Review for Exam 2. Review for Exam 2.

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Review for Exam 2. 5 or 6 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Exam covers: Regular-singular points (5.5). Euler differential equation (5.). Power series solutions (5.2). Variation of parameters (3.6). Undetermined coefficients (3.5) Constant coefficients, homogeneous, (3.1)-(3.). Review for Exam 2. 5 or 6 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Exam covers: Regular-singular points (5.5). Euler differential equation (5.). Power series solutions (5.2). Variation of parameters (3.6). Undetermined coefficients (3.5) Constant coefficients, homogeneous, (3.1)-(3.).

Regular-singular points (5.5). Summary: Look for solutions y(x) = a n (x x 0 ) (n+r). Recall: Since r 0, holds y = (n+r)a n (x x 0 ) (n+r 1) (n+r)a n (x x 0 ) (n+r 1), n=1 Find the indicial equation for r, the recurrence relation for a n. Introduce the larger root r + of the indicial polynomial into the recurrence relation and solve for a n. (a) If (r + r ) is not an integer, then each r + and r define linearly independent solutions. (b) If (r + r ) is an integer, then both r + and r define proportional solutions. Regular-singular points (5.5). ( ) Consider the equation x 2 y + x 2 + 1 y = 0. Use a power series centered at the regular-singular point x 0 = 0 to find the three first terms of the solution corresponding to the larger root of the indicial polynomial. Solution: y = a n x (n+r), y = (n + r)(n + r 1)a n x (n+r 2), x 2 y = (n + r)(n + r 1)a n x (n+r) We also need to compute ( x 2 + 1 ) y = a n x (n+r+2) + 1 a nx (n+r),

Regular-singular points (5.5). ( ) Consider the equation x 2 y + x 2 + 1 y = 0. Use a power series centered at the regular-singular point x 0 = 0 to find the three first terms of the solution corresponding to the larger root of the indicial polynomial. Solution: ( x 2 + 1 ) y = a n x (n+r+2) + 1 a nx (n+r). Re-label m = n + 2 in the first term and then switch back to n, ( x 2 + 1 ) y = a (n 2) x (n+r) 1 + a nx (n+r), n=2 The equation is (n+r)(n+r 1)a n x (n+r) + a (n 2) x (n+r) + n=2 1 a nx (n+r) = 0. Regular-singular points (5.5). ( ) Consider the equation x 2 y + x 2 + 1 y = 0. Use a power series centered at the regular-singular point x 0 = 0 to find the three first terms of the solution corresponding to the larger root of the indicial polynomial. Solution: (n+r)(n+r 1)a n x (n+r) + a (n 2) x (n+r) + n=2 1 a nx (n+r) = 0. r(r 1) + 1 ] a 0 x r + (r + 1)r + 1 ] a 1 x (r+1) + (n + r)(n + r 1)a n + a (n 2) + 1 ] a n x (n+r) = 0. n=2

Regular-singular points (5.5). ( ) Consider the equation x 2 y + x 2 + 1 y = 0. Use a power series centered at the regular-singular point x 0 = 0 to find the three first terms of the solution corresponding to the larger root of the indicial polynomial. Solution: r(r 1) + 1 ] a 0 = 0, (r + 1)r + 1 ] a 1 = 0, (n + r)(n + r 1) + 1 ] a n + a (n 2) = 0. The indicial equation r 2 r + 1 = 0 implies r ± = 1 2. The indicial equation r 2 + r + 1 = 0 implies r ± = 1 2. Choose r = 1 2. That implies a 0 arbitrary and a 1 = 0. Regular-singular points (5.5). ( ) Consider the equation x 2 y + x 2 + 1 y = 0. Use a power series centered at the regular-singular point x 0 = 0 to find the three first terms of the solution corresponding to the larger root of the indicial polynomial. Solution: r = 1 2, a 1 = 0, (n + r)(n + r 1) + 1 ] a n = a (n 2). ( n+ 1 )( n 1 ) + 1 ] a n = a 2 2 (n 2) n 2 1 + 1 ] a n = a (n 2) n 2 a n = a (n 2) a n = a (n 2) n 2 a 2 = a 0, a = a 2 16 = a 0 6.

