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Method of Frobenius Example First Solution Second Solution (Fails)

Method of Frobenius – A ProblematicCase

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?

1. The method of Frobenius works for differential equationsof the form y′′+P(x)y′+Q(x)y = 0 in which P or Q is notanalytic at the point of expansion x0.

2. But P and Q cannot be arbitrary: (x− x0)P(x) and(x− x0)2Q(x) must be analytic at x0.

3. Instead of a series solution y =∞

∑n=0

cn(x− x0)n, we obtain a

solution of the form y =∞

∑n=0

cn(x− x0)n+r.

4. As for series solutions, we substitute the series and itsderivatives into the equation to obtain an equation for r anda set of equations for the cn.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?1. The method of Frobenius works for differential equations

of the form y′′+P(x)y′+Q(x)y = 0 in which P or Q is notanalytic at the point of expansion x0.

2. But P and Q cannot be arbitrary: (x− x0)P(x) and(x− x0)2Q(x) must be analytic at x0.

3. Instead of a series solution y =∞

∑n=0

cn(x− x0)n, we obtain a

solution of the form y =∞

∑n=0

cn(x− x0)n+r.

4. As for series solutions, we substitute the series and itsderivatives into the equation to obtain an equation for r anda set of equations for the cn.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?1. The method of Frobenius works for differential equations

of the form y′′+P(x)y′+Q(x)y = 0 in which P or Q is notanalytic at the point of expansion x0.

2. But P and Q cannot be arbitrary: (x− x0)P(x) and(x− x0)2Q(x) must be analytic at x0.

3. Instead of a series solution y =∞

∑n=0

cn(x− x0)n, we obtain a

solution of the form y =∞

∑n=0

cn(x− x0)n+r.

4. As for series solutions, we substitute the series and itsderivatives into the equation to obtain an equation for r anda set of equations for the cn.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?1. The method of Frobenius works for differential equations

of the form y′′+P(x)y′+Q(x)y = 0 in which P or Q is notanalytic at the point of expansion x0.

2. But P and Q cannot be arbitrary: (x− x0)P(x) and(x− x0)2Q(x) must be analytic at x0.

3. Instead of a series solution y =∞

∑n=0

cn(x− x0)n, we obtain a

solution of the form y =∞

∑n=0

cn(x− x0)n+r.

4. As for series solutions, we substitute the series and itsderivatives into the equation to obtain an equation for r anda set of equations for the cn.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?1. The method of Frobenius works for differential equations

of the form y′′+P(x)y′+Q(x)y = 0 in which P or Q is notanalytic at the point of expansion x0.

2. But P and Q cannot be arbitrary: (x− x0)P(x) and(x− x0)2Q(x) must be analytic at x0.

3. Instead of a series solution y =∞

∑n=0

cn(x− x0)n, we obtain a

solution of the form y =∞

∑n=0

cn(x− x0)n+r.

4. As for series solutions, we substitute the series and itsderivatives into the equation to obtain an equation for r anda set of equations for the cn.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?

5. These equations will allow us to compute r and the cn.6. For each value of r (typically there are two), we can

compute the solution just like for series.7. The method of Frobenius is guaranteed to produce one

solution. But when the two values for r differ by an integer,it may not produce two linearly independent solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?5. These equations will allow us to compute r and the cn.

6. For each value of r (typically there are two), we cancompute the solution just like for series.

7. The method of Frobenius is guaranteed to produce onesolution. But when the two values for r differ by an integer,it may not produce two linearly independent solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?5. These equations will allow us to compute r and the cn.6. For each value of r (typically there are two), we can

compute the solution just like for series.

7. The method of Frobenius is guaranteed to produce onesolution. But when the two values for r differ by an integer,it may not produce two linearly independent solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?5. These equations will allow us to compute r and the cn.6. For each value of r (typically there are two), we can

compute the solution just like for series.7. The method of Frobenius is guaranteed to produce one

solution.

