The solution of the initial value problem y ″ + 3 x y ′ + 2 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 by power series method and find the value of the sum at x = 0.5 . The sum of the initial value problem is s 5 ( x ) = 1 + x − x 2 − 5 6 x 3 + 2 3 x 4 + 11 24 x 5 _ . The sum of the initial value problem at x = 0.5 is 1.202 _ . Definition used: Power series: “The general power series is an infinite series of the form, ( 1 ) ∑ m = 0 ∞ a m ( x − x 0 ) m = a 0 + a 1 ( x − x 0 ) + a 2 ( x − x 0 ) 2 + ⋯ . Here, x is a variable. a 0 , a 1 , a 2 , ⋯ are constants, called the coefficients of the series. x 0 is a constant, called the center of the series. In particular, if x 0 = 0 , a power series in powers of x ( 2 ) ∑ m = 0 ∞ a m x m = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ⋯ .” “The nth partial sum of y ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + ⋯ at x 0 = 0 is s n ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + ⋯ + a n x n ”. Calculation: The given ODE is y ″ + 3 x y ′ + 2 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 . Consider the power series, y = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + ⋯ = ∑ m = 0 ∞ a m x m . Differentiate the series with respect to x . y ′ = a 1 + 2 a 2 x + 3 a 3 x 2 + 4 a 4 x 3 + 5 a 5 x 4 + ⋯ = ∑ m = 1 ∞ m a m x m − 1 Again differentiate the series with respect to x . y ″ = 2 a 2 + 2 ⋅ 3 a 3 x + 3 ⋅ 4 a 4 x 2 + 4 ⋅ 5 a 5 x 3 + ⋯ = ∑ m = 2 ∞ m ( m − 1 ) a m x m − 2 Substitute the values of y ″ and y ′ in the given ODE. y ″ + 3 x y ′ + 2 y = 0 [ ( 2 a 2 + 2 ⋅ 3 a 3 x + 3 ⋅ 4 a 4 x 2 + 4 ⋅ 5 a 5 x 3 + ⋯ ) + 3 x ( a 1 + 2 a 2 x + 3 a 3 x 2 + 4 a 4 x 3 + 5 a 5 x 4 + ⋯ ) + 2 ( a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + ⋯ ) ] = 0 [ ( 2 a 2 + 6 a 3 x + 12 a 4 x 2 + 20 a 5 x 3 + ⋯ ) + ( 3 a 1 x + 6 a 2 x 2 + 9 a 3 x 3 + 12 a 4 x 4 + 15 a 5 x 5 + ⋯ ) + ( 2 a 0 + 2 a 1 x + 2 a 2 x 2 + 2 a 3 x 3 + 2 a 4 x 4 + 2 a 5 x 5 + ⋯ ) ] = 0 [ ( 2 a 2 + 2 a 0 ) + ( 2 a 1 + 3 a 1 + 6 a 3 ) x + ( 2 a 2 + 6 a 2 + 12 a 4 ) x 2 + ( 2 a 3 + 9 a 3 + 20 a 5 ) x 3 + ⋯ ] = 0 Equate the coefficients of x , x 2 , x 3 and constants to 0. 2 a 2 + 2 a 0 = 0 ⋯ ⋯ ( 3 ) 5 a 1 + 6 a 3 = 0 ⋯ ⋯ ( 4 ) 8 a 2 + 12 a 4 = 0 ⋯ ⋯ ( 5 ) 11 a 3 + 20 a 5 = 0 ⋯ ⋯ ( 6 ) Solve the above equations. From equation (3), the value of a 2 = − a 0 . From equation (4), the value of a 3 as follows. 