MA-207 Differential Equations II Ronnie Sebastian Department of - PowerPoint PPT Presentation

First solution in regular singular case Theorem x 2 y ′′ + xb ( x ) y ′ + c ( x ) y = 0 Consider the ODE ( ∗ ) where b ( x ) and c ( x ) are analytic at 0 . Then x = 0 is a regular singular point of ODE. 9 / 33

First solution in regular singular case Theorem x 2 y ′′ + xb ( x ) y ′ + c ( x ) y = 0 Consider the ODE ( ∗ ) where b ( x ) and c ( x ) are analytic at 0 . Then x = 0 is a regular singular point of ODE. Then ( ∗ ) has a solution of the form a n x n a 0 � = 0 , y ( x ) = x r � r ∈ C ( ∗∗ ) n ≥ 0 The solution ( ∗∗ ) is called Frobenius solution or fractional power series solution. 9 / 33

First solution in regular singular case Theorem x 2 y ′′ + xb ( x ) y ′ + c ( x ) y = 0 Consider the ODE ( ∗ ) where b ( x ) and c ( x ) are analytic at 0 . Then x = 0 is a regular singular point of ODE. Then ( ∗ ) has a solution of the form a n x n a 0 � = 0 , y ( x ) = x r � r ∈ C ( ∗∗ ) n ≥ 0 The solution ( ∗∗ ) is called Frobenius solution or fractional power series solution. a n x n converges on ( − ρ, ρ ) , where ρ is the � The power series n ≥ 0 minimum of the radius of convergence of b ( x ) and c ( x ) . We will consider the solution y ( x ) in the open interval (0 , ρ ) . 9 / 33

Second solution in regular singular case The analysis now breaks into the following three cases r 1 − r 2 / ∈ Z r 1 = r 2 0 � = r 1 − r 2 ∈ Z 10 / 33

Second solution: r 1 − r 2 / ∈ Z In this case, because of the assumption that r 1 − r 2 / ∈ Z , it follows that I ( n + r 2 ) � = 0 for any n ≥ 1 . 11 / 33

Second solution: r 1 − r 2 / ∈ Z In this case, because of the assumption that r 1 − r 2 / ∈ Z , it follows that I ( n + r 2 ) � = 0 for any n ≥ 1 . Thus, as before, the second solution is given by � a n ( r 2 ) x n + r 2 y 2 ( x ) = n ≥ 0 11 / 33

Second solution: r 1 − r 2 / ∈ Z In this case, because of the assumption that r 1 − r 2 / ∈ Z , it follows that I ( n + r 2 ) � = 0 for any n ≥ 1 . Thus, as before, the second solution is given by � a n ( r 2 ) x n + r 2 y 2 ( x ) = n ≥ 0 Example x 2 y ′′ − x 2 y ′ + (1+ x ) Consider the ODE y = 0 2 11 / 33

Second solution: r 1 − r 2 / ∈ Z In this case, because of the assumption that r 1 − r 2 / ∈ Z , it follows that I ( n + r 2 ) � = 0 for any n ≥ 1 . Thus, as before, the second solution is given by � a n ( r 2 ) x n + r 2 y 2 ( x ) = n ≥ 0 Example x 2 y ′′ − x 2 y ′ + (1+ x ) Consider the ODE y = 0 2 Observe that x = 0 is a regular singular point. 11 / 33

Second solution: r 1 − r 2 / ∈ Z In this case, because of the assumption that r 1 − r 2 / ∈ Z , it follows that I ( n + r 2 ) � = 0 for any n ≥ 1 . Thus, as before, the second solution is given by � a n ( r 2 ) x n + r 2 y 2 ( x ) = n ≥ 0 Example x 2 y ′′ − x 2 y ′ + (1+ x ) Consider the ODE y = 0 2 Observe that x = 0 is a regular singular point. I ( r ) = r ( r − 1) − 1 2 r + 1 2 = (2 r ( r − 1) − r + 1) / 2 = (2 r 2 − 3 r + 1) / 2 = ( r − 1)(2 r − 1) / 2 Roots of I ( r ) = 0 are r 1 = 1 and r 2 = 1 / 2 11 / 33

Second solution: r 1 − r 2 / ∈ Z 2 x 2 y ′′ − xy ′ + (1 + x ) y = 0 ) Example (continues . . . Their difference r 1 − r 2 = 1 / 2 is not an integer. 12 / 33

Second solution: r 1 − r 2 / ∈ Z 2 x 2 y ′′ − xy ′ + (1 + x ) y = 0 ) Example (continues . . . Their difference r 1 − r 2 = 1 / 2 is not an integer. The equation defining a n , for n ≥ 1 , is I ( n + r ) a n + 1 2 a n − 1 = 0 12 / 33

Second solution: r 1 − r 2 / ∈ Z 2 x 2 y ′′ − xy ′ + (1 + x ) y = 0 ) Example (continues . . . Their difference r 1 − r 2 = 1 / 2 is not an integer. The equation defining a n , for n ≥ 1 , is I ( n + r ) a n + 1 2 a n − 1 = 0 Thus, a n − 1 ( r ) a n ( r ) = − ( n + r − 1)(2 n + 2 r − 1) 12 / 33

Second solution: r 1 − r 2 / ∈ Z 2 x 2 y ′′ − xy ′ + (1 + x ) y = 0 ) Example (continues . . . Their difference r 1 − r 2 = 1 / 2 is not an integer. The equation defining a n , for n ≥ 1 , is I ( n + r ) a n + 1 2 a n − 1 = 0 Thus, a n − 1 ( r ) a n ( r ) = − ( n + r − 1)(2 n + 2 r − 1) a n − 1 Thus, a n ( r 1 ) = a n (1) = − n (2 n + 1) 1 = ( − 1) n n !((2 n + 1)(2 n − 1) . . . 3 12 / 33

Second solution: r 1 − r 2 / ∈ Z 2 x 2 y ′′ − xy ′ + (1 + x ) y = 0 ) Example (continues . . .   ( − 1) n x n � y 1 ( x ) = x  1 +  n !(2 n + 1)(2 n − 1) . . . 3 n ≥ 1 13 / 33

Second solution: r 1 − r 2 / ∈ Z 2 x 2 y ′′ − xy ′ + (1 + x ) y = 0 ) Example (continues . . .   ( − 1) n x n � y 1 ( x ) = x  1 +  n !(2 n + 1)(2 n − 1) . . . 3 n ≥ 1 a n − 1 a n ( r 2 ) = − n (2 n − 1) 1 = ( − 1) n n !(2 n − 1)(2 n − 3) . . . 1   ( − 1) n x n � y 2 ( x ) = x 1 / 2  1 +  n !(2 n − 1)(2 n − 3) . . . 1 n ≥ 1 13 / 33

Second solution: r 1 − r 2 / ∈ Z 2 x 2 y ′′ − xy ′ + (1 + x ) y = 0 ) Example (continues . . .   ( − 1) n x n � y 1 ( x ) = x  1 +  n !(2 n + 1)(2 n − 1) . . . 3 n ≥ 1 a n − 1 a n ( r 2 ) = − n (2 n − 1) 1 = ( − 1) n n !(2 n − 1)(2 n − 3) . . . 1   ( − 1) n x n � y 2 ( x ) = x 1 / 2  1 +  n !(2 n − 1)(2 n − 3) . . . 1 n ≥ 1 1 Since | a n | are smaller that n ! , it is clear that both solutions converge on (0 , ∞ ) . 13 / 33

Second solution: r 1 = r 2 Consider the function of two variables � a n ( r ) x n + r ψ ( r, x ) := n ≥ 0 14 / 33

Second solution: r 1 = r 2 Consider the function of two variables � a n ( r ) x n + r ψ ( r, x ) := n ≥ 0 Consider the differential operator L := x 2 d 2 dx 2 + xb ( x ) d dx + c ( x ) We have already computed the coefficient of x n + r in Lψ ( r, x ) . Recall that this is given by 1 The coefficient of x r is I ( r ) a 0 14 / 33

