Solutions To Power Series (Differential Equation ) Problems

 1.  Give the singular points for each of the following:

(a) (x - 3) y" + (x + 1)y = 0

(b) (x² + 1) y"'  + y"   -  x² y  =  0

Solutions:

 Rewrite each in standard form:

(a)  (x - 3) y" + (x + 1)y = 0

y"   =    - (x + 1) y /  (x - 3)  

The singular point is x = 3

(b)  (x² + 1) y"'  + y"   -  x² y  =  0

 y"'  =    - y" / (x + i)  (x - i)   + x² y  /(x + i)  (x - i) 

  The singular points are x = + i

2. Determine whether x = 0 is an ordinary point of the differential equation:

x² y " + 2 y'  +  xy = 0 

Soln:  Rewrite in standard form:

y " =   -  2 y' / x²  -   xy  / x² 

y " =   -  (2 / x²) y'-   y  / x   

P(x)  =  2 / x²     

Q (x) =  1/x   

Neither function is analytic at x= 0 so x = 0 is a singular point of the differential equation because the denominators go to 0.

3. Find the power series solution for the differential equation:

x y" +  x³ y   - 3 xy = 0   

That satisfies: y = 0 and y' = 2  at x = 1

Soln.:

Rewrite in standard form:

 y" =  -  x² y   +  3 y

Where:  P(x) =   x³ / x =  x ²   and Q (x) =  3x/ x = 3

Here P(x)  =  x ²  and  Q(x) = 3  are both polynomials and hence are analytic everywhere.  Therefore every value of x is an ordinary point.

Differentiating the DE in successive steps:

 y" =  -   x ² y'   +  3 y 

y"' =  -  x ² y" - 2x y'  +  y'    +  y  

y ^iv  =  -  x ² y"' - 2 xy"  - 2y'   +  y

Evaluate each derivative at x =1 :

y"(1) =  -2  = - (1)² (2) + 3(0)

y"'(1) =  0  =  - (1)² (-2) - 2(1)(2)  + (2) - 0

y ^iv (1) = 0  =  - (1)² (0) - 2(1)(-2) - 2(2)  + 0

Rem:

[(x - x ₁ ) ⁿ  D ⁿ  +  (x - x ₁ )   b1 (x) D ^n-1  + 

(x - x ₁  ) ^{n- 2}  b1 (x) D ^n-2  +.  .. .....(x - x ₁  ) ^{n- 1}  b _{n- 1}  (x) D + . . . 

Then:

y(x) = 2(x - 1) -  (x - 1) ²  +  ..   (x - 1) ¹   + . . .