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Consider the Second-Order Differential Equation \neq 0
Which of These Is the Recurrence Relation for the of Frobenius

Question 27

Multiple Choice

Consider the second-order differential equation  Consider the second-order differential equation   . Suppose the method of Frobenius is used to determine a power series solution of this equation. The indicial equation has r = 0 as a double root. So, one of the solutions can be represented as the power series   . Assume a<sub>0</sub>  \neq  0. Which of these is the recurrence relation for the coefficients? A)    a_{1}=0, a_{n}=-\frac{64 \cdot a_{n-2}}{2^{n}}, n \geq 2   B)    a_{1}=1, a_{n}=\frac{64 \cdot a_{n-2}}{2^{n}}, n \geq 2   C)    a_{1}=1, a_{n}=\frac{64 \cdot a_{n-2}}{n^{2}}, n \geq 2   D)    a_{1}=0, a_{n}=-\frac{64 \cdot a_{n-2}}{n^{2}}, n \geq 2   E)    a_{1}=0, a_{n}=-\frac{64 \cdot a_{n-2}}{n}, n \geq 2 .
Suppose the method of Frobenius is used to determine a power series solution of this equation. The indicial equation has r = 0 as a double root. So, one of the solutions can be represented as the power series  Consider the second-order differential equation   . Suppose the method of Frobenius is used to determine a power series solution of this equation. The indicial equation has r = 0 as a double root. So, one of the solutions can be represented as the power series   . Assume a<sub>0</sub>  \neq  0. Which of these is the recurrence relation for the coefficients? A)    a_{1}=0, a_{n}=-\frac{64 \cdot a_{n-2}}{2^{n}}, n \geq 2   B)    a_{1}=1, a_{n}=\frac{64 \cdot a_{n-2}}{2^{n}}, n \geq 2   C)    a_{1}=1, a_{n}=\frac{64 \cdot a_{n-2}}{n^{2}}, n \geq 2   D)    a_{1}=0, a_{n}=-\frac{64 \cdot a_{n-2}}{n^{2}}, n \geq 2   E)    a_{1}=0, a_{n}=-\frac{64 \cdot a_{n-2}}{n}, n \geq 2 .
Assume a0 \neq 0.
Which of these is the recurrence relation for the coefficients?


A) a1=0,an=64an22n,n2 a_{1}=0, a_{n}=-\frac{64 \cdot a_{n-2}}{2^{n}}, n \geq 2
B) a1=1,an=64an22n,n2 a_{1}=1, a_{n}=\frac{64 \cdot a_{n-2}}{2^{n}}, n \geq 2
C) a1=1,an=64an2n2,n2 a_{1}=1, a_{n}=\frac{64 \cdot a_{n-2}}{n^{2}}, n \geq 2
D) a1=0,an=64an2n2,n2 a_{1}=0, a_{n}=-\frac{64 \cdot a_{n-2}}{n^{2}}, n \geq 2
E) a1=0,an=64an2n,n2 a_{1}=0, a_{n}=-\frac{64 \cdot a_{n-2}}{n}, n \geq 2

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