Consider the second-order differential equation .
Suppose the method of Frobenius is used to determine the general solution of this differential equation.
Which of the following is the form of a pair of linearly independent solution of this differential
A) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $
B) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n+1}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $
C) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n}, y_{2}(x)=x^{\frac{3}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $
D) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n+1}, y_{2}(x)=x^{\frac{3}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $
E) $ y_{1}(x)=\ln (x) \sum_{n=0}^{\infty} a_{n} x^{n+1}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $

Question

Consider the second-order differential equation  .
Suppose the method of Frobenius is used to determine the general solution of this differential equation.
Which of the following is the form of a pair of linearly independent solution of this differential
A) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $ 
B) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n+1}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $ 
C) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n}, y_{2}(x)=x^{\frac{3}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $ 
D) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n+1}, y_{2}(x)=x^{\frac{3}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $ 
E) $ y_{1}(x)=\ln (x) \sum_{n=0}^{\infty} a_{n} x^{n+1}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $

Quizplus · Accepted Answer

The Answer of Consider the second-order differential equation  .
Suppose the method of Frobenius is used to determine the general solution of this differential equation.
Which of the following is the form of a pair of linearly independent solution of this differential
A) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $ 
B) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n+1}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $ 
C) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n}, y_{2}(x)=x^{\frac{3}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $ 
D) $ y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n+1}, y_{2}(x)=x^{\frac{3}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $ 
E) $ y_{1}(x)=\ln (x) \sum_{n=0}^{\infty} a_{n} x^{n+1}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} $

Consider the Second-Order Differential Equation y1(x)=∑n=0∞anxn,y2(x)=x−14∑n=0∞bnxn y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n} y1​(x)=∑n=0∞​an​xn,y2​(x)=x−41​∑n=0∞​bn​xn

Consider the Second-Order Differential Equation $y_{1}(x)=\sum_{n=0}^{\infty} a_{n} x^{n}, y_{2}(x)=x^{-\frac{1}{4}} \sum_{n=0}^{\infty} b_{n} x^{n}$