MA-207 Di erential Equations II

MA-207 Di erential Equations II Ronnie Sebastian Department of Mathematics Indian Institute of Technology Bombay Powai, Mumbai - 76 1/33...

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MA-207 Differential Equations II Ronnie Sebastian

Department of Mathematics Indian Institute of Technology Bombay Powai, Mumbai - 76

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Ordinary and singular points

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Ordinary and singular points Definition Consider the second-order linear ODE in standard form y 00 + p(x)y 0 + q(x)y = 0

(∗)

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Ordinary and singular points Definition Consider the second-order linear ODE in standard form y 00 + p(x)y 0 + q(x)y = 0 1

(∗)

x0 ∈ R is called an ordinary point of (∗) if p(x) and q(x) are analytic at x0

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Ordinary and singular points Definition Consider the second-order linear ODE in standard form y 00 + p(x)y 0 + q(x)y = 0

(∗)

1

x0 ∈ R is called an ordinary point of (∗) if p(x) and q(x) are analytic at x0

2

x0 ∈ R is called regular singular point if (x − x0 )p(x) and (x − x0 )2 q(x) are analytic at x0 .

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Ordinary and singular points Definition Consider the second-order linear ODE in standard form y 00 + p(x)y 0 + q(x)y = 0

(∗)

1

x0 ∈ R is called an ordinary point of (∗) if p(x) and q(x) are analytic at x0

2

x0 ∈ R is called regular singular point if (x − x0 )p(x) and (x − x0 )2 q(x) are analytic at x0 . This is equivalent to saying that there are functions b(x) and c(x) which are analytic at x0 such that p(x) =

b(x) (x − x0 )

q(x) =

c(x) (x − x0 )2

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Ordinary and singular points Definition Consider the second-order linear ODE in standard form y 00 + p(x)y 0 + q(x)y = 0

(∗)

1

x0 ∈ R is called an ordinary point of (∗) if p(x) and q(x) are analytic at x0

2

x0 ∈ R is called regular singular point if (x − x0 )p(x) and (x − x0 )2 q(x) are analytic at x0 . This is equivalent to saying that there are functions b(x) and c(x) which are analytic at x0 such that p(x) =

3

b(x) (x − x0 )

q(x) =

c(x) (x − x0 )2

If x0 ∈ R is not ordinary or regular singular, then we call it irregular singular. 2 / 33

Ordinary and singular points Example x = 0 is an irregular singular point of x3 y 00 + xy 0 + y = 0

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Ordinary and singular points Example x = 0 is an irregular singular point of x3 y 00 + xy 0 + y = 0 Let us write the ODE in standard form y 00 +

1 0 1 y + 3y = 0 2 x x

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Ordinary and singular points Example x = 0 is an irregular singular point of x3 y 00 + xy 0 + y = 0 Let us write the ODE in standard form y 00 + Then p(x) =

1 0 1 y + 3y = 0 2 x x 1 x2

q(x) =

1 x3

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Ordinary and singular points Example x = 0 is an irregular singular point of x3 y 00 + xy 0 + y = 0 Let us write the ODE in standard form y 00 + Then p(x) =

1 0 1 y + 3y = 0 2 x x 1 x2

q(x) =

1 x3

Clearly, 1 1 x2 q(x) = x x are not analytic at 0. Thus, x = 0 is an irregular singular point. xp(x) =

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Example Consider the Cauchy-Euler equation x2 y 00 + b0 xy 0 + c0 y = 0

b0 , c0 ∈ R

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Example Consider the Cauchy-Euler equation x2 y 00 + b0 xy 0 + c0 y = 0

b0 , c0 ∈ R

x = 0 is a regular singular point, since we can write the ODE as y 00 +

b0 0 c0 y + 2y = 0 x x

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Example Consider the Cauchy-Euler equation x2 y 00 + b0 xy 0 + c0 y = 0

b0 , c0 ∈ R

x = 0 is a regular singular point, since we can write the ODE as y 00 +

b0 0 c0 y + 2y = 0 x x

All x 6= 0 are ordinary points.

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Example Consider the Cauchy-Euler equation x2 y 00 + b0 xy 0 + c0 y = 0

b0 , c0 ∈ R

x = 0 is a regular singular point, since we can write the ODE as y 00 +

b0 0 c0 y + 2y = 0 x x

All x 6= 0 are ordinary points. Assume x > 0

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Example Consider the Cauchy-Euler equation x2 y 00 + b0 xy 0 + c0 y = 0

b0 , c0 ∈ R

x = 0 is a regular singular point, since we can write the ODE as y 00 +

b0 0 c0 y + 2y = 0 x x

All x 6= 0 are ordinary points. Assume x > 0 Note that y = xr solves the equation iff r(r − 1) + b0 r + c0 = 0 ⇐⇒ r2 + (b0 − 1)r + c0 = 0

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Example Consider the Cauchy-Euler equation x2 y 00 + b0 xy 0 + c0 y = 0

b0 , c0 ∈ R

x = 0 is a regular singular point, since we can write the ODE as y 00 +

b0 0 c0 y + 2y = 0 x x

All x 6= 0 are ordinary points. Assume x > 0 Note that y = xr solves the equation iff r(r − 1) + b0 r + c0 = 0 ⇐⇒ r2 + (b0 − 1)r + c0 = 0 Let r1 and r2 denote the roots of this quadratic equation. 4 / 33

Example (continues . . .)

