Higher Engineering Mathematics - Chapter 13 PDF

13 Linear Differential Equations 1. Definitions. 2. Complete solution. 3. Operator D. 4. Rules for finding the Complementary function. 5. Inverse operator. 6. Rules for finding the particular integral. 7. Working procedure. 8. Two other methods of finding P.I.— Method of variation of parameter...

13 Linear Differential Equations 1. Definitions. 2. Complete solution. 3. Operator D. 4. Rules for finding the Complementary function. 5. Inverse operator. 6. Rules for finding the particular integral. 7. Working procedure. 8. Two other methods of finding P.I.— Method of variation of parameters ; Method of undetermined coefficients. 9. Cauchy’s and Legendre’s linear equations. 10. Linear dependence of solutions. 11. Simultaneous linear equations with constant coefficients. 12. Objective Type of Questions. 13.1 DEFINITIONS Linear differential equations are those in which the dependent variable and its derivatives occur only in the first degree and are not multiplied together. Thus the general linear differential equation of the nth order is of the form n n −1 n−2 d y d y d y n + p1 n −1 + p2 +...+ pny = X, n−2 dx dx dx where p1, p2,..., pn and X are functions of x only. Linear differential equations with constant co-efficients are of the form n n −1 n−2 d y d y d y n + k1 +... + kny = X n −1 + k2 n−2 dx dx dx where k1, k2,..., kn are constants. Such equations are most important in the study of electro-mechanical vibra- tions and other engineering problems. 13.2 (1) THEOREM If y1, y2 are only two solutions of the equation n n −1 n−2 d y d y d y n + k1 n −1 + k2 n−2 +... + kny = 0...(1) dx dx dx then c1y1 + c2y2 (= u) is also its solution. Since y = y1 and y = y2 are solutions of (1). n n −1 n−2 d y1 d y1 d y1 ∴ n + k1 n −1 + k2 n −2 +... + kny1 = 0...(2) dx dx dx d n y2 d n −1 y2 d n − 2 y2 and + k1 + k2 +... + kny2 = 0...(3) dxn dx n −1 dx n − 2 If c1, c2 be two arbitrary constants, then d n (c1 y1 + c2 y2 ) d n −1 (c1 y1 + c2 y2 ) + k1 +... + kn(c1y1 + c2y2) dx n dxn −1 471  472 HIGHER ENGINEERING MATHEMATICS ⎛ d n y1 n −1 d y1 ⎞ ⎛ d n y2 n −1 d y2 ⎞ = c1 ⎜ n + k1 n −1 +... + kn y1 ⎟ + c2 ⎜ n + k1 n −1 +... + kn y2 ⎟ ⎝ dx dx ⎠ ⎝ dx dx ⎠ = c1(0) + c2(0) = 0 [By (2) and (3)] d nu d n −1u i.e., n + k1 n −1 +... + kn y = 0...(4) dx dx This proves the theorem. (2) Since the general solution of a differential equation of the nth order contains n arbitrary constants, it follows, from above, that if y1, y2, y3,..., yn, are n independent solutions of (1), then c1y1 + c2y2 +... + cnyn (= u) is its complete solution. (3) If y = v be any particular solution of n n −1 d y d y n + k1 n −1 +... + kny = X...(5) dx dx then d nv d n −1v +... + k v = X...