Problems in Linear Differential Equations

PROBLEMS IN LINEAR DIFFERENTIAL EQUATIONS SEBASTIAN VATTAMATTAM 1. Homogeneous Equations Example 1.1. Solve d2y dx2 + ady dx + by = 0; a; b 2 R::::(1) Suppose y = emx::::(2) is a solution of (1). Then, m2emx + amemx + bemx = 0 , m2 + am + b = 0::::(3) So, (2) is a solution of (1) im is a root of (3). Let m1;m2 be the roots of (3). Case 1: m1;m2 are real and distinct. y = c1em1x + c2em2x Case 2: m1 = p + iq;m2 = p 􀀀 iq y = epx(c1 cos qx + c2 sin qx) Case 3: m1 = m2 = m (c1 + c2x)emx Example 1.2. Solve y00 + y0 􀀀 2y = 0 The auxiliary equation is m2 + m 􀀀 2 = 0 ) m1 = 􀀀1;m2 = 2 Date: 15 June 2010. Key words and phrases. Complementary Function, Particular Integral, Complete Solution. 12 SEBASTIAN VATTAMATTAM General Solution is y = c1e􀀀x + c2e2x Example 1.3. Solve y00 􀀀 2y0 + 10y = 0 The auxiliary equation is m2 􀀀 2m + 10 = 0 ) m1 = 1 + 3i;m2 = 1 􀀀 3i General Solution is y = ex(c1 cos 3x + c2 sin 3x Example 1.4. Solve y00 􀀀 2y0 + 10y = 0 given that y(0) = 4; y0(0) = 1 The auxiliary equation is m2 􀀀 2m + 10 = 0 ) m1 = 1 + 3i;m2 = 1 􀀀 3i General Solution is y(x) = ex(c1 cos 3x + c2 sin 3x) y(0) = c1 ) c1 = 4::::(1) y0(x) = ex(c1 cos 3x + c2 sin 3x 􀀀 3c1 sin 3x + 3c2 cos 3x) Therefore y0(0) = c1 + 3c2 ) c1 + 3c2 = 1::::(2) From (1) and (2), c1 = 4; c2 = 􀀀1 Therefore, the solution is y(x) = ex(4 cos 3x 􀀀 sin 3x) Remember d dxexf(x) = ex[f(x) + f0(x)] Example 1.5. Solve y00 + n2y = 0 The auxiliary equation is m2 + n2 = 0 ) m1 = ni;m2 = 􀀀niORDINARY DIFFERENTIAL EQUATIONS 3 General Solution is y = c1 cos nx + c2 sin nx Example 1.6. Solve y00 + 8y0 + 16y = 0 The auxiliary equation is m2 + 8m + 16 = 0 ) m1 = 􀀀4;􀀀4 General Solution isy(x) = (c1 + c2x)e􀀀4x Example 1.7. Solve y00 􀀀 4y0 + 4y = 0; y(0) = 3; y0(0) = 1 General Solution isy(x) = (c1 + c2x)e2x y(0) = c1 ) c1 = 3::::(1) y0(x) = c2e2x + 2(c1 + c2x)e2x Therefore y0(0) = c2 + 2c1 ) c2 + 2c1 = 1::::(2) From (1) and (2), c1 = 3; c2 = 􀀀5 Therefore, the solution is y(x) = (3 􀀀 5x)e2x 2. Non-homogeneous Equations Inverse Operators (1) 1D(x) = Z (x)dx (2) 1 D 􀀀 