MATH 374 Lecture 20 Variation of Parameters 7

MATH 374 Lecture 20 Variation of Parameters

7. 2: Variation of Parameters l 2 Given the equation: y’’ + p(x) y’ + q(x) y = R(x), (1) suppose that y 1 and y 2 are linearly independent solutions of the corresponding homogeneous equation: y’’ + p(x) y’ + q(x) y = 0, (2) on a<x<b.

Variation of Parameters – Second Order case l l 3 Since we know linear combinations of y 1 and y 2 solve (2), maybe a solution of (1) can be put in the form: y = A(x)y 1 + B(x)y 2 (3) for some functions A(x) and B(x). Let’s assume such a solution exists and try to find A and B!

y’’ + p(x) y’ + q(x) y = R(x) (1) y = A(x)y 1 + B(x)y 2 (3) Variation of Parameters – Second Order case Differentiating (3), y’ = A’y 1 + Ay 1’ + B’y 2 + By 2’ (4) y’’ = A’’y 1 + 2 A’y 1’+ Ay 1’’ + B’’y 2 + 2 B’y 2’ + By 2’’. (5) Substituting (3), (4), and (5) into (1) yields (A’’y 1 + 2 A’y 1’ + Ay 1’’ + B’’y 2 + 2 B’y 2’ + By 2’’) + + p(x)(A’y 1 + Ay 1’ + B’y 2 + By 2’) + + q(x)(Ay 1 + By 2) = R(x) l 4

y’’ + p(x) y’ + q(x) y = R(x) (1) y = A(x)y 1 + B(x)y 2 (3) Variation of Parameters – Second Order case Differentiating (3), y’ = A’y 1 + Ay 1’ + B’y 2 + By 2’ (4) y’’ = A’’y 1 + 2 A’y 1’+ Ay 1’’ + B’’y 2 + 2 B’y 2’ + By 2’’. (5) Substituting (3), (4), and (5) into (1) yields (A’’y 1 + 2 A’y 1’ + Ay 1’’ + B’’y 2 + 2 B’y 2’ + By 2’’) + + p(x)(A’y 1 + Ay 1’ + B’y 2 + By 2’) + + q(x)(Ay 1 + By 2) = R(x) l 5

y’’ + p(x) y’ + q(x) y = R(x) (1) y’’ + p(x) y’ + q(x) y = 0, (2) Variation of Parameters – Second Order case (A’’y 1 + 2 A’y 1’ + Ay 1’’ + B’’y 2 + 2 B’y 2’ + By 2’’) + + p(x) (A’y 1 + Ay 1’ + B’y 2 + By 2’) + + (Ay 1 + By 2) = R(x) Rearranging, we get: = 0 since y 1 solves (2) (A’’y 1+B’’y 2) + 2(A’y 1’+B’y 2’) + A(y 1’’+p(x)y 1’+q(x)y 1) + B(y 2’’+p(x)y 2’+q(x)y 2) + p(x)(A’y 1+B’y 2) = R(x) = 0 since y 2 solves (2) 6

y’’ + p(x) y’ + q(x) y = R(x) (1) y = A(x)y 1 + B(x)y 2 (3) Variation of Parameters – Second Order case (A’’y 1+B’’y 2) + 2(A’y 1’+B’y 2’) + + p(x)(A’y 1+B’y 2) = R(x). l Thus, (A’’y 1+B’’y 2) + 2(A’y 1’+B’y 2’) + p(x)(A’y 1+B’y 2) = R(x) (6) must hold. l Note that if A and B are found to make (6) hold, then reversing the above argument shows that (3) solves (1). l To find A and B, let’s try to find two equations that relate A and B. 7

y = A(x)y 1 + B(x)y 2 (3) (A’’y 1+B’’y 2) + 2(A’y 1’+B’y 2’) + p(x)(A’y 1+B’y 2) = R(x) (6) Variation of Parameters – Second Order case l l 8 Notice that if A’y 1 + B’y 2 = 0, (7) then since (7) implies 0 = (A’y 1 + B’y 2)’ = A’’y 1 + B’’y 2 + A’y 1’ + B’y 2’ (8) it follows from (7) and (8) that for (6) to hold, we’d need: A’y 1’ + B’y 2’ = R(x). (9) Hence, if we can find A and B that solve system (7), (9), equation (6) will hold and (1) is solved by a solution of form (3)!!

A’y 1 + B’y 2 = 0, (7) A’y 1’ + B’y 2’ = R(x). (9) Variation of Parameters – Second Order case l 9 Now, a solution to system (7), (9) will exist, provided W(x) = y 1 y 2’ – y 1’y 2 0. Since y 1 and y 2 are linearly independent on (a, b), W(x) is never zero on (a, b) (see HW problems 23 and 24 for proof).

A’y 1 + B’y 2 = 0, (7) A’y 1’ + B’y 2’ = R(x). (9) Variation of Parameters – Second Order case l l 10 Solving system (7), (9) for A’ and B’ gives: A’ = - (y 2 R(x))/W(x) B’ = (y 1 R(x))/W(x), from which it follows that A(x) = -s (y 2 R(x))/W(x) dx B(x) = s (y 1 R(x))/W(x) dx. (10) Thus, the solution to (1) is y = A(x)y 1 + B(x)y 2 with A and B given by (10).

Variation of Parameters – nth Order case l 11 Note: For linear equations of order n, a similar technique will work – Look for a solution of the form y = A 1(x)y 1 + … + An(x)yn with y 1, y 2, … , yn linearly independent solutions of the corresponding homogeneous problem.

Example 1 l l 12 Use Variation of Parameters to solve y’’ – y – 2 y = e-x. (11) Solution: The auxiliary equation for y’’ – y’ – 2 y = 0 is m 2 – m – 2 = 0. ) (m-2)(m+1) = 0 ) roots are m = 2, -1. Thus, yc = c 1 e 2 x + c 2 e-x is the general solution to the homogeneous problem corresponding to (11).

A(x) = -s (y 2 R(x))/W(x) dx B(x) = s (y 1 R(x))/W(x) dx. (10) Example 1 (continued) l 13 Taking y 1 = e 2 x and y 2 = e-x, which are linearly independent solutions of the homogeneous problem corresponding to (11), (10) implies, with W(x) = y 1 y 2’ – y 1’y 2, = (e 2 x)(-e-x) – (2 e 2 x)(e-x) = -3 ex, A(x) = - s (y 2 R(x))/W(x) dx = - s (e-x e-x)/(-3 ex) dx = - s (e-2 x)/(-3 ex) dx = s 1/3 e-3 x dx = -1/9 e-3 x + C 1

A(x) = -s (y 2 R(x))/W(x) dx B(x) = s (y 1 R(x))/W(x) dx. (10) Example 1 (continued) and l 14 B(x) = s (y 1 R(x))/W(x) dx = s (e 2 x e-x)/(-3 ex) dx = s ex/(-3 ex) dx = s -1/3 dx = -1/3 x + C 2 Therefore, the solution to (11) is: y = A(x)y 1 + B(x)y 2 = (-1/9 e-3 x + C 1)e 2 x + (-1/3 x + C 2)e-x = C 1 e 2 x + C 2 e-x -1/9 e-x -1/3 x e-x (Re-solve (11) using the Method of Undetermined Coefficients!)