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1. Let be the solution of the given differential equation. 2. Find 3. 4. Substitute the expressions of y, etc. in the given differential equations. 5. Calculate Coefficients of various powers of x by equating the coefficient to zero. 6. Put the values of In the differential equation to get the required series solution. 
Here we have Let the solution of the given differential equation be Since x = 0 is the ordinary point of the given equation Put these values in the given differential equation Equating the coefficients of various powers of x to zero, we get Therefore, the solution is 
Here we have Let us suppose Since x = 0 is the ordinary point of (1) Then And Put these values in equation (1) We get
Equating to zero the coefficients of the various powers of x, we get And so on…. In general, we can write Now putting n = 5, Put n = 6 Put n = 7, Put n = 8, Put n = 9, Put n = 10, Put the above values in equation (1), we get 
……..(1) 
Case1: when roots m1 and m2 are distinct and these are not differing by an integer The complete solution in this case will be Case2: when roots m1 and m2 are equal Case3: when roots are distinct but differ by an integer Case4: Roots are distinct and differing by an integer, making some coefficient indeterminate 
Here we have ………… (1) Since x = 0 is a regular singular point, we assume the solution in the form So that
Substituting for y, in equation (1), we get …..(2)
The coefficient of the lowest degree term in (2) is obtained by putting k = 0 in first summation only and equating it to zero. Then the indicial equation is Since The coefficient of next lowest degree termin (2) is obtained by putting k = 1 in first summation and k = 0 in the second summation and equating it to zero. Equating to zero the coefficient of the recurrence relation is given by Or Which gives
Hence for Form m = 1/3 Hence for m = 1/3, the second solution will be The complete solution will be

If x = 0 is a regular singularity of the equation. ……..(1) Then the series solution is Which is called Frobenius series. 
If we put n = 0 then Bessel function becomes Now if n = 1, then 
As we know that Now put n = 1/2 in equation (1), then we get Hence proved. 
Put n = 1/2 in equation (1) of the above question, we get 
As we know that On differentiating with respect to x, we obtain Putting r – 1 = s 
We have Differentiating w.r.t. x, we get 
We know that from formula first and second Now adding these two, we get Or 
We know that On subtracting, we get 
We know that Multiply this by we get I.e. Or 
We know that Multiply by we get Or 
We know that The recurrence formula On differentiating, we get Now replace n by n 1 and n by n+1 in (1), we have Put the values of and from the above equations in (2), we get 
We know that from recurrence formula On integrating we get On taking n = 2 in (1), we get Again Put the value of from equation (2) and (3), we get By equation (1), when n = 1 
2. General solution of Bessel equation 
............(1) Where is a parameter which can be reduced Bessel’s diff equation of order p in t. ..............(2) Where For p nonintegral, the general solution of Equation (2) is Thus the general solution of Equation (1) is When p is nonintegral. 
The Legendre’s equations is Now the solution of the given equation is the series of descending powers of x is Here is an arbitrary constant. If n is a positive integer and The above solution is So that Here is called the Legendre’s function of first kind. Note Legendre’s equations of second kind is and can be defined as
The general solution of Legendre’s equation is Here A and B are arbitrary constants. 
Rodrigue’s formula can be defined as 
We know that by Rodrigue formula If n = 0, then it becomes If n = 1, If n = 2, Now putting n =3, 4, 5……..n we get ………………………………….. Where N = n/2 if n is even and N = 1/2 (n1) if n is odd.

By equating the coefficients of like powers of x, we get Put these values in equation (1), we get

We know that On integrating by parts, we get
Now integrate (n – 2) times by parts, we get 
Formula1: Fromula2: Formula3: Formula4: Formula5: Formula6: 
Now coefficient of in Coefficient of in Coefficient of in And so on. Coefficient of in the expansion of equation (1) The coefficients of etc. in (1) are Therefore

We know that
Equating the coefficients of both sides, we have 
is a solution of …………………. (1) And is a solution of ……………. (2) Now multiply (1) by z and (2) by y and subtracting, we have
Now integrate from 1 to +1, we get 
On integrating by parts, we get Now integrating m – 2 times, we get 
2. The general solution of Legendre’s equation is Here A and B are arbitrary constants. 3. Rodrigue’s formula can be defined as 4. Orthogonality of Legendre polynomials 