5 a 1 + 6 a 3 = 0 6 a 3 = − 5 a 1 a 3 = − 5 6 a 1 From equation (5), the value of a 5 as follows. 8 a 2 + 12 a 4 = 0 8 a 2 + 12 a 4 = 0 12 a 4 = 8 a 0 a 4 = 2 a 0 3 From equation (6), the value of a 4 as follows. 11 a 3 + 20 a 5 = 0 11 ( − 5 6 a 1 ) + 20 a 5 = 0 20 a 5 = 55 6 a 1 a 5 = 11 24 a 1 Substitute the co-efficient values in the general solution of the ODE. y ( x ) = ( a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + ⋯ ) = ( a 0 + a 1 x + ( − a 0 ) x 2 + ( − 5 a 1 6 ) x 3 + ( 2 a 0 3 ) x 4 + ( 11 24 a 1 ) x 5 + ⋯ ) Differentiate with respect to x as follows. y ′ ( x ) = ( 0 + a 1 ( 1 ) + 2 ( − a 0 ) x + 2 ( − 5 a 1 6 ) x 2 + 4 ( 2 a 0 3 ) x 3 + 5 ( 11 24 a 1 ) x 4 + ⋯ ) = ( a 1 − ( 2 a 0 ) x + ( − 5 a 1 3 ) x 3 + ( 8 a 0 3 ) x 4 + ( 55 24 a 1 ) x 5 + ⋯ ) Use initial condition y ( 0 ) = 1 , y ′ ( 0 ) = 1 to obtain a 0 as follows. y ( 0 ) = ( a 0 + a 1 x + ( − a 0 ) x 2 + ( − 5 a 1 6 ) x 3 + ( 2 a 0 3 ) x 4 + ( 11 24 a 1 ) x 5 + ⋯ ) 1 = ( a 0 ) a 0 = 1 Use initial condition y ( 0 ) = 1 , y ′ ( 0 ) = 1 to obtain a 1 as follows. y ′ ( 0 ) = ( a 1 − ( 2 a 0 ) ( 0 ) + ( − 5 a 1 3 ) ( 0 ) 3 + ( 8 a 0 3 ) ( 0 ) 4 + ( 55 24 a 1 ) ( 0 ) 5 + ⋯ ) 1 = ( a 1 ) a 1 = 1 Substitute a 0 in general solution and obtain the sum as follows. s n ( x ) = ( a 0 + a 1 x + ( − a 0 ) x 2 + ( − 5 a 1 6 ) x 3 + ( 2 a 0 3 ) x 4 + ( 11 24 a 1 ) x 5 + ⋯ + a n x n ) s 5 ( x ) = ( a 0 + a 1 x + ( − a 0 ) x 2 + ( − 5 a 1 6 ) x 3 + ( 2 a 0 3 ) x 4 + ( 11 24 a 1 ) x 5 ) s 5 ( x ) = ( 1 + x − x 2 − 5 6 x 3 + 2 3 x 4 + 11 24 x 5 ) s 5 ( x ) = 1 + x − x 2 − 5 6 x 3 + 2 3 x 4 + 11 24 x 5 Thus, the sum of the initial value problem is s 5 ( x ) = 1 + x − x 2 − 5 6 x 3 + 2 3 x 4 + 11 24 x 5 _ . Draw the graph of the sum s 5 ( x ) = 1 + x − x 2 − 5 6 x 3 + 2 3 x 4 + 11 24 x 5 as shown below. From Figure 1, it is observed that the y -intercept is 1. Substitute x = 0.5 in s 5 ( x ) = 1 + x − x 2 − 5 6 x 3 + 2 3 x 4 + 11 24 x 5 and obtain s 5 ( 0.5 ) as follows. s 5 ( x ) = 1 + x − x 2 − 5 6 x 3 + 2 3 x 4 + 11 24 x 5 s 5 ( 0.5 ) = 1 + 0.5 − ( 0.5 ) 2 − 5 6 ( 0.5 ) 3 + 2 3 ( 0.5 ) 4 + 11 24 ( 0.5 ) 5 = 1 + 0.5 − 0.25 − 0.1041 + 0.04167 + 0.014323 ≈ 1.202 Thus, the sum of the initial value problem at x = 0.5 is 1.202 _ .

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