To determine The general solution of the initial value problem ( 9 x 2 D 2 + 3 x D + I ) y = 0 , y ( 1 ) = 1 , y ′ ( 1 ) = 0 . Explanation Result used: The general solution of Euler-Cauchy equation: “Suppose that, the linear ordinary differential equation of the form x 2 y ″ + a x y ′ + b y = 0 . In order to obtain the auxiliary equation, substitute y = x m , y ′ = m x m − 1 and y ″ = m ( m − 1 ) x m − 2 . Then the auxiliary equation becomes m 2 + ( a − 1 ) m + b = 0 . The general solution of the equation is of the form: Case 1: If the roots are real distinct m 1 and m 2 , then the general solution is, y h ( x ) = c 1 x m 1 + c 2 x m 2 . Case 2: If the roots are real double roots m 1 and m 1 , then the general solution is, y h ( x ) = ( c 1 + c 2 ln x ) x m 1 . Case 3: If the roots are complex a ± i b , then the general solution is, y h ( x ) = x a ( A cos ( b ln x ) + B sin ( b ln x ) ) . Calculation: The corresponding homogeneous linear ordinary differential equation of the given equation ( 9 x 2 D 2 + 3 x D + I ) y = 0 is 9 x 2 y ″ + 3 x y ′ + y = 0 . Rewrite the homogeneous linear ordinary differential equation 9 x 2 y ″ + 3 x y ′ + y = 0 as shown below. 9 x 2 y ″ + 3 x y ′ + y = 0 9 ( x 2 y ″ + 3 x y ′ 9 + y 9 ) = 0 9 ( x 2 y ″ + x y ′ 3 + y 9 ) = 0 Compute the general solution y h of the homogeneous linear ordinary differential equation as follows. The auxiliary equation of x 2 y ″ + x y ′ 3 + y 9 is m 2 + ( 1 3 − 1 ) m + 1 9 = 0 . Hence the auxiliary equation of 9 ( x 2 y ″ + x y ′ 3 + y 9 ) = 0 becomes 9 ( m 2 + ( 1 3 − 1 ) m + 1 9 ) = 0 Obtain the roots of 9 ( m 2 + ( 1 3 − 1 ) m + 1 9 ) = 0 as shown below. 9 ( m 2 + ( 1 3 − 1 ) m + 1 9 ) = 0 9 ( m 2 − 2 3 m + 1 9 ) = 0 9 ( m − 1 3 ) 2 = 0 ( m − 1 3 ) 2 = 0 m = 1 3 ( double root ) Thus, the roots of the auxiliary equation 9 ( m 2 + ( 1 3 − 1 ) m + 1 9 ) = 0 are 1 3 , 1 3

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Chapter 2.5, Problem 18P

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The general solution of the initial value problem (9x2D2+3xD+I)y=0, y(1)=1, y′(1)=0.

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Chapter 2.5, Problem 17P

Chapter 2.5, Problem 19P

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