Second solution: r 1 = r 2 Consider the function of two variables � a n ( r ) x n + r ψ ( r, x ) := n ≥ 0 Consider the differential operator L := x 2 d 2 dx 2 + xb ( x ) d dx + c ( x ) We have already computed the coefficient of x n + r in Lψ ( r, x ) . Recall that this is given by 1 The coefficient of x r is I ( r ) a 0 2 The coefficient of x n + r , for n ≥ 1 , is n − 1 n − 1 � � I ( n + r ) a n ( r ) + b n − i ( i + r ) a i ( r ) + c n − i a i ( r ) i =0 i =0 14 / 33

Second solution: r 1 = r 2 Consider the functions a n ( r ) , defined inductively using the equations a 0 ( r ) := 1 15 / 33

Second solution: r 1 = r 2 Consider the functions a n ( r ) , defined inductively using the equations a 0 ( r ) := 1 and for n ≥ 1 n − 1 n − 1 � � I ( n + r ) a n ( r ) + b n − i ( i + r ) a i ( r ) + c n − i a i ( r ) = 0 i =0 i =0 15 / 33

Second solution: r 1 = r 2 Consider the functions a n ( r ) , defined inductively using the equations a 0 ( r ) := 1 and for n ≥ 1 n − 1 n − 1 � � I ( n + r ) a n ( r ) + b n − i ( i + r ) a i ( r ) + c n − i a i ( r ) = 0 i =0 i =0 With these definitions, it follows that Lψ ( r, x ) = I ( r ) x r 15 / 33

Second solution: r 1 = r 2 If r 1 − r 2 / ∈ Z then the second solution is given by y 2 ( x ) = x r 2 � a n ( r 2 ) x n n ≥ 0 16 / 33

Second solution: r 1 = r 2 If r 1 − r 2 / ∈ Z then the second solution is given by y 2 ( x ) = x r 2 � a n ( r 2 ) x n n ≥ 0 Now let us consider the case when I has repeated roots 16 / 33

Second solution: r 1 = r 2 If r 1 − r 2 / ∈ Z then the second solution is given by y 2 ( x ) = x r 2 � a n ( r 2 ) x n n ≥ 0 Now let us consider the case when I has repeated roots Since I has repeated roots r 1 = r 2 , it follows that, for every n ≥ 1 , the polynomial � n i =1 I ( i + r ) does not vanish at r = r 1 Consequently, it is clear that the a n ( r ) are analytic in a small neighborhood around r = r 1 = r 2 . 16 / 33

Second solution: r 1 = r 2 d Now let us apply the differential operator dr on both sides of the equation Lψ ( r, x ) = I ( r ) x r . Clearly the operators L and d dr commute with each other, and so we get 17 / 33

Second solution: r 1 = r 2 d Now let us apply the differential operator dr on both sides of the equation Lψ ( r, x ) = I ( r ) x r . Clearly the operators L and d dr commute with each other, and so we get drLψ ( r, x ) = L d d drψ ( r, x ) = d n ( r ) x n + r + a n ( r ) x n + r log x � a ′ drI ( r ) x r � � = L n ≥ 0 = I ′ ( r ) x r + I ( r ) x r log x 17 / 33

Second solution: r 1 = r 2 d Now let us apply the differential operator dr on both sides of the equation Lψ ( r, x ) = I ( r ) x r . Clearly the operators L and d dr commute with each other, and so we get drLψ ( r, x ) = L d d drψ ( r, x ) = d n ( r ) x n + r + a n ( r ) x n + r log x � a ′ drI ( r ) x r � � = L n ≥ 0 = I ′ ( r ) x r + I ( r ) x r log x Thus, if we plug in r = r 1 = r 2 in the above, then we get � � � n ( r 2 ) x n + r 2 + a n ( r 2 ) x n + r 2 log x a ′ L = 0 n ≥ 0 17 / 33

Second solution: r 1 = r 2 Theorem (Second solution: r 1 = r 2 ) A second solution to the differential equation is given by 18 / 33