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Example (continues . . .) If the roots r1 6= r2 are real, then x r1

and xr2

are two independent solutions.

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Example (continues . . .) If the roots r1 6= r2 are real, then x r1

and xr2

are two independent solutions. If the roots r1 = r2 are real, then xr1

and (log x)xr1

are two independent solutions.

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Example (continues . . .) If the roots r1 6= r2 are real, then x r1

and xr2

are two independent solutions. If the roots r1 = r2 are real, then xr1

and (log x)xr1

are two independent solutions. If the roots are complex (written as a ± ib), then xa cos(b log x)

and xa sin(b log x)

are two independent solutions.

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Example (continues . . .) If the roots r1 6= r2 are real, then x r1

and xr2

are two independent solutions. If the roots r1 = r2 are real, then xr1

and (log x)xr1

are two independent solutions. If the roots are complex (written as a ± ib), then xa cos(b log x)

and xa sin(b log x)

are two independent solutions. This example motivates us to look for solutions which involve xr . 5 / 33

First solution in regular singular case

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First solution in regular singular case Consider

x2 y 00 + xb(x)y 0 + c(x)y = 0 with X X b(x) = bi xi c(x) = ci xi i≥0

i≥0

analytic functions in a small neighborhood of 0.

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First solution in regular singular case Consider

x2 y 00 + xb(x)y 0 + c(x)y = 0 with X X b(x) = bi xi c(x) = ci xi i≥0

i≥0

analytic functions in a small neighborhood of 0. x = 0 is a regular singular point.

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First solution in regular singular case Consider

x2 y 00 + xb(x)y 0 + c(x)y = 0 with X X b(x) = bi xi c(x) = ci xi i≥0

i≥0

analytic functions in a small neighborhood of 0. x = 0 is a regular singular point. Define the indicial equation I(r) := r(r − 1) + b0 r + c0

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First solution in regular singular case Consider

x2 y 00 + xb(x)y 0 + c(x)y = 0 with X X b(x) = bi xi c(x) = ci xi i≥0

i≥0

analytic functions in a small neighborhood of 0. x = 0 is a regular singular point. Define the indicial equation I(r) := r(r − 1) + b0 r + c0 Look for solution of the type y(x) =

X

an xn+r

n≥0

by substituting this into the differential equation and setting the coefficient of xn+r to 0. 6 / 33

First solution in regular singular case We get the following 1 The coefficient of xr is I(r)a , thus we need I(r)a = 0 0 0

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First solution in regular singular case We get the following 1 The coefficient of xr is I(r)a , thus we need I(r)a = 0 0 0 2 The coefficient of xn+r , for n ≥ 1, is I(n + r)an +

n−1 X i=0

bn−i (i + r)ai +

n−1 X

cn−i ai

i=0

We need this to be 0

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First solution in regular singular case We get the following 1 The coefficient of xr is I(r)a , thus we need I(r)a = 0 0 0 2 The coefficient of xn+r , for n ≥ 1, is I(n + r)an +

n−1 X i=0

bn−i (i + r)ai +

n−1 X

cn−i ai

i=0

We need this to be 0 Let r1 and r2 be roots of I(r) = 0. Assume r1 and r2 are real and r1 ≥ r2 .

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First solution in regular singular case We get the following 1 The coefficient of xr is I(r)a , thus we need I(r)a = 0 0 0 2 The coefficient of xn+r , for n ≥ 1, is I(n + r)an +

n−1 X i=0

bn−i (i + r)ai +

n−1 X

cn−i ai

i=0

We need this to be 0 Let r1 and r2 be roots of I(r) = 0. Assume r1 and r2 are real and r1 ≥ r2 . Define a0 = 1.

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First solution in regular singular case We get the following 1 The coefficient of xr is I(r)a , thus we need I(r)a = 0 0 0 2 The coefficient of xn+r , for n ≥ 1, is I(n + r)an +

n−1 X i=0

bn−i (i + r)ai +

n−1 X

cn−i ai

i=0

We need this to be 0 Let r1 and r2 be roots of I(r) = 0. Assume r1 and r2 are real and r1 ≥ r2 . Define a0 = 1. Set r = r1 in the above equation and define an , for n ≥ 1, inductively by the equation Pn−1 Pn−1 i=0 bn−i (i + r1 )ai + i=0 cn−i ai an (r1 ) = − I(n + r1 ) 7 / 33

First solution in regular singular case

Since I(n + r1 ) 6= 0 for n ≥ 1, an (r1 ) is a well defined real number.

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First solution in regular singular case

Since I(n + r1 ) 6= 0 for n ≥ 1, an (r1 ) is a well defined real number. Thus, y1 (x) =

X

an (r1 )xn+r1

n≥0

is a possible solution to the above differential equation.

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First solution in regular singular case Theorem Consider the ODE

x2 y 00 + xb(x) y 0 + c(x) y = 0

(∗)

where b(x) and c(x) are analytic at 0. Then x = 0 is a regular singular point of ODE.