a(x) = eax Z (x)e􀀀axdx4 SEBASTIAN VATTAMATTAM (3) 1 f(D)eax = 1 f(a)eax; if f(a) 6= 0 (4) If f(a) = 0; then 1 f(D)eax = x 1 f0(a)eax; if f0(a) 6= 0 (5) If f0(a) = 0; then 1 f(D)eax = x2 1 f00(a)eax; if f00(a) 6= 0 (6)1 f(D2) sin(ax + b) = 1 f(􀀀a2) sin(ax + b); if f(􀀀a2) 6= 0 (7) If f(􀀀a2) = 0; then 1 f(D2) sin(ax + b) = x 1 f0(􀀀a2) sin(ax + b); if f0(􀀀a2) 6= 0 Example 2.1. Solve (D 􀀀 2)2y = 8(e2x + sin 2x + x2) (1) Find the Complementary Function (CF) (m 􀀀 2)2 = 0 ) m = 2; 2 CF = (c1 + c2x)e2x Example 2.2. Solve d2x dt2 􀀀 2dx dt + x = et m2 􀀀 2m + 1 = 0 ) m = 1; 1 CF = (A + Bt)et PI = 1 D2 􀀀 2D + 1et = t 1 2D 􀀀 2et = t212et = t2et 2 Note: f(1) = f0(1) = 0 CS = CF + PIORDINARY DIFFERENTIAL EQUATIONS 5 Example 2.3. Solve (D2 􀀀 3D + 2)y = sin 3x m2 􀀀 3m + 2 = 0 ) m = 1; 2 CF = Aex + Be2x PI = 1 D2 􀀀 3D + 2 sin 3x = 1 􀀀9 􀀀 3D + 2 sin 3x = 3D 􀀀 7 􀀀(3D + 7) sin 3x = (3D 􀀀 7) sin 3x 9D2 􀀀 49 = 􀀀1 􀀀81 􀀀 49(3 3 cos 3x 􀀀 7 sin 3x) CS = CF + PI Example 2.4. Solve (D2 + 4D + 5)y = e2x + cos 4x m2 + 4m + 5 = 0 ) m = 􀀀2 i CF = e􀀀2x(Acos x + B sin x) PI1 = 1 D2 + 4D + 5e2x = 1 22 + 4 2 + 5e2x = 1 17e2x PI2 = 1 D2 + 4D + 5 cos 4x = 1 􀀀16 + 4D + 5 cos 4x = 1 4D 􀀀 11 cos 4x = 4D + 11 (4D 􀀀 11)(4D + 11) cos 4x = 4D + 11 16D2 􀀀 121 cos 4x = 4D + 11 16(􀀀16) 􀀀 121 cos 4x = 􀀀1 377[4(􀀀sin 4x)4+11 cos 4x] = 16 377 sin 4x􀀀 11 377 cos 4x CS = CF + PI Example 2.5. Solve d2x dt2 + 2dx dt + 5x = et sin t; x(0) = 0; x0(0) = 1 m2 + 2m + 5 = 0 ) m = 􀀀1 2i CF = e􀀀t(Acos 2t + B sin 2t)6 SEBASTIAN VATTAMATTAM PI = 1 D2 + 2D + 5et sin t = et 1 (D + 1)2 + 2(D + 1) + 5 sin t = et 1 D2 + 4D + 8 sin t = et 1 􀀀1 + 4D + 8 sin t = et 4D 􀀀 7 (4D + 7)(4D 􀀀 7) sin t = et 1 16D2 􀀀 49(4 cos t􀀀7 sin t) = et 􀀀16 􀀀 49(4 cos t􀀀7 sin t) CS is y = CF + PI Example 2.6. Solve (D2 + 3D + 2)y = x2 + sin x m2 + 3m + 2 = 0 ) m = 􀀀1;􀀀2 CF = Ae􀀀x + Be􀀀2x PI1 = 1 D2 + 3D + 2x2 = 12(1 + 3D2 + D2 2 )􀀀1x2 = 12[1 􀀀 (3D2 + D2 2 ) + (3D2 + D2 2 )r 􀀀 :::]x2 = 12[1 􀀀 3D2 􀀀 D2 2 + 9D2 4 ::]x2 = 12[x2 􀀀 3x 􀀀 1 + 9=2] PI2 = 1 D2 + 3D + 2 sin x = 1 􀀀1 + 3D + 2 sin x = 3D 􀀀 1 (3D + 1)(3D 􀀀 1) sin x = 3D 􀀀 1 9D2 􀀀 1 sin x = 1 􀀀10(3D 􀀀 1) sin x = 􀀀1 10 (3 cos x 􀀀 sin x) CS is y = CF + PI1 + PI2ORDINARY DIFFERENTIAL EQUATIONS 7 Example 2.7. Solved2y dx2 + 2dy dx + y = e3x m2 + 2m + 1 = 0 ) m = 􀀀1;􀀀1 CF = (A + Bx)e􀀀x PI = 1 D2 + 2D + 1e3x = 1 9 + 6 + 1e3x CS is y = CF + PI Example 2.8. Solve d2y dx2 􀀀 4y = x sinh x (1) CF m2 􀀀 4 = 0 ) m = 2 CF = Ae2x + Be􀀀2x (2) PI (D2 􀀀 4)y = xex 􀀀 e􀀀x 2 = xex 2 􀀀 xe􀀀x 2 (a) PI1 = 1 2(D2 􀀀 4)(xex) = 12ex 1 (D + 1)2 􀀀 4x = 12ex 1 D2 + 2D 􀀀 3(x) = 12ex 1 􀀀3(1 􀀀 2D3 􀀀 D2 3 )x = 12􀀀ex 3 [1􀀀(2D3 +D2 3 )]􀀀1x = 􀀀ex 6 [1+2D3 +D2 3 ]x = 􀀀ex 6 (x + 2=3)8 SEBASTIAN VATTAMATTAM (b) PI2 = 􀀀1 2(D2 􀀀 4)(xe􀀀x) = 􀀀1 2 e􀀀x 1 (D 􀀀 1)2 􀀀 4x = 􀀀1 2 e􀀀x 1 D2 􀀀 2D 􀀀 3(x) = 12e􀀀x 1 3(1 + 2D3 􀀀 D2 3 )x = 12􀀀e􀀀x 3 [1+(2D3 􀀀D2 3 )]􀀀1x = 􀀀ex 6 [1􀀀2D3 +D2 3 ]x = e􀀀x 6 (x 􀀀 2=3) (c) PI = PI1􀀀PI2 = 􀀀1 6 [xex+23ex􀀀xe􀀀x+23e􀀀x] = 􀀀1 6 [2x(ex 􀀀 e􀀀x 2 )+43(ex + e􀀀x 2 )] = 􀀀1 3 x sinh x 􀀀 29 cosh x Complete Solution is y = CF + PI 3. Equations Reducible to Linear Differential Equations Note Let x = ez ) z = log x: dy dx = dy dz dz dx = dy dz 1x ) xdy dx = Dy;D = d dz x2 d2y dx2 = D(D 􀀀 1)y x3 d3y dx3 = D(D 􀀀 1)(D 􀀀 2)y Example 3.1. Solve x2 d2y dx2 􀀀 xdy dx + y = log x::::(1) Let x = ez ) z = log x: Then xdy dx = Dy;D = d dzORDINARY DIFFERENTIAL EQUATIONS 9 x2 d2y dx2 = D(D 􀀀 1)y x3 d3y dx3 = D(D 􀀀 1)(D 􀀀 2)y (1) becomes D(D 􀀀 1)y 􀀀 Dy + y = z (D 􀀀 1)2y = z (m 􀀀 1)2 = 0 ) m = 1; 1 CF = (A + Bz)ez PI = 1 (D 􀀀 1)2 z = (1􀀀D)􀀀2z = (1+2D+3D2+:::)z = z+2 CS is y = CF + PI = (A + B log x)x + logx + 2 Example 3.2. Solve x2 d2y dx2 􀀀 3xdy dx + 4y = x2 cos(log x):::(1) Let x = ez ) z = log x: Then xdy dx = Dy;D = d dz x2 d2y dx2 = D(D 􀀀 1)y (1) becomesD(D 􀀀 1)y 􀀀 3Dy + 4y = e2z cos z (D2 􀀀 4D + 4)y = e2z cos z m2 􀀀 4m + 4 = 0 ) m = 2; 2 CF = (A + Bz)e2z PI = 1 D2 􀀀 4D + 4e2z cos z = e2z 1 (D + 2)2 􀀀 4(D + 2) + 4 cos z = e2z 1 D2 cos z = e2z 1 􀀀1 cos z = 􀀀e2z cos z CS is y = CF + PIx2(A + B log x) 􀀀 x2 cos(log x)10 SEBASTIAN VATTAMATTAM For any clarication please contact vattamattam@gmail.com Mary Bhavan, Vallikkad Road, Ettumanoor - 686631 E-mail address: vattamattam@gmail.com