Second solution: r 1 = r 2 Theorem (Second solution: r 1 = r 2 ) A second solution to the differential equation is given by n ( r 2 ) x n + r 2 + � � a n ( r 2 ) x n + r 2 log x a ′ n ≥ 0 n ≥ 0 18 / 33

Second solution: r 1 = r 2 Example Consider the ODE x 2 y ′′ + 3 xy ′ + (1 − 2 x ) y = 0 19 / 33

Second solution: r 1 = r 2 Example Consider the ODE x 2 y ′′ + 3 xy ′ + (1 − 2 x ) y = 0 This has a regular singularity at x = 0 . 19 / 33

Second solution: r 1 = r 2 Example Consider the ODE x 2 y ′′ + 3 xy ′ + (1 − 2 x ) y = 0 This has a regular singularity at x = 0 . I ( r ) = r ( r − 1) + 3 r + 1 = r 2 + 2 r + 1 has a repeated roots − 1 , − 1 . 19 / 33

Second solution: r 1 = r 2 Example Consider the ODE x 2 y ′′ + 3 xy ′ + (1 − 2 x ) y = 0 This has a regular singularity at x = 0 . I ( r ) = r ( r − 1) + 3 r + 1 = r 2 + 2 r + 1 has a repeated roots − 1 , − 1 . Let us find the Frobenius solution directly by putting y = x r � a n ( r ) x n a 0 = 1 n ≥ 0 19 / 33

Second solution: r 1 = r 2 Example Consider the ODE x 2 y ′′ + 3 xy ′ + (1 − 2 x ) y = 0 This has a regular singularity at x = 0 . I ( r ) = r ( r − 1) + 3 r + 1 = r 2 + 2 r + 1 has a repeated roots − 1 , − 1 . Let us find the Frobenius solution directly by putting y = x r � a n ( r ) x n a 0 = 1 n ≥ 0 y ′ = � ( n + r ) a n ( r ) x n + r − 1 n ≥ 0 19 / 33

Second solution: r 1 = r 2 Example Consider the ODE x 2 y ′′ + 3 xy ′ + (1 − 2 x ) y = 0 This has a regular singularity at x = 0 . I ( r ) = r ( r − 1) + 3 r + 1 = r 2 + 2 r + 1 has a repeated roots − 1 , − 1 . Let us find the Frobenius solution directly by putting y = x r � a n ( r ) x n a 0 = 1 n ≥ 0 y ′ = � ( n + r ) a n ( r ) x n + r − 1 n ≥ 0 ∞ y ′′ = � ( n + r )( n + r − 1) a n ( r ) x n + r − 2 n ≥ 0 19 / 33

Second solution: r 1 = r 2 Example (continues . . . ) x 2 y ( x, r ) ′′ + 3 xy ( x, r ) ′ + (1 − 2 x ) y ( x, r ) ∞ � [( n + r )( n + r − 1) + 3( n + r ) + 1] a n ( r ) x n + r = n =0 ∞ � 2 a n ( r ) x n + r +1 − n =0 20 / 33

Second solution: r 1 = r 2 Example (continues . . . ) x 2 y ( x, r ) ′′ + 3 xy ( x, r ) ′ + (1 − 2 x ) y ( x, r ) ∞ � [( n + r )( n + r − 1) + 3( n + r ) + 1] a n ( r ) x n + r = n =0 ∞ � 2 a n ( r ) x n + r +1 − n =0 Recursion relations for n ≥ 1 are 20 / 33

Second solution: r 1 = r 2 Example (continues . . . ) x 2 y ( x, r ) ′′ + 3 xy ( x, r ) ′ + (1 − 2 x ) y ( x, r ) ∞ � [( n + r )( n + r − 1) + 3( n + r ) + 1] a n ( r ) x n + r = n =0 ∞ � 2 a n ( r ) x n + r +1 − n =0 Recursion relations for n ≥ 1 are 2 a n − 1 ( r ) a n ( r ) = ( n + r )( n + r − 1) + 3( n + r ) + 1 2 a n − 1 ( r ) = ( n + r + 1) 2 2 n = [( n + r + 1)( n + r ) . . . ( r + 2)] 2 a 0 20 / 33