Regular-singular points (5.5). ( ) Consider the equation x 2 y + x 2 + 1 y = 0. Use a power series centered at the regular-singular point x 0 = 0 to find the three first terms of the solution corresponding to the larger root of the indicial polynomial. Solution: r = 1 2, a 1 = 0, a 2 = a 0, and a = a 0 6. Then, y(x) = x r ( a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a x + ). Recall: a 1 = 0 and the recurrence relation imply a n = 0 for n odd. Therefore, y(x) = a 0 x 1/2( 1 1 x 2 + 1 6 x + ). Review for Exam 2. 5 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Exam covers: Regular-singular points (5.5). Euler differential equation (5.). Power series solutions (5.2). Variation of parameters (3.6). Undetermined coefficients (3.5) Constant coefficients, homogeneous, (3.1)-(3.).

Euler differential equation (5.). Summary: (x x 0 ) 2 y + (x x 0 )p 0 y + q 0 y = 0. Find r ± solutions of r(r 1) + p 0 r + q 0 = 0. If r + r and both are real, then fundamental solutions are y + = x x 0 r +, y = x x 0 r. If r ± = α ± iβ, then real-valued fundamental solutions are y + = x x 0 α cos ( β ln x x 0 ), y = x x 0 α sin ( β ln x x 0 ). If r + = r and both are real, then fundamental solutions are y + = x x 0 r +, y = x x 0 r + ln x x 0. Euler differential equation (5.). Find real-valued fundamental solutions of (x 2) 2 y + 5(x 2) y + 8 y = 0. Solution: This is an Euler equation. Find r solution of r(r 1) + 5r + 8 = 0, that is, r 2 + r + 8 = 0, r ± = 1 ] ± 16 32 r ± = 2 ± 2i. 2 Real valued fundamental solutions are y + (x) = x 2 2 cos ( 2 ln x 2 ), y (x) = x 2 2 sin ( 2 ln x 2 ).

Review for Exam 2. 5 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Exam covers: Regular-singular points (5.5). Euler differential equation (5.). Power series solutions (5.2). Variation of parameters (3.6). Undetermined coefficients (3.5) Constant coefficients, homogeneous, (3.1)-(3.). Power series solutions (5.2). Using a power series centered at x 0 = 0 find the three first terms of the general solution of ( x 2 ) y + 2y = 0. Solution: We look for solutions y = y = a n x n. Therefore, n(n 1)a n x (n 2) The differential equation is then given by ( x 2 ) n(n 1)a n x (n 2) + 2 n(n 1)a n x (n 2) a n x n = 0, n(n 1)a n x n + 2a n x n = 0.

Power series solutions (5.2). Using a power series centered at x 0 = 0 find the three first terms of the general solution of ( x 2 ) y + 2y = 0. Solution: n(n 1)a n x (n 2) n=2 n(n 1)a n x n + 2a n x n = 0. Re-label the first sum, m = n 2 and then switch back to n (n + 2)(n + 1)a n+2 x n n(n 1)a n x n + 2a n x n = 0. ] (n + 2)(n + 1)an+2 n(n 1)a n + 2a n x n = 0. (n + 2)(n + 1)a n+2 + ( n 2 + n + 2)a n = 0. Power series solutions (5.2). Using a power series centered at x 0 = 0 find the three first terms of the general solution of ( x 2 ) y + 2y = 0. Solution: (n + 2)(n + 1)a n+2 + ( n 2 + n + 2)a n = 0. Notice: n 2 + n + 2 = (n 2)(n + 1), hence (n+2)(n+1)a n+2 (n 2)(n+1)a n = 0 a n+2 = (n 2)a n (n + 2). For n even the power series terminates at n = 2, since a 2 = 2a 0 8, a = 0, a 6 = 0, For n odd: a 3 = a 1 12, a 5 = a 3 20 = a 1 (12)(20), y = a 0 1 1 x 2] + a 1 x 1 12 x 3 1 (12)(20) x 5 + ].