(6) n + k1 n −1 n dx dx d n (u + v) d n −1 (u + v) Adding (4) and (6), we have +... + kn(u + v) = X + k1 dx n dx n −1 This shows that y = u + v is the complete solution of (5). The part u is called the complementary function (C.F.) and the part v is called the particular integral (P.I.) of (5). ∴ the complete solution (C.S.) of (5) is y = C.F. + P.I. Thus in order to solve the question (5), we have to first find the C.F., i.e., the complete solution of (1), and then the P.I., i.e. a particular solution of (5). 13.3 OPERATOR D 2 3 Denoting d , d , d etc. by D, D2, D3 etc., so that dx dx2 dx3 2 dy 3 = Dy, d y = D2y, d x = D3y etc., the equation (5) above can be written in the symbolic form (Dn + k1Dn 1 + dx dx 2 dx 3... + kn)y = X, i.e., f (D)y = X, where f (D) = Dn + k1Dn 1 +... + kn, i.e., a polynomial in D. Thus the symbol D stands for the operation of differentiation and can be treated much the same as an algebraic quantity i.e., f (D) can be factorised by ordinary rules of algebra and the factors may be taken in any order. For instance 2 d y dy +2 3y = (D2 + 2D 3) y = (D + 3)(D 1)y or (D 1)(D + 3) y. dx 2 dx 13.4 RULES FOR FINDING THE COMPLEMENTARY FUNCTION n n −1 n−2 d y d y d y To solve the equation n + k1 n −1 + k2 n−2 +... + kn y = 0...(1) dx dx dx where k s are constants. The equation (1) in symbolic form is (Dn + k1Dn 1 + k2Dn 2 +... + kn)y = 0...(2) Its symbolic co-efficient equated to zero i.e. Dn + k1Dn 1 + k2Dn 2 +... + kn = 0 is called the auxiliary equation (A.E.). Let m1, m2,..., mn be its roots. Case I. If all the roots be real and different, then (2) is equivalent to (D m1) (D m2)... (D mn)y = 0...(3)  LINEAR DIFFERENTIAL EQUATIONS 473 dy Now (3) will be satisfied by the solution of (D mn)y = 0, i.e., by mn y = 0. dx − mn x This is a Leibnitz s linear and I.F. = e −m x m x ∴ its solution is y e n = cn, i.e., y = cne n Similarly, since the factors in (3) can be taken in any order, it will be satisfied by the solutions of (D m1) y = 0, (D m2)y = 0 etc. i.e., by y = c1em1 x , y = c2 em2 x etc. mn x Thus the complete solution of (1) is y = c1em1 x + c2 em2 x +... + cne...(4) Case II. If two roots are equal (i.e., m1 = m2), then (4) becomes m3 x mn x y = (c1 + c2) em1 x + c3 e +... + cn e mx m x m x y = C e 1 + c3 e 3 +... + cn e n [∵ c1 + c2 = one arbitrary constant C] It has only n 1 arbitrary constants and is, therefore, not the complete solution of (1). In this case, we proceed as follows : The part of the complete solution corresponding to the repeated root is the complete solution of (D m1) (D m1)y = 0 Putting (D m1)y = z, it becomes (D m1) z = 0 or dz m1 z = 0 dx − m1 x m1 x This is a Leibnitz s linear in z and I.F. = e. ∴ its solution is ze − m1 x = c1 or z = c 1e m1 x dy m1 x Thus (D m1)y = z = c1e or m1 y = c1 e...