Second solution: r 1 = r 2 Example (continues . . . ) Setting r = − 1 (and a 0 = 1 ) yields the fractional power series solution 2 n y 1 ( x ) = 1 � ( n !) 2 x n x n ≥ 0 The power series converges on (0 , ∞ ) . 21 / 33

Second solution: r 1 = r 2 Example (continues . . . ) Setting r = − 1 (and a 0 = 1 ) yields the fractional power series solution 2 n y 1 ( x ) = 1 � ( n !) 2 x n x n ≥ 0 The power series converges on (0 , ∞ ) . The second solution is y 2 ( x ) = y 1 ( x ) log x + x − 1 � n ( − 1) x n a ′ n ≥ 1 21 / 33

Second solution: r 1 = r 2 Example (continues . . . ) Setting r = − 1 (and a 0 = 1 ) yields the fractional power series solution 2 n y 1 ( x ) = 1 � ( n !) 2 x n x n ≥ 0 The power series converges on (0 , ∞ ) . The second solution is y 2 ( x ) = y 1 ( x ) log x + x − 1 � n ( − 1) x n a ′ n ≥ 1 where 2 n a n ( r ) = [( n + r + 1)( n + r ) . . . ( r + 2)] 2 n ( r ) = − 2 . 2 n [( n + r + 1)( n + r ) . . . ( r + 2)] ′ a ′ [( n + r + 1)( n + r ) . . . ( r + 2)] 3 21 / 33

Second solution: r 1 = r 2 Example (continued) � 1 1 1 � = − 2 a n ( r ) n + r + 1 + n + r + · · · + r + 2 22 / 33

Second solution: r 1 = r 2 Example (continued) � 1 1 1 � = − 2 a n ( r ) n + r + 1 + n + r + · · · + r + 2 Putting r = − 1 , we get n ( − 1) = − 2 n +1 H n a ′ ( n !) 2 where H n = 1 + 1 2 + · · · + 1 n (These are the partial sums of the harmonic series.) 22 / 33

Second solution: r 1 = r 2 Example (continued) � 1 1 1 � = − 2 a n ( r ) n + r + 1 + n + r + · · · + r + 2 Putting r = − 1 , we get n ( − 1) = − 2 n +1 H n a ′ ( n !) 2 where H n = 1 + 1 2 + · · · + 1 n (These are the partial sums of the harmonic series.) So the second solution is 2 n +1 H n y 2 ( x ) = y 1 ( x )log x − 1 � x n ( n !) 2 x n ≥ 1 22 / 33