Review for Exam 2. 5 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Exam covers: Regular-singular points (5.5). Euler differential equation (5.). Power series solutions (5.2). Variation of parameters (3.6). Undetermined coefficients (3.5) Constant coefficients, homogeneous, (3.1)-(3.). Variation of parameters (3.6). Use the variation of parameters to find the general solution of y + y + y = x 2 e 2x. Solution: We find the solutions of the homogeneous equation, r 2 + r + = 0 r ± = 1 ] ± 16 16 r ± = 2. 2 Fundamental solutions of the homogeneous equations are y 1 = e 2x, y 2 = x e 2x. We now compute their Wronskian, W = y 1 y 2 y 1 y 2 = e 2x x e 2x 2e 2x (1 2x) e 2x = (1 2x) e x + 2x e x. Hence W = e x.

Variation of parameters (3.6). Use the variation of parameters to find the general solution of y + y + y = x 2 e 2x. Solution: y 1 = e 2x, y 2 = x e 2x, g = x 2 e 2x, W = e x. Now we find the functions u 1 and u 2, u 1 = y 2g W = x e 2x x 2 e 2x e x = 1 x u 1 = ln x. u 2 = y 1g W = e 2x x 2 e 2x e x = x 2 u 2 = 1 x. y p = u 1 y 1 + u 2 y 2 = ln x e 2x 1 x xe 2x = (1 + ln x ) e 2x. Since ỹ p = ln x e 2x is solution, y = (c 1 + c 2 x ln x ) e 2x. Review for Exam 2. 5 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Exam covers: Regular-singular points (5.5). Euler differential equation (5.). Power series solutions (5.2). Variation of parameters (3.6). Undetermined coefficients (3.5) Constant coefficients, homogeneous, (3.1)-(3.).

Undetermined coefficients (3.5) Use the undetermined coefficients to find the general solution of y + y = 3 sin(2x) + e 3x Solution: Find the solutions of the homogeneous problem, r 2 + = 0 r ± = ±2i. y 1 = cos(2x), y 2 = sin(2x). Start with the first source, f 1 (x) = 3 sin(2x). The function ỹ p1 = k 1 sin(2x) + k 2 cos(2x) is the wrong guess, since it is solution of the homogeneous equation. We guess: y p = x k 1 sin(2x) + k 2 cos(2x) ]. y p = k 1 sin(2x) + k 2 cos(2x) ] + 2x k 1 cos(2x) k 2 sin(2x) ]. y p = k 1 cos(2x) k 2 sin(2x) ] + x k 1 sin(2x) k 2 cos(2x) ]. Undetermined coefficients (3.5) Use the undetermined coefficients to find the general solution of y + y = 3 sin(2x) + e 3x. Solution: Recall: y 1 = sin(2x), and y 2 = cos(2x). k 1 cos(2x) k 2 sin(2x) ] + x k 1 sin(2x) k 2 cos(2x) ] + x k 1 sin(2x) + k 2 cos(2x) ] = 3 sin(2x), Therefore, k 1 cos(2x) k 2 sin(2x) ] = 3 sin(2x). Evaluating at x = 0 and x = π/ we get k 1 = 0, k 2 = 3 k 1 = 0, k 2 = 3. Therefore, y p1 = 3 x cos(2x).

Undetermined coefficients (3.5) Use the undetermined coefficients to find the general solution of y + y = 3 sin(2x) + e 3x. Solution: Recall: y p1 = 3 x cos(2x). We now compute y p2 for f 2 (x) = e 3x. We guess: y p2 = k e 3x. Then, y p 2 = 9 e 3x, (9 + )ke 3x = e 3x k = 1 13 Therefore, the general solution is y(x) = c 1 sin(2x) + (c 2 3 ) x y p2 = 1 13 e3x. cos(2x) + 1 13 e3x.

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