(5) dx Its I.F. being e − m1 x , the solution of (5) is y e − m1 x = ∫ ce m1 x m x 1 dx + c2 = c1x + c2 or y = (c1x + c2) e 1 m x Thus the complete solution of (1) is y = (c1x + c2) em1 x + c3 em3 x +... + cn e n If, however, the A.E. has three equal roots (i.e., m1 = m2 = m3), then the complete solution is m x y = (c1x2 + c2x + c3) em1 x + c4 em4 x +... + cn e n Case III. If one pair of roots be imaginary, i.e., m1 = α + iβ, m2 = α iβ, then the complete solution is m3 x mn x y = c1e(α + iβ)x + c2e(α iβ)x + c3 e +... + cn e m3 x mn x = eαx(c1eiβx + c2e iβx) + c3 e +... + cn e m x m x = eαx[c1(cos βx + i sin βx) + c2 (cos βx i sin βx)] + c3 e 3 +... + cn e n [∵ by Euler s Theorem, eiθ = cos θ + i sin θ] m3 x mn x = eax (C1 cos βx + C2 sin βx) + c3 e +... + cn e where C1 = c1 + c2 and C2 = i(c1 c2). Case IV. If two points of imaginary roots be equal i.e., m1 = m2 = α + iβ, m3 = m4 = α iβ, then by case II, the complete solution is mn x y = eax[(c1x + c2) cos βx + (c3x + c4) sin βx] +... + cn e. 2 Example 13.1. Solve d x + 5 dx + 6x = 0, given x(0) = 0, dx (0) = 15. (V.T.U., 2010) dt 2 dt dt Solution. Given equation in symbolic form is (D2 + 5D + 6) x = 0. Its A.E. is D2 + 5D + 6 = 0, i.e., (D + 2) (D + 3) = 0 whence D = 2, 3. 2t 3t dx 2t 3t ∴ C.S. is x = c1e + c2e and = 2ae 3c2e dt When t = 0, x = 0. ∴ 0 = c1 + c2 (i) When t = 0, dx/dt = 15 ∴ 15 = 2c1 3c2...(ii)  474 HIGHER ENGINEERING MATHEMATICS Solving (i) and (ii), c1 = 15, c2 = 15. Hence the required solution is x = 15 (e 2t e 3t). 2 Example 13.2. Solve d x + 6 dx + 9x = 0. dt 2 dt Solution. Given equation in symbolic form is (D2 + 6D + 9) = 0 ∴ A.E. is D2 + 6D + 9 = 0, i.e., (D + 3)2 = 0 whence D = 3, 3. Hence the C.S. is x = (c1 + c2t) e 3t. Example 13.3. Solve (D3 + D2 + 4D + 4) = 0. Solution. Here the A.E. is D3 + D2 + 4D + 4 = 0 i.e., (D2 + 4) (D + 1) = 0 ∴ D= 1, ± 2i. Hence the C.S. is y = c1e x + e0x (c2 cos 2x + c3 sin 2x) i.e., y = c1e x + c2 cos 2x + c3 sin 2x. Example 13.4. Solve (i) (D4 4D + 4) y = 0 (Bhopal, 2008) (ii) (D2 + 1)3 y = 0 where D ≡ d/dx. Solution. (i) The A.E. equation is D4 4D2 + 4 = 0 or (D2 2)2 = 0 ∴ D2 = 2, 2 i.e., D = ± 2,± 2. 2x − 2x Hence the C.S. is ((c1 + c2x) e + (c3 + c4x) e ) [Roots being repeated] 2 3 (ii) The A.E. equation is (D + 1) = 0 ∴ D = ± i, ± i, ± i. Hence the C.S. is y = e [(c1 + c2x + c3x2) cos x + (c4 + c5x + c6x2) sin x] ox i.e., y = (c1 + c2 + c3x2) cos x + (c4 + c5 x + c6x2) sin x. 4 Example 13.5. Solve d x + 4x = 0. 4 dt Solution. Given equation in symbolic form is (D4 + 4) x = 0 ∴ A.E. is D4 + 4 = 0 or (D4 + 4D2 + 4) 4D2 = 0 or (D2 + 2)2 (2D)2 = 0 or (D2 + 2D + 2) (D2 2D + 2) = 0 ∴ either D2 + 2D + 2 = 0 or D2 2D + 2 = 0 − 2 ± (− 4) 2 ± (− 4) whence D= and i.e., D = 1 ± i and 1 ± i. 2 2 Hence the required solution is x = e t (c1 cos t + c2 sin t) + et (c3 cos t + c4 sin t). PROBLEMS 13.1 Solve : d2 x dx dx(0) 1. −4 dt 2 dt + 13x = 0, x(0), dt = 2. (V.T.U., 2008) d3 y d2 y dy 2. y″ 2y′ + 10y = 0, y (0) = 4, y′ (0) = 1. 