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First solution in regular singular case Theorem Consider the ODE

x2 y 00 + xb(x) y 0 + c(x) y = 0

(∗)

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 X y(x) = xr an xn a0 6= 0, r ∈ C (∗∗) n≥0

The solution (∗∗) is called Frobenius solution or fractional power series solution.

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First solution in regular singular case Theorem Consider the ODE

x2 y 00 + xb(x) y 0 + c(x) y = 0

(∗)

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 X y(x) = xr an xn a0 6= 0, r ∈ C (∗∗) n≥0

The solution (∗∗) is called Frobenius solution or fractional power series solution. X The power series an xn converges on (−ρ, ρ), where ρ is the 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 r1 − r2 ∈ /Z r1 = r2 0 6= r1 − r2 ∈ Z

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Second solution: r1 − r2 ∈ /Z In this case, because of the assumption that r1 − r2 ∈ / Z, it follows that I(n + r2 ) 6= 0 for any n ≥ 1.

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Second solution: r1 − r2 ∈ /Z In this case, because of the assumption that r1 − r2 ∈ / Z, it follows that I(n + r2 ) 6= 0 for any n ≥ 1. Thus, as before, the second solution is given by X y2 (x) = an (r2 )xn+r2 n≥0

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Second solution: r1 − r2 ∈ /Z In this case, because of the assumption that r1 − r2 ∈ / Z, it follows that I(n + r2 ) 6= 0 for any n ≥ 1. Thus, as before, the second solution is given by X y2 (x) = an (r2 )xn+r2 n≥0

Example Consider the ODE

x2 y 00 − x2 y 0 +

(1+x) 2 y

=0

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Second solution: r1 − r2 ∈ /Z In this case, because of the assumption that r1 − r2 ∈ / Z, it follows that I(n + r2 ) 6= 0 for any n ≥ 1. Thus, as before, the second solution is given by X y2 (x) = an (r2 )xn+r2 n≥0

Example Consider the ODE

x2 y 00 − x2 y 0 +

(1+x) 2 y

=0

Observe that x = 0 is a regular singular point.

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Second solution: r1 − r2 ∈ /Z In this case, because of the assumption that r1 − r2 ∈ / Z, it follows that I(n + r2 ) 6= 0 for any n ≥ 1. Thus, as before, the second solution is given by X y2 (x) = an (r2 )xn+r2 n≥0

Example Consider the ODE

x2 y 00 − x2 y 0 +

(1+x) 2 y

=0

Observe that x = 0 is a regular singular point. I(r) = r(r − 1) − 12 r +

1 2

= (2r(r − 1) − r + 1)/2 = (2r2 − 3r + 1)/2 = (r − 1)(2r − 1)/2 Roots of I(r) = 0 are r1 = 1 and r2 = 1/2 11 / 33

Second solution: r1 − r2 ∈ /Z Example (continues . . .

2x2 y 00 − xy 0 + (1 + x)y = 0)

Their difference r1 − r2 = 1/2 is not an integer.

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Second solution: r1 − r2 ∈ /Z Example (continues . . .

2x2 y 00 − xy 0 + (1 + x)y = 0)

Their difference r1 − r2 = 1/2 is not an integer. The equation defining an , for n ≥ 1, is 1 I(n + r)an + an−1 = 0 2

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Second solution: r1 − r2 ∈ /Z Example (continues . . .

2x2 y 00 − xy 0 + (1 + x)y = 0)

Their difference r1 − r2 = 1/2 is not an integer. The equation defining an , for n ≥ 1, is 1 I(n + r)an + an−1 = 0 2 Thus, an (r) = −

an−1 (r) (n + r − 1)(2n + 2r − 1)

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Second solution: r1 − r2 ∈ /Z Example (continues . . .

2x2 y 00 − xy 0 + (1 + x)y = 0)

Their difference r1 − r2 = 1/2 is not an integer. The equation defining an , for n ≥ 1, is 1 I(n + r)an + an−1 = 0 2 Thus, an−1 (r) (n + r − 1)(2n + 2r − 1) an−1 an (r1 ) = an (1) = − n(2n + 1) 1 = (−1)n n!((2n + 1)(2n − 1) . . . 3 an (r) = −

Thus,

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Second solution: r1 − r2 ∈ /Z 2x2 y 00 − xy 0 + (1 + x)y = 0)  n n X (−1) x  y1 (x) = x 1 + n!(2n + 1)(2n − 1) . . . 3

Example (continues . . . 

n≥1

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Second solution: r1 − r2 ∈ /Z 2x2 y 00 − xy 0 + (1 + x)y = 0)  n n X (−1) x  y1 (x) = x 1 + n!(2n + 1)(2n − 1) . . . 3 n≥1 an−1 an (r2 ) = − n(2n − 1) 1 = (−1)n n!(2n − 1)(2n − 3) . . . 1   n n X (−1) x  y2 (x) = x1/2 1 + n!(2n − 1)(2n − 3) . . . 1

Example (continues . . . 

n≥1

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Second solution: r1 − r2 ∈ /Z 2x2 y 00 − xy 0 + (1 + x)y = 0)  n n X (−1) x  y1 (x) = x 1 + n!(2n + 1)(2n − 1) . . . 3 n≥1 an−1 an (r2 ) = − n(2n − 1) 1 = (−1)n n!(2n − 1)(2n − 3) . . . 1   n n X (−1) x  y2 (x) = x1/2 1 + n!(2n − 1)(2n − 3) . . . 1

Example (continues . . . 

n≥1

Since |an | are smaller that converge on (0, ∞).

