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

Question 71

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 explicit formula for the coefficients ? A)    a_{2 n}=\frac{(-1) ^{n} 2^{3 n}}{n !} a_{0, n} \geq 1   B)    a_{2 n}=\frac{(-1) ^{n} 2^{5 n}}{n !} a_{0}, n \geq 1   C)    a_{2 n}=\frac{(-1) ^{n} 2^{3 n}}{(n !) ^{2}} a_{0}, n \geq 1   D)    a_{2 n}=\frac{(-1) ^{n} 2^{5 n}}{(n !) ^{2}} a_{0}, n \geq 1 .
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 explicit formula for the coefficients ? A)    a_{2 n}=\frac{(-1) ^{n} 2^{3 n}}{n !} a_{0, n} \geq 1   B)    a_{2 n}=\frac{(-1) ^{n} 2^{5 n}}{n !} a_{0}, n \geq 1   C)    a_{2 n}=\frac{(-1) ^{n} 2^{3 n}}{(n !) ^{2}} a_{0}, n \geq 1   D)    a_{2 n}=\frac{(-1) ^{n} 2^{5 n}}{(n !) ^{2}} a_{0}, n \geq 1 .
Assume a0 \neq 0.
Which of these is the explicit formula for the coefficients ?


A) a2n=(1) n23nn!a0,n1 a_{2 n}=\frac{(-1) ^{n} 2^{3 n}}{n !} a_{0, n} \geq 1
B) a2n=(1) n25nn!a0,n1 a_{2 n}=\frac{(-1) ^{n} 2^{5 n}}{n !} a_{0}, n \geq 1
C) a2n=(1) n23n(n!) 2a0,n1 a_{2 n}=\frac{(-1) ^{n} 2^{3 n}}{(n !) ^{2}} a_{0}, n \geq 1
D) a2n=(1) n25n(n!) 2a0,n1 a_{2 n}=\frac{(-1) ^{n} 2^{5 n}}{(n !) ^{2}} a_{0}, n \geq 1

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