3. 3 + − y = 0. (Calicut, 2013) dx dx2 dx 3 3 2 d y 4. + y = 0. (C.S.V.T.U., 2012 ; V.T.U., 2000 S) 5. d y − 3 d y + 3 dy y = 0. (V.T.U., 2014) 3 3 2 dx dx dx dx d4 y d2 y 6. 4 +8 2 + 16y = 0. (J.N.T.U., 2005) 7. (4D4 8D3 7D2 + 11D + 6)y = 0. (V.T.U., 2008) dx dx 8. (D2 + 1)2 (D 1)y = 0. 9. (D2 + D + 1)2 (D 2) y = 0. (Delhi, 2012) 4 d x 10. If = m4x, show that x = c1 cos mt + c2 sin mt + c3 cosh mt + c4 sinh mt. dt 4  LINEAR DIFFERENTIAL EQUATIONS 475 13.5 INVERSE OPERATOR 1 (1) Definition. X is that function of x, not containing arbitrary constants which when operated f ( D) upon by f (D) gives X. ⎧ 1 ⎫ i.e., f (D) ⎨ X⎬ = X ⎩ f ( D) ⎭ 1 Thus X satisfies the equation f (D)y = X and is, therefore, its particular integral. f ( D) Obviously, f (D) and 1/f (D) are inverse operators. 1 (2) D X = Xdx ∫ 1 Let X =y...(i) D 1 dy Operating by D, D X = Dy i.e., X = D dx Integrating both sides w.r.t. x, y = ∫ X dx, no constant being added as (i) does not contain any constant. 1 Thus D X=∫ X dx. 1 (3) X = e ∫ Xe ax ax dx. D a 1 Let X = y...(ii) D−a 1 Operating by D a, (D a). X = ( D − a) y. D−a dy or X= − ay , i.e., dy − ay = X which is a Leibnitz s linear equation. dx dx ∴ I.F. being e ax, its solution is ∫ Xe − ax ye ax = dx, no constant being added as (ii) doesn t contain any constant. 1 X = y = eax ∫ Xe − ax Thus dx. D−a 13.6 RULES FOR FINDING THE PARTICULAR INTEGRAL dn y d n −1 y d n−2 y Consider the equation + k1 + k2 +... + kn y = X dx n dx n−1 dx n −2 which is symbolic form of (D + k1Dn n 1 + k Dn 2 +... + k )y = X. 2 n 1 ∴ P.I. = n X. D + k1 Dn −1 + k2 Dn−2 +... + kn Case I. When X = eax Since Deax = aeax D2eax = a2eax................................................ Dneax = aneax ∴ (D + k1D +... + kn)eax = (an + k1an n n 1 1 +... + kn)eax, i.e., f (D)eax = f (a)eax  476 HIGHER ENGINEERING MATHEMATICS 1 1 1 f (a) eax 1 ax Operating on both sides by , f ( D) eax = or eax = f (a) e f ( D) f ( D) f ( D) f ( D) ∴ dividing by f (a), 1 1 ax eax = e provided f (a) ≠ 0...(1) f(D) f(a) If f(a) = 0, the above rule fails and we proceed further. Since a is a root of A.E. f (D) = Dn + k1Dn 1 +... + kn = 0. ∴ D a is a factor of f (D). Suppose f (D) = (D a) φ (D), where φ (a) ≠ 0. Then 1 1 1 ax 1 1 ax eax =. e =. e [By (1)] f ( D) D − a φ( D) D − a φ(a) 1 1 1 eax =. eax ∫e ax =.. e− ax dx [By §13.5 (3)] φ(a) D − a φ(a) 1 ax 1 1 1 ∫ dx = x φ(a) e eax = x ax = e i.e., eax...(2) φ(a) f(D) f ′(a) ⎡ ∵ f ′( D) = ( D − a) φ′( D) + 1. φ( D) ⎢ ∴ f ′( a) = 0 × φ′( a) + φ( a) ⎣ 1 1 If f ′(a) = 0, then applying (2) again, we get eax = x 2 eax , provided f′(a) ≠ 0...