Explanation Theorem used: Let r 1 and r 2 be the roots of the indicial equation F ( r ) = r ( r − 1 ) + p 0 r + q 0 with r 1 ≥ r 2 if r 1 and r 2 are real. Then in either the interval − ρ < x < 0 or the interval 0 < x < ρ , there exists a solution of the form y 1 ( x ) = | x | r 1 [ 1 + ∑ n = 1 ∞ a n ( r 1 ) x n ] where a n ( r 1 ) are the given by recurrence relation with a 0 = 1 and r = r 1 . If r 1 − r 2 is not zero or a positive integer, then in either the interval − ρ < x < 0 or the interval 0 < x < ρ , there exists a second solution of the form y 2 ( x ) = | x | r 2 [ 1 + ∑ n = 1 ∞ a n ( r 2 ) x n ] a n ( r 2 ) are also determined by the recurrence relation with a 0 = 1 and r = r 2 . The power series converges at least for | x | < ρ . If r 1 = r 2 , then the second solution is y 2 ( x ) = y 1 ( x ) ln | x | + | x | r 1 [ 1 + ∑ n = 1 ∞ b n ( r 1 ) x n ] . If r 1 − r 2 = N , positive integer, then y 2 ( x ) = a y 1 ( x ) ln | x | + | x | r 2 [ 1 + ∑ n = 1 ∞ c n ( r 2 ) x n ] . Calculation: The given differential equation is given as x 2 y " + 3 x y ' + ( 1 + x ) y = 0 . Compare the equation x 2 y " + 3 x y ' + ( 1 + x ) y = 0 with P ( x ) y " + Q ( x ) y ' + R ( x ) y = 0 here P ( x ) = x 2 , Q ( x ) = 3 x and R ( x ) = 1 + x . The singular points occur when P ( x 0 ) = 0 thus the singular point of the given equation is x 0 = 0 . If x 0 = 0 is a regular singular point, check if the following limits are finite. p 0 = lim x → 0 x Q ( x ) P ( x ) and q 0 = lim x → 0 x 2 R ( x ) P ( x ) For x 0 = 0 , therefore the value of p 0 = lim x → 0 x 3 x x 2 = 3 and q 0 = lim x → 0 x 2 ( 1 + x ) x 2 = 1 . Since both are finite, x 0 = 0 is a regular singular point of the given differential equation. Because p 0 = 3 and q 0 = 1 , the corresponding indicial equation is as follows: F ( r ) = r ( r − 1 ) + p 0 r + q 0 Substitute the value of p 0 = 3 and q 0 = 1 in F ( r ) = r ( r − 1 ) + p 0 r + q 0 = 0 . F ( r ) = r ( r − 1 ) + 3 r + 1 = ( r + 1 ) 2 Let F ( r ) = 0 in the above equation it becomes as follows: ( r + 1 ) 2 = 0 r 1 = r 2 = − 1 Therefore, the roots of the equation ( r + 1 ) 2 = 0 is r 1 = − 1 , r 2 = − 1 Hence, x 0 is a regular singular point. From the above Theorem, the first solution is given as y 1 ( x ) = | x | r 1 [ 1 + ∑ n = 1 ∞ a n ( r 1 ) x n ] Substitute the value of r 1 = − 1 in y 1 ( x ) = | x | r 1 [ 1 + ∑ n = 1 ∞ a n ( r 1 ) x n ] y 1 ( x ) = x − 1 [ 1 + ∑ n = 1 ∞ a n ( − 1 ) x n ] Differentiate the equation y 1 ( x ) = x − 1 [ 1 + ∑ n = 1 ∞ a n ( − 1 ) x n ] . y 1 ' ( x ) = − x − 2 + ∑ n = 1 ∞ n a n + 1 ( − 1 ) x n − 1 The coefficients a n ( − 1 ) , apart from a 0 ( − 1 ) = 1 are determined by the recurrence relation F ( r + n ) a n + ∑ k = 1 n − 1 a k [ ( r + k ) p n − k + q n − k ] = 0 in case of r = r 1 . n 2 a n ( 0 ) + ∑ k = 1 n − 1 a k ( 0 ) [ k p n − k + q n − k ] = 0 , where ∑ n = 0 ∞ p n x n = x Q ( x ) P ( x ) = 3 and ∑ n = 0 ∞ q n x n = x 2 R ( x ) P ( x ) = 1 + x . Hence the value of p 0 = 3 , p n = 0 , ∀ n ∈ N and q 1 = 1 , q n = 1 , ∀ n ∈ N 0 . Factoring out a n ( − 1 ) from the equation n 2 a n ( 0 ) + ∑ k = 1 n − 1 a k ( 0 ) [ k p n − k + q n − k ] = 0 is as follows: a n ( − 1 ) = − ∑ k = 0 n − 1 a k ( − 1 ) [ ( k − 1 ) p n − k + q n − k ] n 2 Solving the first couple of elements is as follows: a 1 ( − 1 ) = − a 0 ( − 1 ) [ − p 1 + q 1 ] 1 2 = − 1 a 2 ( − 1 ) = 1 ( 2 ! 2 ) a 3 ( − 1 ) = − 1 ( 3 ! 2 ) Hence, the first solution is y 1 ( x ) = x − 1 ∑ n = 0 ∞ ( − 1 ) n ( n ! ) 2 x n _

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 5.7, Problem 2P

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To show: That the given differential equation has a regular singular point at x=0 is and determine two solutions for x>0 by using the given data.

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Consider the cones K = = {(x1, x2, x3) | € R³ : X3 ≥√√√2x² + 3x² M = = {(21,22,23) (x1, x2, x3) Є R³: x3 > + 2 3 Prove that M = K*. Hint: Adapt the proof from the lecture notes for finding the dual of the Lorentz cone. Alternatively, prove the formula (AL)* = (AT)-¹L*, for any cone LC R³ and any 3 × 3 nonsingular matrix A with real entries, where AL = {Ax = R³ : x € L}, and apply it to the 3-dimensional Lorentz cone with an appropriately chosen matrix A.

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Let M = M₁U M₂ UM3 and K M₁ = {(x1, x2) ER²: 2 ≤ x ≤ 8, 2≤ x ≤8}, M₂ = {(x1, x2)™ € R² : 4 ≤ x₁ ≤ 6, 0 ≤ x2 ≤ 10}, M3 = {(x1, x2) Є R²: 0 ≤ x₁ ≤ 10, 4≤ x ≤ 6}, ¯ = cone {(1, 2), (1,3)†} ≤ R². (a) Determine the set E(M,K) of efficient points of M with respect to K. (b) Determine the set P(M, K) of properly efficient points of M with respect to K.

Chapter 5 Solutions

Elementary Differential Equations

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That the given differential equation has a regular singular point at x = 0 is and determine two solutions for x &gt; 0 by using the given data.

To show: That the given differential equation has a regular singular point at x=0 is and determine two solutions for x>0 by using the given data.

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Chapter 5 Solutions

That the given differential equation has a regular singular point at x = 0 is and determine two solutions for x > 0 by using the given data.