Explanation The given initial value problem is 4 x 2 y ″ + y = 0 , y ( − 1 ) = 2 , y ′ ( − 1 ) = 4 . The above equation can be rewritten as 4 t 2 d 2 y d t 2 + y = 0 with initial conditions y ( 1 ) = 2 and y ′ ( 1 ) = − 4 . Then the higher derivatives becomes y ′ = m t m − 1 and y ″ = m ( m − 1 ) t m − 2 . Substitute the values of y and y ″ in the equation 4 t 2 d 2 y d t 2 + y = 0 and obtain the auxiliary equation as follows. 4 t 2 d 2 y d t 2 + y = 0 4 t 2 m ( m − 1 ) t m − 2 + t m = 0 4 m ( m − 1 ) t m + t m = 0 ( 4 m ( m − 1 ) + 1 ) t m = 0 Solve the above auxiliary equation and obtain the roots of m as shown below. 4 m ( m − 1 ) + 1 = 0 4 m 2 − 4 m + 1 = 0 ( 2 m − 1 ) 2 = 0 m 1 = 1 2 , m 2 = 1 2 Therefore, the factors of the above equation are m 1 = 1 2 , m 2 = 1 2 . Since the roots are repeated, the general solution is given by y = c 1 t m + c 2 t m ln t . Therefore, y = c 1 t 1 2 + c 2 t 1 2 ln t . Also, y ′ = 1 2 c 1 t − 1 2 + 1 2 c 2 t − 1 2 ln t + c 2 t 1 2 1 t . Apply the initial conditions y ( 1 ) = 2 and y ′ ( 1 ) = − 4 and obtain the values of the constants c 1 and c 2 as shown below

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Chapter 3.6, Problem 49E

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The solution of the initial value problem 4x2y″+y=0, y(−1)=2, y′(−1)=4 on the interval (−∞,0) using the substitution t=−x.

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Chapter 3.6, Problem 48E

Chapter 3.6, Problem 50E

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The solution of the initial value problem 4 x 2 y ″ + y = 0 , y ( − 1 ) = 2 , y ′ ( − 1 ) = 4 on the interval ( − ∞ , 0 ) using the substitution t = − x .

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The solution of the initial value problem 4x2y″+y=0, y(−1)=2, y′(−1)=4 on the interval (−∞,0) using the substitution t=−x.

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