Variation of Parameters MATH 374 Lecture 20 Variation of Parameters

7.2: Variation of Parameters Given the equation: y’’ + p(x) y’ + q(x) y = R(x), (1) suppose that y1 and y2 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 Since we know linear combinations of y1 and y2 solve (2), maybe a solution of (1) can be put in the form: y = A(x)y1 + B(x)y2 (3) for some functions A(x) and B(x). Let’s assume such a solution exists and try to find A and B!

Variation of Parameters – Second Order case y’’ + p(x) y’ + q(x) y = R(x) (1) y = A(x)y1 + B(x)y2 (3) Variation of Parameters – Second Order case Differentiating (3), y’ = A’y1 + Ay1’ + B’y2 + By2’ (4) y’’ = A’’y1 + 2A’y1’+ Ay1’’ + B’’y2 + 2B’y2’ + By2’’. (5) Substituting (3), (4), and (5) into (1) yields (A’’y1 + 2A’y1’ + Ay1’’ + B’’y2 + 2B’y2’ + By2’’) + + p(x)(A’y1 + Ay1’ + B’y2 + By2’) + + q(x)(Ay1 + By2) = R(x)

Variation of Parameters – Second Order case y’’ + p(x) y’ + q(x) y = R(x) (1) y = A(x)y1 + B(x)y2 (3) Variation of Parameters – Second Order case Differentiating (3), y’ = A’y1 + Ay1’ + B’y2 + By2’ (4) y’’ = A’’y1 + 2A’y1’+ Ay1’’ + B’’y2 + 2B’y2’ + By2’’. (5) Substituting (3), (4), and (5) into (1) yields (A’’y1 + 2A’y1’ + Ay1’’ + B’’y2 + 2B’y2’ + By2’’) + + p(x)(A’y1 + Ay1’ + B’y2 + By2’) + + q(x)(Ay1 + By2) = R(x)

Variation of Parameters – Second Order case 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’’y1 + 2A’y1’ + Ay1’’ + B’’y2 + 2B’y2’ + By2’’) + + p(x) (A’y1 + Ay1’ + B’y2 + By2’) + + (Ay1 + By2) = R(x) Rearranging, we get: (A’’y1+B’’y2) + 2(A’y1’+B’y2’) + A(y1’’+p(x)y1’+q(x)y1) + B(y2’’+p(x)y2’+q(x)y2) + p(x)(A’y1+B’y2) = R(x) = 0 since y1 solves (2) = 0 since y2 solves (2)

Variation of Parameters – Second Order case y’’ + p(x) y’ + q(x) y = R(x) (1) y = A(x)y1 + B(x)y2 (3) Variation of Parameters – Second Order case (A’’y1+B’’y2) + 2(A’y1’+B’y2’) + + p(x)(A’y1+B’y2) = R(x). Thus, (A’’y1+B’’y2) + 2(A’y1’+B’y2’) + p(x)(A’y1+B’y2) = R(x) (6) must hold. Note that if A and B are found to make (6) hold, then reversing the above argument shows that (3) solves (1). To find A and B, let’s try to find two equations that relate A and B.

Variation of Parameters – Second Order case y = A(x)y1 + B(x)y2 (3) (A’’y1+B’’y2) + 2(A’y1’+B’y2’) + p(x)(A’y1+B’y2) = R(x) (6) Variation of Parameters – Second Order case Notice that if A’y1 + B’y2 = 0, (7) then since (7) implies 0 = (A’y1 + B’y2)’ = A’’y1 + B’’y2 + A’y1’ + B’y2’ (8) it follows from (7) and (8) that for (6) to hold, we’d need: A’y1’ + B’y2’ = 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)!!

Variation of Parameters – Second Order case A’y1 + B’y2 = 0, (7) A’y1’ + B’y2’ = R(x). (9) Variation of Parameters – Second Order case Now, a solution to system (7), (9) will exist, provided W(x) = y1y2’ – y1’y2  0. Since y1 and y2 are linearly independent on (a,b), W(x) is never zero on (a,b) (see HW problems 23 and 24 for proof).

Variation of Parameters – Second Order case A’y1 + B’y2 = 0, (7) A’y1’ + B’y2’ = R(x). (9) Variation of Parameters – Second Order case Solving system (7), (9) for A’ and B’ gives: A’ = - (y2 R(x))/W(x) B’ = (y1 R(x))/W(x), from which it follows that A(x) = -s (y2 R(x))/W(x) dx B(x) = s (y1 R(x))/W(x) dx. (10) Thus, the solution to (1) is y = A(x)y1 + B(x)y2 with A and B given by (10).

Variation of Parameters – nth Order case Note: For linear equations of order n, a similar technique will work – Look for a solution of the form y = A1(x)y1 + … + An(x)yn with y1, y2, … , yn linearly independent solutions of the corresponding homogeneous problem.

Example 1 Use Variation of Parameters to solve y’’ – y – 2y = e-x. (11) Solution: The auxiliary equation for y’’ – y’ – 2y = 0 is m2 – m – 2 = 0. ) (m-2)(m+1) = 0 ) roots are m = 2, -1. Thus, yc = c1e2x + c2e-x is the general solution to the homogeneous problem corresponding to (11).

A(x) = -s (y2 R(x))/W(x) dx B(x) = s (y1 R(x))/W(x) dx. (10) Example 1 (continued) Taking y1 = e2x and y2 = e-x, which are linearly independent solutions of the homogeneous problem corresponding to (11), (10) implies, with W(x) = y1y2’ – y1’y2,= (e2x)(-e-x) – (2e2x)(e-x) = -3ex, A(x) = - s (y2 R(x))/W(x) dx = - s (e-x e-x)/(-3ex) dx = - s (e-2x)/(-3ex) dx = s 1/3 e-3x dx = -1/9 e-3x + C1

Example 1 (continued) and B(x) = s (y1 R(x))/W(x) dx A(x) = -s (y2 R(x))/W(x) dx B(x) = s (y1 R(x))/W(x) dx. (10) Example 1 (continued) and B(x) = s (y1 R(x))/W(x) dx = s (e2x e-x)/(-3ex) dx = s ex/(-3ex) dx = s -1/3 dx = -1/3 x + C2 Therefore, the solution to (11) is: y = A(x)y1 + B(x)y2 = (-1/9e-3x + C1)e2x + (-1/3 x + C2)e-x = C1e2x + C2e-x -1/9e-x -1/3 x e-x (Re-solve (11) using the Method of Undetermined Coefficients!)