1 n! ,

it is clear that both solutions

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Second solution: r1 = r2 Consider the function of two variables X ψ(r, x) := an (r)xn+r n≥0

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Second solution: r1 = r2 Consider the function of two variables X ψ(r, x) := an (r)xn+r n≥0

Consider the differential operator L := x2

d2 d + xb(x) + c(x) 2 dx dx

We have already computed the coefficient of xn+r in Lψ(r, x). Recall that this is given by 1

The coefficient of xr is I(r)a0

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Second solution: r1 = r2 Consider the function of two variables X ψ(r, x) := an (r)xn+r n≥0

Consider the differential operator L := x2

d2 d + xb(x) + c(x) 2 dx dx

We have already computed the coefficient of xn+r in Lψ(r, x). Recall that this is given by 1 2

The coefficient of xr is I(r)a0 The coefficient of xn+r , for n ≥ 1, is I(n + r)an (r) +

n−1 X i=0

bn−i (i + r)ai (r) +

n−1 X

cn−i ai (r)

i=0 14 / 33

Second solution: r1 = r2

Consider the functions an (r), defined inductively using the equations a0 (r) := 1

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Second solution: r1 = r2

Consider the functions an (r), defined inductively using the equations a0 (r) := 1 and for n ≥ 1 I(n + r)an (r) +

n−1 X i=0

bn−i (i + r)ai (r) +

n−1 X

cn−i ai (r) = 0

i=0

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Second solution: r1 = r2

Consider the functions an (r), defined inductively using the equations a0 (r) := 1 and for n ≥ 1 I(n + r)an (r) +

n−1 X

bn−i (i + r)ai (r) +

i=0

n−1 X

cn−i ai (r) = 0

i=0

With these definitions, it follows that Lψ(r, x) = I(r)xr

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Second solution: r1 = r2

If r1 − r2 ∈ / Z then the second solution is given by X y2 (x) = xr2 an (r2 )xn n≥0

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Second solution: r1 = r2

If r1 − r2 ∈ / Z then the second solution is given by X y2 (x) = xr2 an (r2 )xn n≥0

Now let us consider the case when I has repeated roots

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Second solution: r1 = r2

If r1 − r2 ∈ / Z then the second solution is given by X y2 (x) = xr2 an (r2 )xn n≥0

Now let us consider the case when I has repeated roots Since I has repeated Qn roots r1 = r2 , it follows that, for every n ≥ 1, the polynomial i=1 I(i + r) does not vanish at r = r1 Consequently, it is clear that the an (r) are analytic in a small neighborhood around r = r1 = r2 .

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Second solution: r1 = r2 d on both sides of the Now let us apply the differential operator dr d r equation Lψ(r, x) = I(r)x . Clearly the operators L and dr commute with each other, and so we get

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Second solution: r1 = r2 d on both sides of the Now let us apply the differential operator dr d r equation Lψ(r, x) = I(r)x . Clearly the operators L and dr commute with each other, and so we get

d d Lψ(r, x) = L ψ(r, x) dr dr X  d a0n (r)xn+r + an (r)xn+r log x = I(r)xr =L dr n≥0

0

= I (r)xr + I(r)xr log x

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Second solution: r1 = r2 d on both sides of the Now let us apply the differential operator dr d r equation Lψ(r, x) = I(r)x . Clearly the operators L and dr commute with each other, and so we get

d d Lψ(r, x) = L ψ(r, x) dr dr X  d a0n (r)xn+r + an (r)xn+r log x = I(r)xr =L dr n≥0

0

= I (r)xr + I(r)xr log x Thus, if we plug in r = r1 = r2 in the above, then we get X  L a0n (r2 )xn+r2 + an (r2 )xn+r2 log x = 0 n≥0

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Second solution: r1 = r2

Theorem (Second solution: r1 = r2 ) A second solution to the differential equation is given by

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Second solution: r1 = r2

Theorem (Second solution: r1 = r2 ) A second solution to the differential equation is given by X X a0n (r2 )xn+r2 + an (r2 )xn+r2 log x n≥0

n≥0

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Second solution: r1 = r2 Example Consider the ODE x2 y 00 + 3xy 0 + (1 − 2x)y = 0

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Second solution: r1 = r2 Example Consider the ODE x2 y 00 + 3xy 0 + (1 − 2x)y = 0 This has a regular singularity at x = 0.

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Second solution: r1 = r2 Example Consider the ODE x2 y 00 + 3xy 0 + (1 − 2x)y = 0 This has a regular singularity at x = 0. I(r) = r(r − 1) + 3r + 1 = r2 + 2r + 1 has a repeated roots −1, −1.