But when the two values for r differ by an integer,it may not produce two linearly independent solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

What is the Method of Frobenius?5. These equations will allow us to compute r and the cn.6. For each value of r (typically there are two), we can

compute the solution just like for series.7. The method of Frobenius is guaranteed to produce one

solution. But when the two values for r differ by an integer,it may not produce two linearly independent solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2

+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1

+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r

= 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r

+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r

+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2

−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r

= 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r

+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r

+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r

−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r

= 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0

(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr

+((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Frobenius Solution for x2y′′+ xy′+(x2−4

)y = 0

x2y′′+ xy′+(x2−4

)y = 0

x2∞

∑n=0

cn(n+r)(n+r−1)xn+r−2+x∞

∑n=0

cn(n+r)xn+r−1+(x2−4

) ∞

∑n=0

cnxn+r = 0

∞

∑n=0

(n+r)(n+r−1)cnxn+r+∞

∑n=0

(n+ r)cnxn+r+∞

∑n=0

cnxn+r+2−∞

∑n=0

4cnxn+r = 0

∞

∑k=0

(k+r)(k+r−1)ckxk+r+∞

∑k=0

(k+r)ckxk+r+∞

∑k=2

ck−2xk+r−∞

∑k=0

4ckxk+r = 0(r(r−1)c0 + rc0−4c0

)xr +

((r +1)rc1 +(r +1)c1−4c1

)xr+1

+∞

∑k=2

[(k + r)(k + r−1)ck +(k + r)ck + ck−2−4ck

]xk+r = 0

(r2−4

)c0xr+

((r+1)2−4

)c1xr+1+

∞

∑k=2

[((k+r)2−4

)ck+ck−2

]xk+r = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Indicial Roots, Recurrence Relation

r2−4 = 0, r1,2 =±2

For r = 2 we obtain

c1

((2+1)2−4

)= 0, c1 = 0

Recurrence relation for k ≥ 2:((k +2)2−4

)ck + ck−2 = 0

ck =− ck−2

(k +2)2−4= − ck−2

k(k +4)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Indicial Roots, Recurrence Relation

r2−4 = 0

, r1,2 =±2

For r = 2 we obtain

c1

((2+1)2−4

)= 0, c1 = 0

Recurrence relation for k ≥ 2:((k +2)2−4

)ck + ck−2 = 0

ck =− ck−2

(k +2)2−4= − ck−2

k(k +4)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Indicial Roots, Recurrence Relation

r2−4 = 0, r1,2 =±2

For r = 2 we obtain

c1

((2+1)2−4

)= 0, c1 = 0

Recurrence relation for k ≥ 2:((k +2)2−4

)ck + ck−2 = 0

ck =− ck−2

(k +2)2−4= − ck−2

k(k +4)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Indicial Roots, Recurrence Relation

r2−4 = 0, r1,2 =±2

For r = 2 we obtain

c1

((2+1)2−4

)= 0

, c1 = 0

Recurrence relation for k ≥ 2:((k +2)2−4

)ck + ck−2 = 0

ck =− ck−2

(k +2)2−4= − ck−2

k(k +4)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Indicial Roots, Recurrence Relation

r2−4 = 0, r1,2 =±2

For r = 2 we obtain

c1

((2+1)2−4

)= 0, c1 = 0

Recurrence relation for k ≥ 2:((k +2)2−4

)ck + ck−2 = 0

ck =− ck−2

(k +2)2−4= − ck−2

k(k +4)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Indicial Roots, Recurrence Relation

r2−4 = 0, r1,2 =±2

For r = 2 we obtain

c1

((2+1)2−4

)= 0, c1 = 0

Recurrence relation for k ≥ 2:((k +2)2−4

)ck + ck−2 = 0

ck =− ck−2

(k +2)2−4= − ck−2

k(k +4)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Indicial Roots, Recurrence Relation

r2−4 = 0, r1,2 =±2

For r = 2 we obtain

c1

((2+1)2−4

)= 0, c1 = 0

Recurrence relation for k ≥ 2:((k +2)2−4

)ck + ck−2 = 0

ck =− ck−2

(k +2)2−4

= − ck−2

k(k +4)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Indicial Roots, Recurrence Relation

r2−4 = 0, r1,2 =±2

For r = 2 we obtain

c1

((2+1)2−4

)= 0, c1 = 0

Recurrence relation for k ≥ 2:((k +2)2−4

)ck + ck−2 = 0

ck =− ck−2

(k +2)2−4= − ck−2

k(k +4)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Even Numbered Terms

c2n = − c2n−2

2n(2n+4)

