The basis solutions of the differential equation x y ″ + ( 1 − 2 x ) y ′ + ( x − 1 ) y = 0 by power series method. The basis solution of the differential equation x y ″ + ( 1 − 2 x ) y ′ + ( x − 1 ) y = 0 is e x _ . 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 + ⋯ .” Calculation: The given ODE is x y ″ + ( 1 − 2 x ) y ′ + ( x − 1 ) y = 0 . 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. x y ″ + ( 1 − 2 x ) y ′ + ( x − 1 ) y = 0 [ x ( 2 a 2 + 2 ⋅ 3 a 3 x + 3 ⋅ 4 a 4 x 2 + 4 ⋅ 5 a 5 x 3 + ⋯ ) + ( 1 − 2 x ) ( a 1 + 2 a 2 x + 3 a 3 x 2 + 4 a 4 x 3 + 5 a 5 x 4 + ⋯ ) + ( x − 1 ) ( 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 x + 6 a 3 x 2 + 12 a 4 x 3 + 20 a 5 x 4 + ⋯ ) + ( a 1 + 2 a 2 x + 3 a 3 x 2 + 4 a 4 x 3 + 5 a 5 x 4 + ⋯ ) + ( − 2 a 1 x − 4 a 2 x 2 − 6 a 3 x 3 − 8 a 4 x 4 − 10 a 5 x 5 − ⋯ ) + ( a 0 x + a 1 x 2 + a 2 x 3 + a 3 x 4 + a 4 x 5 + a 5 x 6 + ⋯ ) + ( − a 0 − a 1 x − a 2 x 2 − a 3 x 3 − a 4 x 4 − a 5 x 5 − ⋯ ) ] = 0 [ ( a 1 − a 0 ) + ( 2 a 2 + 2 a 2 − 2 a 1 + a 0 − a 1 ) x + ( 6 a 3 + 3 a 3 − 4 a 2 + a 1 − a 2 ) x 2 + ( 12 a 4 + 4 a 4 − 6 a 3 + a 2 − a 3 ) x 3 + ⋯ ] = 0 Equate the coefficients of x , x 2 , x 3 and constants to 0. a 1 − a 0 = 0 ⋯ ⋯ ( 3 ) 4 a 2 − 3 a 1 + a 0 = 0 ⋯ ⋯ ( 4 ) 9 a 3 − 5 a 2 + a 1 = 0 ⋯ ⋯ ( 5 ) 16 a 4 − 7 a 3 + a 2 = 0 ⋯ ⋯ ( 6 ) Solve the above equations. From equation (3), the value of a 1 = a 0 . From equation (4), the value of a 2 as follows. 4 a 2 − 3 a 1 + a 0 = 0 4 a 2 − 3 a 0 + a 0 = 0 4 a 2 − 2 a 0 = 0 a 2 = 1 2 a 0 From equation (5), the value of a 3 as follows. 9 a 3 − 5 a 2 + a 1 = 0 9 a 3 − 5 ( 1 2 a 0 ) + a 0 = 0 9 a 3 = 3 2 a 0 a 3 = 1 6 a 0 From equation (6), the value of a 4 as follows. 16 a 4 − 7 a 3 + a 2 = 0 16 a 4 − 7 ( 1 6 a 0 ) + ( 1 2 a 0 ) = 0 16 a 4 = 4 6 a 0 a 4 = 1 24 a 1 Substitute the co-efficient values in the general solution of the ODE. y = ( a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + ⋯ ) = ( a 0 + a 0 x + ( 1 2 a 0 ) x 2 + ( 1 6 a 0 ) x 3 + ( 1 24 a 0 ) x 4 + ⋯ ) = ( a 0 + a 0 x + ( 1 2 ! a 0 ) x 2 + ( 1 3 ! a 0 ) x 3 + ( 1 4 ! a 0 ) x 4 + ⋯ ) = a 0 ( 1 + x + 1 2 ! x 2 + 1 3 ! x 3 + 1 4 ! x 4 + ⋯ ) From above equation, it is observed that the series 1 + x + 1 2 ! x 2 + 1 3 ! x 3 + 1 4 ! x 4 + ⋯ is e x . Thus, the basis solution of the differential equation x y ″ + ( 1 − 2 x ) y ′ + ( x − 1 ) y = 0 is e x _ .

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