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Second solution: r1 = r2 Example Consider the ODE x2 y 00 + 3xy 0 + (1 − 2x)y = 0 This has a regular singularity at x = 0. I(r) = r(r − 1) + 3r + 1 = r2 + 2r + 1 has a repeated roots −1, −1. Let us find the Frobenius solution directly by putting X y = xr an (r)xn a0 = 1 n≥0

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Second solution: r1 = r2 Example Consider the ODE x2 y 00 + 3xy 0 + (1 − 2x)y = 0 This has a regular singularity at x = 0. I(r) = r(r − 1) + 3r + 1 = r2 + 2r + 1 has a repeated roots −1, −1. Let us find the Frobenius solution directly by putting X y = xr an (r)xn a0 = 1 n≥0

y0 =

X

(n + r)an (r)xn+r−1

n≥0

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Second solution: r1 = r2 Example Consider the ODE x2 y 00 + 3xy 0 + (1 − 2x)y = 0 This has a regular singularity at x = 0. I(r) = r(r − 1) + 3r + 1 = r2 + 2r + 1 has a repeated roots −1, −1. Let us find the Frobenius solution directly by putting X y = xr an (r)xn a0 = 1 n≥0

y0 = y 00 =

X

(n + r)an (r)xn+r−1

n≥0 ∞ X

(n + r)(n + r − 1)an (r)xn+r−2

n≥0 19 / 33

Second solution: r1 = r2 Example (continues . . .) x2 y(x, r)00 + 3xy(x, r)0 + (1 − 2x)y(x, r) ∞ X = [(n + r)(n + r − 1) + 3(n + r) + 1] an (r)xn+r n=0



∞ X

2an (r)xn+r+1

n=0

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Second solution: r1 = r2 Example (continues . . .) x2 y(x, r)00 + 3xy(x, r)0 + (1 − 2x)y(x, r) ∞ X = [(n + r)(n + r − 1) + 3(n + r) + 1] an (r)xn+r n=0



∞ X

2an (r)xn+r+1

n=0

Recursion relations for n ≥ 1 are

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Second solution: r1 = r2 Example (continues . . .) x2 y(x, r)00 + 3xy(x, r)0 + (1 − 2x)y(x, r) ∞ X = [(n + r)(n + r − 1) + 3(n + r) + 1] an (r)xn+r n=0



∞ X

2an (r)xn+r+1

n=0

Recursion relations for n ≥ 1 are 2an−1 (r) an (r) = (n + r)(n + r − 1) + 3(n + r) + 1 2an−1 (r) = (n + r + 1)2 2n = a0 [(n + r + 1)(n + r) . . . (r + 2)]2 20 / 33

Second solution: r1 = r2 Example (continues . . .) Setting r = −1 (and a0 = 1) yields the fractional power series solution 1 X 2n n y1 (x) = x x (n!)2 n≥0

The power series converges on (0, ∞).

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Second solution: r1 = r2 Example (continues . . .) Setting r = −1 (and a0 = 1) yields the fractional power series solution 1 X 2n n y1 (x) = x x (n!)2 n≥0

The power series converges on (0, ∞). The second solution is y2 (x) = y1 (x) log x + x−1

X

a0n (−1)xn

n≥1

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Second solution: r1 = r2 Example (continues . . .) Setting r = −1 (and a0 = 1) yields the fractional power series solution 1 X 2n n y1 (x) = x x (n!)2 n≥0

The power series converges on (0, ∞). The second solution is y2 (x) = y1 (x) log x + x−1

X

a0n (−1)xn

n≥1

where an (r) =

2n [(n + r + 1)(n + r) . . . (r + 2)]2

a0n (r) =

−2.2n [(n + r + 1)(n + r) . . . (r + 2)]0 [(n + r + 1)(n + r) . . . (r + 2)]3 21 / 33

Second solution: r1 = r2 Example (continued)  = −2an (r)

1 1 1 + + ··· + n+r+1 n+r r+2



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Second solution: r1 = r2 Example (continued)  = −2an (r)

1 1 1 + + ··· + n+r+1 n+r r+2



Putting r = −1, we get 2n+1 Hn a0n (−1) = − (n!)2 where 1 1 Hn = 1 + + · · · + 2 n (These are the partial sums of the harmonic series.)

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Second solution: r1 = r2 Example (continued)  = −2an (r)

1 1 1 + + ··· + n+r+1 n+r r+2



Putting r = −1, we get 2n+1 Hn a0n (−1) = − (n!)2 where 1 1 Hn = 1 + + · · · + 2 n (These are the partial sums of the harmonic series.) So the second solution is y2 (x) = y1 (x)log x −

1 X 2n+1 Hn n x x (n!)2 n≥1

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Second solution: r1 = r2 Example (continued)  = −2an (r)

1 1 1 + + ··· + n+r+1 n+r r+2



Putting r = −1, we get 2n+1 Hn a0n (−1) = − (n!)2 where 1 1 Hn = 1 + + · · · + 2 n (These are the partial sums of the harmonic series.) So the second solution is y2 (x) = y1 (x)log x −

1 X 2n+1 Hn n x x (n!)2 n≥1

It is clear that this series converges on (0, ∞). 22 / 33

Second solution: 0 6= r1 − r2 ∈ Z Define N := r1 − r2 Note that each an (r) isQa rational function in r, in fact, the denominator is exactly ni=1 I(i + r).

