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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) |
Case-1: when roots m1 and m2 are distinct and these are not differing by an integer- The complete solution in this case will be- Case-2: when roots m1 and m2 are equal- Case-3: when roots are distinct but differ by an integer- Case-4: 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-
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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 non-integral, the general solution of Equation (2) is Thus the general solution of Equation (1) is When p is non-integral. |
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 (n-1) if n is odd.
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By equating the coefficients of like powers of x, we get- Put these values in equation (1), we get-
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We know that- On integrating by parts, we get-
Now integrate (n – 2) times by parts, we get- |
Formula-1: Fromula-2: Formula-3: Formula-4: Formula-5: Formula-6: |
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-
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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- |