= −c2(n−1)

4n(n+2)

= (−1)2 c2(n−2)

42n(n−1)(n+2)(n+1)

= (−1)3 c2(n−3)

43n(n−1)(n−2)(n+2)(n+1)n...

= (−1)n c0

4nn(n−1) · · ·2 ·1 · (n+2)(n+1) · · ·4 ·3

= (−1)n 24nn!(n+2)!

c0

y1 =∞

∑n=0

(−1)n 24nn!(n+2)!

x2n+2

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Even Numbered Termsc2n = − c2n−2

2n(2n+4)

= −c2(n−1)

4n(n+2)

= (−1)2 c2(n−2)

42n(n−1)(n+2)(n+1)

= (−1)3 c2(n−3)

43n(n−1)(n−2)(n+2)(n+1)n...

= (−1)n c0

4nn(n−1) · · ·2 ·1 · (n+2)(n+1) · · ·4 ·3

= (−1)n 24nn!(n+2)!

c0

y1 =∞

∑n=0

(−1)n 24nn!(n+2)!

x2n+2

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

Even Numbered Termsc2n = − c2n−2

2n(2n+4)

= −c2(n−1)

4n(n+2)

= (−1)2 c2(n−2)

42n(n−1)(n+2)(n+1)

= (−1)3 c2(n−3)

43n(n−1)(n−2)(n+2)(n+1)n...

= (−1)n c0

4nn(n−1) · · ·2 ·1 · (n+2)(n+1) · · ·4 ·3

= (−1)n 24nn!(n+2)!

c0

y1 =∞

∑n=0

(−1)n 24nn!(n+2)!

x2n+2

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Even Numbered Termsc2n = − c2n−2

2n(2n+4)

= −c2(n−1)

4n(n+2)

= (−1)2 c2(n−2)

42n(n−1)(n+2)(n+1)

= (−1)3 c2(n−3)

43n(n−1)(n−2)(n+2)(n+1)n...

= (−1)n c0

4nn(n−1) · · ·2 ·1 · (n+2)(n+1) · · ·4 ·3

= (−1)n 24nn!(n+2)!

c0

y1 =∞

∑n=0

(−1)n 24nn!(n+2)!

x2n+2

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Even Numbered Termsc2n = − c2n−2

2n(2n+4)

= −c2(n−1)

4n(n+2)

= (−1)2 c2(n−2)

42n(n−1)(n+2)(n+1)

= (−1)3 c2(n−3)

43n(n−1)(n−2)(n+2)(n+1)n

...

= (−1)n c0

4nn(n−1) · · ·2 ·1 · (n+2)(n+1) · · ·4 ·3

= (−1)n 24nn!(n+2)!

c0

y1 =∞

∑n=0

(−1)n 24nn!(n+2)!

x2n+2

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Even Numbered Termsc2n = − c2n−2

2n(2n+4)

= −c2(n−1)

4n(n+2)

= (−1)2 c2(n−2)

42n(n−1)(n+2)(n+1)

= (−1)3 c2(n−3)

43n(n−1)(n−2)(n+2)(n+1)n...

= (−1)n c0

4nn(n−1) · · ·2 ·1 · (n+2)(n+1) · · ·4 ·3

= (−1)n 24nn!(n+2)!

c0

y1 =∞

∑n=0

(−1)n 24nn!(n+2)!

x2n+2

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Even Numbered Termsc2n = − c2n−2

2n(2n+4)

= −c2(n−1)

4n(n+2)

= (−1)2 c2(n−2)

42n(n−1)(n+2)(n+1)

= (−1)3 c2(n−3)

43n(n−1)(n−2)(n+2)(n+1)n...