Second solution: r 1 = r 2 Example (continued) � 1 1 1 � = − 2 a n ( r ) n + r + 1 + n + r + · · · + r + 2 Putting r = − 1 , we get n ( − 1) = − 2 n +1 H n a ′ ( n !) 2 where H n = 1 + 1 2 + · · · + 1 n (These are the partial sums of the harmonic series.) So the second solution is 2 n +1 H n y 2 ( x ) = y 1 ( x )log x − 1 � x n ( n !) 2 x n ≥ 1 It is clear that this series converges on (0 , ∞ ) . 22 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Define N := r 1 − r 2 Note that each a n ( r ) is a rational function in r , in fact, the denominator is exactly � n i =1 I ( i + r ) . 23 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Define N := r 1 − r 2 Note that each a n ( r ) is a rational function in r , in fact, the denominator is exactly � n i =1 I ( i + r ) . The polynomial � n i =1 I ( i + r ) evaluated at r 2 vanishes iff n ≥ N . For n ≥ N it vanishes to order exactly 1. Thus, if we define A n ( r ) := a n ( r )( r − r 2 ) then it is clear that for every n ≥ 0 , the function A n ( r ) is analytic in a neighborhood of r 2 . 23 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z In particular, A n ( r 2 ) and A ′ n ( r 2 ) are well defined real numbers. 24 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z In particular, A n ( r 2 ) and A ′ n ( r 2 ) are well defined real numbers. Multiplying the equation Lψ ( r, x ) = I ( r ) x r with r − r 2 we get ( r − r 2 ) Lψ ( r, x ) = L ( r − r 2 ) ψ ( r, x ) = ( r − r 2 ) I ( r ) x r 24 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z In particular, A n ( r 2 ) and A ′ n ( r 2 ) are well defined real numbers. Multiplying the equation Lψ ( r, x ) = I ( r ) x r with r − r 2 we get ( r − r 2 ) Lψ ( r, x ) = L ( r − r 2 ) ψ ( r, x ) = ( r − r 2 ) I ( r ) x r Note that � A n ( r ) x n + r ( r − r 2 ) ψ ( r, x ) = n ≥ 0 24 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z d Now let us apply the differential operator dr on both sides of the equation L ( r − r 2 ) ψ ( r, x ) = ( r − r 2 ) I ( r ) x r to get drL ( r − r 2 ) ψ ( r, x ) = L d d dr ( r − r 2 ) ψ ( r, x ) = d dr ( r − r 2 ) I ( r ) x r = I ( r ) x r + ( r − r 2 ) I ′ ( r ) x r + ( r − r 2 ) I ( r ) x r log x 25 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z d Now let us apply the differential operator dr on both sides of the equation L ( r − r 2 ) ψ ( r, x ) = ( r − r 2 ) I ( r ) x r to get drL ( r − r 2 ) ψ ( r, x ) = L d d dr ( r − r 2 ) ψ ( r, x ) = d dr ( r − r 2 ) I ( r ) x r = I ( r ) x r + ( r − r 2 ) I ′ ( r ) x r + ( r − r 2 ) I ( r ) x r log x Thus we get L d = L d � � A n ( r ) x n + r � � � A n ( r ) x n + r � dr dr n ≥ 0 n ≥ 0 � � n ( r ) x n + r + A n ( r ) x n + r log x � A ′ = L n ≥ 0 = I ( r ) x r + ( r − r 2 ) I ′ ( r ) x r + ( r − r 2 ) I ( r ) x r log x 25 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z If we set r = r 2 into the equation � � n ( r ) x n + r + A n ( r ) x n + r log x � = I ( r ) x r + ( r − r 2 ) I ′ ( r ) x r + A ′ L n ≥ 0 ( r − r 2 ) I ( r ) x r log x 26 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z If we set r = r 2 into the equation � � n ( r ) x n + r + A n ( r ) x n + r log x � = I ( r ) x r + ( r − r 2 ) I ′ ( r ) x r + A ′ L n ≥ 0 ( r − r 2 ) I ( r ) x r log x we get the second solution � � � n ( r 2 ) x n + r 2 + A n ( r 2 ) x n + r 2 log x A ′ L = 0 n ≥ 0 26 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z If we set r = r 2 into the equation � � n ( r ) x n + r + A n ( r ) x n + r log x � = I ( r ) x r + ( r − r 2 ) I ′ ( r ) x r + A ′ L n ≥ 0 ( r − r 2 ) I ( r ) x r log x we get