,∀n∈N0.
Factoring out an(0) from the equation n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0 is as follows:
an(0)=−∑k=0n−1ak(0)[kpn−k+qn−k]n(n+1) with first couple of elements is as follows:
a1(0)=−a0(0)[0+q1]1(1+1)=−11⋅2a2(0)=11⋅22⋅3a3(0)=−11⋅22⋅32⋅4
And the values continues as such.
Hence, the first solution is y1(x)=∑n=0∞(−1)nxnn!(n+1)!_.
According to the same theorem, since r1−r2=1, the second solution is given by y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn].
y2(x)=ay1(x)ln|x|+x−1[1+∑n=1∞cn(−1)xn]=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn
Where a=limr→r2(r−r2)aN(r),N=r−r2 since N=1.
aN(r)=a1(r)=−∑k=0N−1ak(r)[(r+k)pN−k+qN−k](r+N)(r+N+1)
Substitute the limits in the above equation it becomes as follows:
a1(r)=−a0(r)[(r+0)p1+q1](r+2)(r+1)=−[0+1](r+2)(r+1)=−1(r+2)(r+1)
Thus, substitute the value of a=limr→−1(r+1)−1(r+2)(r+1)=−1.
To calculate the value of cn(0) substitute the value of y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn into the initial differential equation.
First differentiate the equation y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn.
y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1
Again, differentiate the equation y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1.
y2''(x)=ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2
Now, the given differential equation becomes as follows:
x2[ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2]+2x[ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1]+x[ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn]=0[2c2(−1)+c1(−1)−3a1(0)]x+∑n=1∞[−(2n+1)an(0)+n(n+1)]cn+1(−1)cn(−1)xn=0
Equating the coefficient on both sides in the above equation.
{2c2(−1)+c1(−1)−3a1(0)=0−(2n+1)an(0)+n(n+1)cn+1(−1)cn(−1)=0
Let c2(−1)=1 in the above equation this implies as follows:
c2(−1)=3a1(0)−c1(−1)2=−H2+H12!c3(−1)=H3+H22⋅3
cn(−1)=(−1)n+1Hn+Hn−1n!(n−1)!
Hence, the second solution is y2(x)=−y1(x)lnx+x−1+[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_.