Post on 25-Jan-2022
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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
logo1
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
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
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?
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
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
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
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)
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
logo1
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
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
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)
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)
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