The basis solutions of the differential equation y ″ + 4 y = 0 by power series method. The basis solutions of the differential equation y ″ + 4 y = 0 are cos 2 x , sin 2 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 y ″ + 4 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. [ ( 2 a 2 + 2 ⋅ 3 a 3 x + 3 ⋅ 4 a 4 x 2 + 4 ⋅ 5 a 5 x 3 + ⋯ ) + 4 ( 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 + 2 ⋅ 3 a 3 x + 3 ⋅ 4 a 4 x 2 + 4 ⋅ 5 a 5 x 3 + ⋯ ) + ( 4 a 0 + 4 a 1 x + 4 a 2 x 2 + 4 a 3 x 3 + 4 a 4 x 4 + 4 a 5 x 5 + ⋯ ) ] = 0 [ ( 2 a 2 + 4 a 0 ) + ( 6 a 3 + 4 a 1 ) x + ( 12 a 4 + 4 a 2 ) x 2 + ( 20 a 5 + 4 a 3 ) x 3 ] = 0 Equate the coefficients of x , x 2 , x 3 and constants to 0. 2 a 2 + 4 a 0 = 0 ⋯ ⋯ ( 3 ) 6 a 3 + 4 a 1 = 0 ⋯ ⋯ ( 4 ) 12 a 4 + 4 a 2 = 0 ⋯ ⋯ ( 5 ) 20 a 5 + 4 a 3 = 0 ⋯ ⋯ ( 6 ) Solve the above equations. From equation (3), obtain the value of a 2 as follows. 2 a 2 + 4 a 0 = 0 2 a 2 = − 4 a 0 a 2 = − 2 a 0 From equation (4), obtain the value of a 3 as follows. 6 a 3 + 4 a 1 = 0 6 a 3 = − 4 a 1 a 3 = − 2 3 a 1 From equation (5), obtain the value of a 4 as follows. 12 a 4 + 4 a 2 = 0 12 a 4 + 4 ( − 2 a 0 ) = 0 12 a 4 = 8 a 0 a 4 = 2 3 a 0 From equation (6), obtain the value of a 5 as follows. 20 a 5 + 4 a 3 = 0 20 a 5 + 4 ( − 2 3 a 1 ) = 0 20 a 5 = 8 3 a 1 a 5 = 2 15 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 1 x + ( − 2 a 0 ) x 2 + ( − 2 3 a 1 ) x 3 + ( 2 3 a 0 ) x 4 + ( 2 15 a 1 ) x 5 + ⋯ ) = ( a 0 + a 1 x − 2 a 0 x 2 − 2 3 a 1 x 3 + 2 3 a 0 x 4 + 2 15 a 1 x 5 + ⋯ ) = a 0 ( 1 − 2 x 2 + 2 3 x 4 − ⋯ ) + a 1 ( x − 2 3 x 3 + 2 15 x 5 − ⋯ ) = a 0 ( 1 − 2 x 2 + 2 3 x 4 − ⋯ ) + a 1 2 ( 2 x − 4 3 x 3 + 4 15 x 5 − ⋯ ) From above equation, it is observed that the series 1 − 2 x 2 + 2 3 x 4 − ⋯ is cos 2 x and the series 2 x − 4 3 x 3 + 4 15 x 5 − ⋯ is sin 2 x . Thus, the basis solutions of the differential equation y ″ + 4 y = 0 are cos 2 x , sin 2 x _ .

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