Solved

Consider the Second-Order Differential Equation \neq 0
Assuming That A0 = 1, One Solution of the of Frobenius

Question 51

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. Assuming that a<sub>0</sub> = 1, one solution of the given differential equation is   Differentiating as needed, which of these relationships is correct? A)    2 y^{\prime}(x) -a_{1}^{*}-128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}-128 a_{n-2}^{*}\right)  x^{n-1}=0   B)    2 y^{\prime}(x) +a_{1}^{*}+128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}+128 a_{n-2}^{*}\right)  x^{n-1}=0   C)    2 y^{\prime}(x) +a_{1}^{*}-128 a_{2}^{*} x-\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}+128 a_{n-2}^{*}\right)  x^{n-1}=0   D)    2 y^{\prime}(x) +a_{1}^{*}+128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}-128 a_{n-2}^{*}\right)  x^{n-1}=0 .
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. Assuming that a<sub>0</sub> = 1, one solution of the given differential equation is   Differentiating as needed, which of these relationships is correct? A)    2 y^{\prime}(x) -a_{1}^{*}-128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}-128 a_{n-2}^{*}\right)  x^{n-1}=0   B)    2 y^{\prime}(x) +a_{1}^{*}+128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}+128 a_{n-2}^{*}\right)  x^{n-1}=0   C)    2 y^{\prime}(x) +a_{1}^{*}-128 a_{2}^{*} x-\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}+128 a_{n-2}^{*}\right)  x^{n-1}=0   D)    2 y^{\prime}(x) +a_{1}^{*}+128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}-128 a_{n-2}^{*}\right)  x^{n-1}=0 Assume a0 \neq 0.
Assuming that a0 = 1, one solution of the given differential equation is  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. Assuming that a<sub>0</sub> = 1, one solution of the given differential equation is   Differentiating as needed, which of these relationships is correct? A)    2 y^{\prime}(x) -a_{1}^{*}-128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}-128 a_{n-2}^{*}\right)  x^{n-1}=0   B)    2 y^{\prime}(x) +a_{1}^{*}+128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}+128 a_{n-2}^{*}\right)  x^{n-1}=0   C)    2 y^{\prime}(x) +a_{1}^{*}-128 a_{2}^{*} x-\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}+128 a_{n-2}^{*}\right)  x^{n-1}=0   D)    2 y^{\prime}(x) +a_{1}^{*}+128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}-128 a_{n-2}^{*}\right)  x^{n-1}=0
Differentiating as needed, which of these relationships is correct?


A) 2y(x) a1128a2x+n=3(n2an128an2) xn1=0 2 y^{\prime}(x) -a_{1}^{*}-128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}-128 a_{n-2}^{*}\right) x^{n-1}=0
B) 2y(x) +a1+128a2x+n=3(n2an+128an2) xn1=0 2 y^{\prime}(x) +a_{1}^{*}+128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}+128 a_{n-2}^{*}\right) x^{n-1}=0
C) 2y(x) +a1128a2xn=3(n2an+128an2) xn1=0 2 y^{\prime}(x) +a_{1}^{*}-128 a_{2}^{*} x-\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}+128 a_{n-2}^{*}\right) x^{n-1}=0
D) 2y(x) +a1+128a2x+n=3(n2an128an2) xn1=0 2 y^{\prime}(x) +a_{1}^{*}+128 a_{2}^{*} x+\sum_{n=3}^{\infty}\left(n^{2} a_{n}^{*}-128 a_{n-2}^{*}\right) x^{n-1}=0

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions