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 _ .

Question

Want to see the full answer?

Answer 1

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Golden Ratio search Method f(x) = 2x^3 - 3x^2 - 12x + 1 Golden ratio search rules 1.If f(x) < f(x2): 1. Eliminate all x values less than x2 2. X2 becomes the new a 3. x, becomes the new x2 4. no change in b If f(x) > f(x2): 1. Eliminate all x values greater than x 2. x, becomes the new b 3. x2 becomes the new x 4. no change in aquesion=Narrow the interval in which the minimizer of the function f is located using the golden search method, starting with the initial interval (0,6], until its width is less than 2. Then, accept the midpoint of this interval as an approximate value of the minimizer of the function fand determine it. (ф=0.62)According to the question above, fill in the table below using the algorithm until the appropriate place.please write every step by step in a verry comprehensive way

In preparing for the upcoming holiday season, Fresh Toy Company (FTC) designed a new doll called The Dougie that teaches children how to dance. The fixed cost to produce the doll is $100,000. The variable cost, which includes material, labor, and shipping costs, is $31 per doll. During the holiday selling season, FTC will sell the dolls for $39 each. If FTC overproduces the dolls, the excess dolls will be sold in January through a distributor who has agreed to pay FTC $10 per doll. Demand for new toys during the holiday selling season is extremely uncertain. Forecasts are for expected sales of 60,000 dolls with a standard deviation of 15,000. The normal probability distribution is assumed to be a good description of the demand. FTC has tentatively decided to produce 60,000 units (the same as average demand), but it wants to conduct an analysis regarding this production quantity before finalizing the decision. (a) Determine the equation for computing FTC's profit for given values of the…

For all integers a and b, (a + b)^4 ≡ a^4 + b^4 (mod 4).