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Second solution: 0 6= r1 − r2 ∈ Z Define N := r1 − r2 Note that each an (r) isQa rational function in r, in fact, the denominator is exactly ni=1 I(i + r). Q The polynomial ni=1 I(i + r) evaluated at r2 vanishes iff n ≥ N . For n ≥ N it vanishes to order exactly 1. Thus, if we define An (r) := an (r)(r − r2 ) then it is clear that for every n ≥ 0, the function An (r) is analytic in a neighborhood of r2 .

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Second solution: 0 6= r1 − r2 ∈ Z

In particular, An (r2 ) and A0n (r2 ) are well defined real numbers.

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Second solution: 0 6= r1 − r2 ∈ Z

In particular, An (r2 ) and A0n (r2 ) are well defined real numbers. Multiplying the equation Lψ(r, x) = I(r)xr with r − r2 we get (r − r2 )Lψ(r, x) = L(r − r2 )ψ(r, x) = (r − r2 )I(r)xr

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Second solution: 0 6= r1 − r2 ∈ Z

In particular, An (r2 ) and A0n (r2 ) are well defined real numbers. Multiplying the equation Lψ(r, x) = I(r)xr with r − r2 we get (r − r2 )Lψ(r, x) = L(r − r2 )ψ(r, x) = (r − r2 )I(r)xr

Note that (r − r2 )ψ(r, x) =

X

An (r)xn+r

n≥0

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Second solution: 0 6= r1 − r2 ∈ Z d on both sides of the Now let us apply the differential operator dr equation L(r − r2 )ψ(r, x) = (r − r2 )I(r)xr to get d d L(r − r2 )ψ(r, x) = L (r − r2 )ψ(r, x) dr dr d = (r − r2 )I(r)xr dr = I(r)xr + (r − r2 )I 0 (r)xr + (r − r2 )I(r)xr log x

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Second solution: 0 6= r1 − r2 ∈ Z d on both sides of the Now let us apply the differential operator dr equation L(r − r2 )ψ(r, x) = (r − r2 )I(r)xr to get d d L(r − r2 )ψ(r, x) = L (r − r2 )ψ(r, x) dr dr d = (r − r2 )I(r)xr dr = I(r)xr + (r − r2 )I 0 (r)xr + (r − r2 )I(r)xr log x

Thus we get   d X d X L An (r)xn+r = L An (r)xn+r dr dr n≥0 n≥0 X  =L A0n (r)xn+r + An (r)xn+r log x n≥0 r

= I(r)x + (r − r2 )I 0 (r)xr + (r − r2 )I(r)xr log x 25 / 33

Second solution: 0 6= r1 − r2 ∈ Z If we set r = r2 into the equation X  L A0n (r)xn+r + An (r)xn+r log x = I(r)xr + (r − r2 )I 0 (r)xr + n≥0

(r − r2 )I(r)xr log x

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Second solution: 0 6= r1 − r2 ∈ Z If we set r = r2 into the equation X  L A0n (r)xn+r + An (r)xn+r log x = I(r)xr + (r − r2 )I 0 (r)xr + n≥0

(r − r2 )I(r)xr log x we get the second solution X  L A0n (r2 )xn+r2 + An (r2 )xn+r2 log x = 0 n≥0

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Second solution: 0 6= r1 − r2 ∈ Z If we set r = r2 into the equation X  L A0n (r)xn+r + An (r)xn+r log x = I(r)xr + (r − r2 )I 0 (r)xr + n≥0

(r − r2 )I(r)xr log x we get the second solution X  L A0n (r2 )xn+r2 + An (r2 )xn+r2 log x = 0 n≥0

Theorem (Second solution: 0 6= r1 − r2 ∈ Z) A second solution to the differential equation is given by

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Second solution: 0 6= r1 − r2 ∈ Z If we set r = r2 into the equation X  L A0n (r)xn+r + An (r)xn+r log x = I(r)xr + (r − r2 )I 0 (r)xr + n≥0

(r − r2 )I(r)xr log x we get the second solution X  L A0n (r2 )xn+r2 + An (r2 )xn+r2 log x = 0 n≥0

Theorem (Second solution: 0 6= r1 − r2 ∈ Z) A second solution to the differential equation is given by X X A0n (r2 )xn+r2 + An (r2 )xn+r2 log x n≥0

n≥0 26 / 33

Second solution: 0 6= r1 − r2 ∈ Z Example Consider the ODE

xy 00 − (4 + x)y 0 + 2y = 0

(∗)

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Second solution: 0 6= r1 − r2 ∈ Z Example Consider the ODE

xy 00 − (4 + x)y 0 + 2y = 0

(∗)

Multiplying (∗) with x, we get x = 0 is a regular singular point. I(r) = r(r − 1) − 4r + 0 = r(r − 5) = 0 with the roots differing by a positive integer.