the second solution � � � n ( r 2 ) x n + r 2 + A n ( r 2 ) x n + r 2 log x A ′ L = 0 n ≥ 0 Theorem (Second solution: 0 � = r 1 − r 2 ∈ Z ) A second solution to the differential equation is given by 26 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z If we set r = r 2 into the equation � � n ( r ) x n + r + A n ( r ) x n + r log x � = I ( r ) x r + ( r − r 2 ) I ′ ( r ) x r + A ′ L n ≥ 0 ( r − r 2 ) I ( r ) x r log x we get the second solution � � � n ( r 2 ) x n + r 2 + A n ( r 2 ) x n + r 2 log x A ′ L = 0 n ≥ 0 Theorem (Second solution: 0 � = r 1 − r 2 ∈ Z ) A second solution to the differential equation is given by n ( r 2 ) x n + r 2 + � � A n ( r 2 ) x n + r 2 log x A ′ n ≥ 0 n ≥ 0 26 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Example xy ′′ − (4 + x ) y ′ + 2 y = 0 Consider the ODE ( ∗ ) 27 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Example xy ′′ − (4 + x ) y ′ + 2 y = 0 Consider the ODE ( ∗ ) Multiplying ( ∗ ) with x , we get x = 0 is a regular singular point. I ( r ) = r ( r − 1) − 4 r + 0 = r ( r − 5) = 0 with the roots differing by a positive integer. 27 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Example xy ′′ − (4 + x ) y ′ + 2 y = 0 Consider the ODE ( ∗ ) Multiplying ( ∗ ) with x , we get x = 0 is a regular singular point. I ( r ) = r ( r − 1) − 4 r + 0 = r ( r − 5) = 0 with the roots differing by a positive integer. ∞ � Put y ( x, r ) = x r a n ( r ) x n , a 0 ( r ) = 1 , into the ODE to get n =0 27 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Example xy ′′ − (4 + x ) y ′ + 2 y = 0 Consider the ODE ( ∗ ) Multiplying ( ∗ ) with x , we get x = 0 is a regular singular point. I ( r ) = r ( r − 1) − 4 r + 0 = r ( r − 5) = 0 with the roots differing by a positive integer. ∞ � Put y ( x, r ) = x r a n ( r ) x n , a 0 ( r ) = 1 , into the ODE to get n =0 � ( n + r )( n + r − 1) a n ( r ) x n + r − 2 x n ≥ 0 ( n + r ) a n ( r ) x n + r − 1 + 2 a n ( r ) x n + r = 0 � � − (4 + x ) n ≥ 0 n ≥ 0 27 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Example xy ′′ − (4 + x ) y ′ + 2 y = 0 Consider the ODE ( ∗ ) Multiplying ( ∗ ) with x , we get x = 0 is a regular singular point. I ( r ) = r ( r − 1) − 4 r + 0 = r ( r − 5) = 0 with the roots differing by a positive integer. ∞ � Put y ( x, r ) = x r a n ( r ) x n , a 0 ( r ) = 1 , into the ODE to get n =0 � ( n + r )( n + r − 1) a n ( r ) x n + r − 2 x n ≥ 0 ( n + r ) a n ( r ) x n + r − 1 + 2 a n ( r ) x n + r = 0 � � − (4 + x ) n ≥ 0 n ≥ 0 the coefficient of x n + r − 1 for n ≥ 1 gives 27 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Example (continues . . . ) ( n + r )( n + r − 1) a n ( r ) − 4( n + r ) a n ( r ) − ( n + r − 1) a n − 1 ( r ) +2 a n − 1 ( r ) = 0 28 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Example (continues . . . ) ( n + r )( n + r − 1) a n ( r ) − 4( n + r ) a n ( r ) − ( n + r − 1) a n − 1 ( r ) +2 a n − 1 ( r ) = 0 For n ≥ 1 , 28 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Example (continues . . . ) ( n + r )( n + r − 1) a n ( r ) − 4( n + r ) a n ( r ) − ( n + r − 1) a n − 1 ( r ) +2 a n − 1 ( r ) = 0 For n ≥ 1 , ( n + r )( n + r − 5) a n = ( n + r − 3) a n − 1 28 / 33

Second solution: 0 � = r 1 − r 2 ∈ Z Example (continues . . . ) ( n + r )( n + r − 1) a n ( r ) − 4( n + r ) a n ( r ) − ( n + r − 1) a n − 1 ( r ) +2 a n − 1 ( r ) = 0 For n ≥ 1 , ( n + r )( n + r − 5) a n = ( n + r − 3) a n − 1 ( n + r − 3) a n ( r ) = ( n + r )( n + r − 5) a n − 1 ( n + r − 3) . . . ( r − 2) = ( n + r ) . . . (1 + r )( n + r − 5) . . . ( r − 4) a 0 28 / 33