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4.6 Variation of Parameters: Finding a Particular Solution d2 y dy a2(x) 2 + a1(x) + a0(x)y = f (x) dx dx (1)Divide by a2(x) to put in standard form d2 y dy + q(x)y = g(x) + p(x) 2 dx dx (2)Find the general solution of the homogeneous equation: d2 y dy + p(x) + q(x)y = 0 dx2 dx yh(x) = c1y1(x) + c2y2(x) where y1(x), y2(x) are two solutions of the homogenous equation. 1 (3) Look for a particular solution of dy d2 y + q(x)y = g(x) + p(x) 2 dx dx of the form C1 C2 yp (x) = V1(x)y1(x) + V2(x)y2(x) Our goal is to find V1(x), V2(x) so that this is a solution of the D.E. 2 (4) Compute yp′ yp = V1y1 + V2y2 yp′ = (V1y1′ + V1′y1) + (V2y2′ + V2′y2) Regroup yp′ = (V1y1′ + V2y2′ ) + (V1′y1 + V2′y2) To simplify, we will require V1′y1 + V2′y2 = 0 Equation ♯ 1 (5) Compute yp′′ yp′ = (V1y1′ + V2y2′ ) + (0) ↑ F rom Eq ♯1 yp′′ = V1y1′′ + V1′y1′ + V2y2′′ + V2′y2′ Substitute into the D.E. 3 yp′′ + p(x)yp′ + q(x)yp = g(x) (V1y1′′ + V1′y1′ + V2y2′′ + V2′y2′ ) + p(x)(V1y1′ + V2y2′ ) + q(x)(V1y1 + V2y2) = g(x) Regroup 0 0 ↓ ↓ V1(y1′′ + p(x)y1′ + q(x)y1) + V2(y2′′ + p(x)y2′ + q(x)y2) +(V1′y1′ + V2′y2′ ) = g(x) y1, y2 solve the hom.eq y1′′ + p(x)y1′ + q(x)y1 y2′′ + p(x)y2′ + q(x)y2 So ց = 0 = 0 V1′y1′ + V2′y2′ = g(x) Equation ♯2 4 (6) Solve the two equations ♯1,♯ 2 for V1′, V2′ y1V1′ + y2V2′ = 0 y1′ V1′ + y2′ V2′ = g(x) (7) Integrate V1′ and V2′ to find V1 and V2. (8) The yp = V1y1 + V2y2 5 Summary d2 y dy + p(x) + q(x)y = g(x) (Std Form) dx2 dx (1) Find homogeneous solution yh = C1y1 + C2y2 (2)Taking y1, y2, form the equations y1V1′ + y2V2′ = 0 y1′ V1′ + y2′ V2′ = g(x) ↑ F romD.E. in std f orm and solve for V1′ and V2′ (3) Integrate V1′, V2′ to find V1, V2. Then yp = V1y1 + V2y2 6 Ex y ′′ + y = sec(x) (Std. Form) (1) Homogenous soln. y ′′ + y = 0 m2 + 1 = 0 m2 m −1 √ = ± −1 = ±i = yh = C1 cos(x) + C2 sin(x) (2) Look for a particular solution of the form yp = V1(x) cos(x) + V2(x) sin(x) ↓ Equations ♯ 1, 2 ↓ y1V1′ + y2V2′ =0 y1′ V1′ + y2′ V2′ = g(x) cos(x)V1′ + sin(x)V2′ =0 − sin(x)V1′ + cos(x)V2′ = sec(x) 7 Using Cramois Rule to find V1′, V2′: 1 ↓ cos(x) 0 sin(x) sec(x) cos(x) 0 − sec(x) sin(x) = V1′ = 2(x) − (− sin2 (x)) cos(x) sin(x) cos − sin(x) cos(x) − tan(x) = − tan(x) = 1 R V1 = − tan(x)dx = − ln |sec(x)| + d1 V1′ (Pick d1 = 0; only need 1 particular soln.) cos(x) 0 − sin(x) sec(x) cos(x) sec(x) − 0 1 = = =1 V2′ = cos(x) sin(x) cos2(x) + sin2(x) 1 − sin(x) cos(x) R V2 = 1dx = x + d2 (Pick d2 = 0; only need 1 particular Soln.) 8 yp = V1(x) cos(x) + V2(x) sin(x) = (− ln | sec(x)|) cos(x) + (x) sin(x) (3) The general solution of the D.E. is y = C1 cos(x) + C2 sin(x) + (− ln | sec(x)|) cos(x) + x sin(x) Note: V1 = − ln | sec(x)| + d1 V2 = x + d2 If we keep the constants of integration here,the y = (− ln |sec(x)| + d1) cos(x) + (x + d2) sin(x) = d1 cos(x) + d2 sin(x) + (− ln | sec(x)|) cos(x) +x sin(x) is the general solution of the D.E.! 9 Ex x2y ′′ − 4xy ′ + 6y = 1 4 ′ 6 1 y − y + 2 y = 2 (StdF orm) x x x ′′ Two solutions of the homogeneous equation are y1 = x2 and y2 = x3. Constrict a particular solution to the nonhomogenous equation. yp = V1y1 + V2y2 yp = V1x2 + V2x3 y1V1′ + y2V2′ = 0 y1′ V1′ + y2′ V2′ = 1 x2 x2V1′ + x3V2′ = 0 2xV1′ + 3x2V2′ = 10 1 x2 x3 0 1 x3 2 0− 2 2 3x −x −1 ′ x x = V1 = = 4 = 3 x2 x3 3x4 − 2x4 x x 2x 3x2 Z 1 x−2 −3 + d1 = 2 + d1 V1 = −x dx = − −2 2x 2 0 x 1 2x 2 1 1 ′ x = = V2 = x2 x3 3x4 − 2x4 x4 2x 3x2 V2 = Z −3 −1 x −4 + d2 = 3 + d2 x dx = −3 3x −1 3 1 1 1 1 2 So yp = ( 2 )x + ( 3 )x = − = 2x 3x 2 3 6 A particular solution of x2y ′′ − 4xy ′ + 6y = 1 is yp = 1 6 11