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Second solution: 0 6= r1 − r2 ∈ Z Example Consider the ODE

xy 00 − (4 + x)y 0 + 2y = 0

(∗)

Multiplying (∗) with x, we get x = 0 is a regular singular point. I(r) = r(r − 1) − 4r + 0 = r(r − 5) = 0 with the roots differing by a positive integer. ∞ X r Put y(x, r) = x an (r)xn , a0 (r) = 1, into the ODE to get n=0

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Second solution: 0 6= r1 − r2 ∈ Z Example Consider the ODE

xy 00 − (4 + x)y 0 + 2y = 0

(∗)

Multiplying (∗) with x, we get x = 0 is a regular singular point. I(r) = r(r − 1) − 4r + 0 = r(r − 5) = 0 with the roots differing by a positive integer. ∞ X r Put y(x, r) = x an (r)xn , a0 (r) = 1, into the ODE to get n=0

x

X

(n + r)(n + r − 1)an (r)xn+r−2

n≥0

−(4 + x)

X

(n + r)an (r)xn+r−1 + 2

n≥0

X

an (r)xn+r = 0

n≥0

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Second solution: 0 6= r1 − r2 ∈ Z Example Consider the ODE

xy 00 − (4 + x)y 0 + 2y = 0

(∗)

Multiplying (∗) with x, we get x = 0 is a regular singular point. I(r) = r(r − 1) − 4r + 0 = r(r − 5) = 0 with the roots differing by a positive integer. ∞ X r Put y(x, r) = x an (r)xn , a0 (r) = 1, into the ODE to get n=0

x

X

(n + r)(n + r − 1)an (r)xn+r−2

n≥0

−(4 + x)

X

(n + r)an (r)xn+r−1 + 2

n≥0

the coefficient of

X

an (r)xn+r = 0

n≥0

xn+r−1

for n ≥ 1 gives

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) (n + r)(n + r − 1)an (r) − 4(n + r)an (r) − (n + r − 1)an−1 (r) +2an−1 (r) = 0

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) (n + r)(n + r − 1)an (r) − 4(n + r)an (r) − (n + r − 1)an−1 (r) +2an−1 (r) = 0 For n ≥ 1,

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) (n + r)(n + r − 1)an (r) − 4(n + r)an (r) − (n + r − 1)an−1 (r) +2an−1 (r) = 0 For n ≥ 1, (n + r)(n + r − 5)an = (n + r − 3)an−1

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) (n + r)(n + r − 1)an (r) − 4(n + r)an (r) − (n + r − 1)an−1 (r) +2an−1 (r) = 0 For n ≥ 1, (n + r)(n + r − 5)an = (n + r − 3)an−1 (n + r − 3) an (r) = an−1 (n + r)(n + r − 5) (n + r − 3) . . . (r − 2) a0 = (n + r) . . . (1 + r)(n + r − 5) . . . (r − 4)

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) (n + r)(n + r − 1)an (r) − 4(n + r)an (r) − (n + r − 1)an−1 (r) +2an−1 (r) = 0 For n ≥ 1, (n + r)(n + r − 5)an = (n + r − 3)an−1 (n + r − 3) an (r) = an−1 (n + r)(n + r − 5) (n + r − 3) . . . (r − 2) a0 = (n + r) . . . (1 + r)(n + r − 5) . . . (r − 4) For the first solution, set r = r1 = 5 (and a0 = 1), we get

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) (n + r)(n + r − 1)an (r) − 4(n + r)an (r) − (n + r − 1)an−1 (r) +2an−1 (r) = 0 For n ≥ 1, (n + r)(n + r − 5)an = (n + r − 3)an−1 (n + r − 3) an (r) = an−1 (n + r)(n + r − 5) (n + r − 3) . . . (r − 2) a0 = (n + r) . . . (1 + r)(n + r − 5) . . . (r − 4) For the first solution, set r = r1 = 5 (and a0 = 1), we get (n + 2) . . . (3) an (5) = (n + 5) . . . 6(n) . . . 1

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) (n + r)(n + r − 1)an (r) − 4(n + r)an (r) − (n + r − 1)an−1 (r) +2an−1 (r) = 0 For n ≥ 1, (n + r)(n + r − 5)an = (n + r − 3)an−1 (n + r − 3) an (r) = an−1 (n + r)(n + r − 5) (n + r − 3) . . . (r − 2) a0 = (n + r) . . . (1 + r)(n + r − 5) . . . (r − 4) For the first solution, set r = r1 = 5 (and a0 = 1), we get (n + 2) . . . (3) an (5) = (n + 5) . . . 6(n) . . . 1 (n + 2)!/2 = (n!)(n + 5)!/5! 28 / 33

Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) =

60 n!(n + 5)(n + 4)(n + 3)

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) = Thus y1 (x) =

60 n!(n + 5)(n + 4)(n + 3) X 60 n≥0

n!(n + 5)(n + 4)(n + 3)

xn+5

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) = Thus y1 (x) =

60 n!(n + 5)(n + 4)(n + 3) X 60 n≥0

n!(n + 5)(n + 4)(n + 3)

xn+5

Recall N = r1 − r2 = 5 − 0 is integer, so the second solution is X X y2 (x) = A0n (r2 )xn+r2 + An (r2 )xn+r2 log x n≥0

n≥0

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) = Thus y1 (x) =

60 n!(n + 5)(n + 4)(n + 3) X 60 n≥0

n!(n + 5)(n + 4)(n + 3)

xn+5

Recall N = r1 − r2 = 5 − 0 is integer, so the second solution is X X y2 (x) = A0n (r2 )xn+r2 + An (r2 )xn+r2 log x n≥0

n≥0

where, for n ≥ 0 An (r) = (r − r2 )an (r)