= (−1)n c0

4nn(n−1) · · ·2 ·1 · (n+2)(n+1) · · ·4 ·3

= (−1)n 24nn!(n+2)!

c0

y1 =∞

∑n=0

(−1)n 24nn!(n+2)!

x2n+2

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Even Numbered Termsc2n = − c2n−2

2n(2n+4)

= −c2(n−1)

4n(n+2)

= (−1)2 c2(n−2)

42n(n−1)(n+2)(n+1)

= (−1)3 c2(n−3)

43n(n−1)(n−2)(n+2)(n+1)n...

= (−1)n c0

4nn(n−1) · · ·2 ·1 · (n+2)(n+1) · · ·4 ·3

= (−1)n 24nn!(n+2)!

c0

y1 =∞

∑n=0

(−1)n 24nn!(n+2)!

x2n+2

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

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Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0

, c1 = 0((k−2)2−4

)ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0

((k−2)2−4

)ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4

=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)

=14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)

= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4

·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

r=-2

c0 := 1

c1

((−2+1)2−4

)= 0, c1 = 0(

(k−2)2−4)

ck + ck−2 = 0

ck = − ck−2

(k−2)2−4=− ck−2

k(k−4)

c2 = − c0

2(−2)=

14

c3 = − c1

3(−1)= 0

c4 = − c2

4 ·0???

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Trying y2 =1x2 +

14

x2y′′2 + xy′2 +(

x2−4)

y2?= 0

x2(

6x4

)+ x

(− 2

x3

)+

(x2−4

)(1x2 +

14

)?= 0

6x2 −

2x2 +1+

14

x2− 4x2 −1 ?= 0

14

x2 ?= 0

NO!

(Reduction of Order can help.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Trying y2 =1x2 +

14

x2y′′2 + xy′2 +(

x2−4)

y2?= 0

x2(

6x4

)+ x

(− 2

x3

)+

(x2−4

)(1x2 +

14

)?= 0

6x2 −

2x2 +1+

14

x2− 4x2 −1 ?= 0

14

x2 ?= 0

NO!

(Reduction of Order can help.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Trying y2 =1x2 +

14

x2y′′2 + xy′2 +(

x2−4)

y2?= 0

x2(

6x4

)+ x

(− 2

x3

)+

(x2−4

)(1x2 +

14

)?= 0

6x2 −

2x2 +1+

14

x2− 4x2 −1 ?= 0

14

x2 ?= 0

NO!

(Reduction of Order can help.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Trying y2 =1x2 +

14

x2y′′2 + xy′2 +(

x2−4)

y2?= 0

x2(

6x4

)+ x

(− 2

x3

)+

(x2−4

)(1x2 +

14

)?= 0

6x2 −

2x2 +1+

14

x2− 4x2 −1 ?= 0

14

x2 ?= 0

NO!

(Reduction of Order can help.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Trying y2 =1x2 +

14

x2y′′2 + xy′2 +(

x2−4)

y2?= 0

x2(

6x4

)+ x

(− 2

x3

)+

(x2−4

)(1x2 +

14

)?= 0

6x2 −

2x2 +1+

14

x2− 4x2 −1 ?= 0

14

x2 ?= 0

NO!

(Reduction of Order can help.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Trying y2 =1x2 +

14

x2y′′2 + xy′2 +(

x2−4)

y2?= 0

x2(

6x4

)+ x

(− 2

x3

)+

(x2−4

)(1x2 +

14

)?= 0

6x2 −

2x2 +1+

14

x2− 4x2 −1 ?= 0

14

x2 ?= 0

NO!

(Reduction of Order can help.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case

logo1

Method of Frobenius Example First Solution Second Solution (Fails)

Trying y2 =1x2 +

14

x2y′′2 + xy′2 +(

x2−4)

y2?= 0

x2(

6x4

)+ x

(− 2

x3

)+

(x2−4

)(1x2 +

14

)?= 0

6x2 −

2x2 +1+

14

x2− 4x2 −1 ?= 0

14

x2 ?= 0

NO!

(Reduction of Order can help.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Method of Frobenius – A Problematic Case