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Second solution: 0 6= r1 − r2 ∈ Z Example (continues . . .) = Thus y1 (x) =

60 n!(n + 5)(n + 4)(n + 3) X 60 n≥0

n!(n + 5)(n + 4)(n + 3)

xn+5

Recall N = r1 − r2 = 5 − 0 is integer, so the second solution is X X y2 (x) = A0n (r2 )xn+r2 + An (r2 )xn+r2 log x n≥0

n≥0

where, for n ≥ 0 An (r) = (r − r2 )an (r) Since r2 = 0, the above becomes An (r) = ran (r) 29 / 33

Second solution: 0 6= r1 − r2 ∈ Z

Example In this example, we can easily check that none of the an (r) have a singularity at r = 0.

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Second solution: 0 6= r1 − r2 ∈ Z

Example In this example, we can easily check that none of the an (r) have a singularity at r = 0. Thus, An (0) = 0 for all n ≥ 0; and A0n (0) = an (0) for all n ≥ 0.

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Second solution: 0 6= r1 − r2 ∈ Z

Example In this example, we can easily check that none of the an (r) have a singularity at r = 0. Thus, An (0) = 0 for all n ≥ 0; and A0n (0) = an (0) for all n ≥ 0. a1 (0) = 21 ; a2 (0) =

1 12 ;

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Second solution: 0 6= r1 − r2 ∈ Z

Example In this example, we can easily check that none of the an (r) have a singularity at r = 0. Thus, An (0) = 0 for all n ≥ 0; and A0n (0) = an (0) for all n ≥ 0. a1 (0) = 21 ; a2 (0) =

1 12 ;

It is easily checked that for n ≥ 3 an (r) =

(n + r − 3)(n + r − 4) n!12

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Second solution: 0 6= r1 − r2 ∈ Z

Example In this example, we can easily check that none of the an (r) have a singularity at r = 0. Thus, An (0) = 0 for all n ≥ 0; and A0n (0) = an (0) for all n ≥ 0. a1 (0) = 21 ; a2 (0) =

1 12 ;

It is easily checked that for n ≥ 3 an (r) =

(n + r − 3)(n + r − 4) n!12

Thus, a3 (0) = a4 (0) = 0.

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Second solution: 0 6= r1 − r2 ∈ Z Example Therefore a second solution is x x2 X (n − 3)(n − 4) n y2 (x) = 1 + + + x 2 12 n!12 n≥5

=1+

x2

x + + 2 12

X k≥0

1 xk+5 k!(k + 5)(k + 4)(k + 3)12

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Second solution: 0 6= r1 − r2 ∈ Z Example Therefore a second solution is x x2 X (n − 3)(n − 4) n y2 (x) = 1 + + + x 2 12 n!12 n≥5

=1+

x2

x + + 2 12

Since X k≥0

X k≥0

1 xk+5 k!(k + 5)(k + 4)(k + 3)12

1 xk+5 k!(k + 5)(k + 4)(k + 3)12

is a multiple of y1 (x), we get that a second solution is y2 (x) = 1 +

x x2 + . 2 12 31 / 33

Summary

While solving an ODE around a regular singular point by the Frobenius method, the cases encountered are

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Summary

While solving an ODE around a regular singular point by the Frobenius method, the cases encountered are roots not differing by an integer

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Summary

While solving an ODE around a regular singular point by the Frobenius method, the cases encountered are roots not differing by an integer repeated roots

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Summary

While solving an ODE around a regular singular point by the Frobenius method, the cases encountered are roots not differing by an integer repeated roots roots differing by a positive integer

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Summary

While solving an ODE around a regular singular point by the Frobenius method, the cases encountered are roots not differing by an integer repeated roots roots differing by a positive integer The larger root always yields a fractional power series solution.

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Summary

While solving an ODE around a regular singular point by the Frobenius method, the cases encountered are roots not differing by an integer repeated roots roots differing by a positive integer The larger root always yields a fractional power series solution. In the first case, the smaller root also yields a fractional power series solution.

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Summary

While solving an ODE around a regular singular point by the Frobenius method, the cases encountered are roots not differing by an integer repeated roots roots differing by a positive integer The larger root always yields a fractional power series solution. In the first case, the smaller root also yields a fractional power series solution. In the second and third cases, the second solution may involve a log term.

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Let us write down some classical ODE’s.

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Let us write down some classical ODE’s. (Euler equation)

αx2 y 00 + βxy 0 + γy = 0

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Let us write down some classical ODE’s. (Euler equation)

αx2 y 00 + βxy 0 + γy = 0

(Bessel equation) x2 y 00 + xy 0 + (x2 − ν 2 )y = 0. We will next look at this case more closely.

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Let us write down some classical ODE’s. (Euler equation)

αx2 y 00 + βxy 0 + γy = 0

(Bessel equation) x2 y 00 + xy 0 + (x2 − ν 2 )y = 0. We will next look at this case more closely. (Laguerre equation)

xy 00 + (1 − x)y 0 + λy = 0

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