(3) f(D) f ″ (a) and so on. Example 13.6. Find the P.I. of (D2 + 5D + 6)y = ex. 1 1 ex Solution. P.I. = e x [Put D = 1] = ex =. D2 + 5 D + 6 12 + 5.1 + 6 12 Example 13.7. Find the P.I. of (D + 2) (D 1) 2y = e 2x + 2 sinh x. 1 1 Solution. P.I. = [ e−2 x + 2 sinh x ] = [ e −2 x + e x − e − x ] 2 2 (D + 2)( D − 1) ( D + 2)( D − 1) Let us evaluate each of these terms separately. 1 1 ⎡ 1 ⎤ e−2 x =.⎢ e −2 x ⎥ (D + 2)( D − 1)2 D + 2 ⎣⎢ ( D − 1)2 ⎦⎥ 1 1 1 1 =. e −2 x =. e −2 x D + 2 (−2 − 1) 2 9 D+2 1 1 x ⎡ d ⎤ =. x. e−2 x = e−2 x ⎢⎣∵ dD ( D + 2) = 1⎥⎦ 9 1 9 1 1 1 1 1 x2 x ⎡ d2 ⎤ ex =. e x =. x2. e x = e ⎢∵ ( D − 1)2 = 2⎥ ( D + 2)( D − 1)2 1 + 2 ( D − 1)2 3 2 6 ⎣ dD 2 ⎦ 1 e− x and 2 e− x = 1 e− x = ( D + 2)( D − 1) (−1 + 2)(− 1 − 1)2 4 x −2 x x2 x 1 − x Hence, P.I. = e + e + e. 9 6 4 Case II. When X = sin (ax + b) or cos (ax + b). Since D sin (ax + b) = a cos (ax + b) D2 sin (ax + b) = a2 sin (ax + b) D3 sin (ax + b) = a3 cos (ax + b)  LINEAR DIFFERENTIAL EQUATIONS 477 D4 sin (ax + b) = a4 sin (ax + b) i.e., D2 sin (ax + b) = ( a2) sin (ax + b) (D2)2 sin (ax + b) = ( a2)2 sin (ax + b) In general (D2)r sin (ax + b) = ( a2)r sin (ax + b) ∴ f (D2) sin (ax + b) = f ( a2) sin (ax + b) Operating on both sides 1/f (D2), 1 1 2. f ( D2 ) sin (ax + b) = f ( a2) sin (ax + b) f (D ) f ( D2 ) 1 or sin (ax + b) = f ( a2) sin (ax + b) f ( D2 ) 1 1 ∴ Dividing by f ( a2). sin (ax + b) provided f ( a2) ≠ 0 sin (ax + b) =...(4) f(D ) f( a 2 ) 2 If f( a2) = 0, the above rule fails and we proceed further. Since cos (ax + b) + i sin (ax + b) = ei (ax + b) [Euler s theorem] 1 1 ∴ sin (ax + b) = I.P. of ei (ax + b) [Since f ( a2) = 0 ∴ by (2)] f (D ) 2 f ( D2 ) 1 = I.P. of x 2 ei ( ax + b) where D 2 = a2 f ′( D ) 1 1 ∴ sin (ax + b) = x sin (ax + b) provided f ′( a2) ≠ 0...(5) 2 f(D ) f ′( −a 2 ) 1 1 If f ′( a2) = 0, 2. sin (ax + b) = x2 sin (ax + b), provided f ″( a2) ≠ 0, and so on. f(D ) f ′′( − a 2 ) 1 1 Similarly, cos (ax + b) = cos (ax + b), provided f ( a2) ≠ 0 f (D 2 ) f( a ) 2 1 1 If f ( a2) = 0, 2 cos (ax + b) = x. cos (ax + b), provided f ′( a2) ≠ 0. f(D ) f ′( a 2 ) 1 1 If f ′( a2) = 0, 2 cos (ax + b) = x2 cos (ax + b), provided f ′′( a2) ≠ 0 and so on. f(D ) f ′′( a 2 ) Example 13.8. Find the P.I. of (D3 + 1)y = cos (2x 1). 1 Solution. P.I. = cos (2x 1) [Put D2 = 22 = 4] 3 D +1 1 = cos (2x 1) [Multiply and divide by 1 + 4D] D(−4) + 1 (1 + 4 D) 1 = cos (2 x − 1) = (1 + 4D). cos (2x 1) [Put D2 = 22 = 4] (1 − 4 D)(1 + 4 D) 2 1 − 16D 1 1 = (1 + 4D) cos (2 x − 1) = [cos (2x 1) + 4D cos (2x 1)] 1 − 16(−4) 65 1 = [cos (2x 1) 8 sin (2x 1)]. 65 d3 y dy Example 13.9. Find the P.I. of +4 = sin 2x. (C.S.V.T.U., 2011) 3 dx dx Solution. Given equation in symbolic form is (D3 + 4D)y = sin 2x  478 HIGHER ENGINEERING MATHEMATICS 1 ∴ P.I. = sin 2 x [∵ D2 + 4 = 0 for D2 = 22, ∴ Apply (5) 477] D( D2 + 4) ⎡ d 3 2 ⎤ = x 1 sin 2 x ⎢⎣∵ dD [ D + 4 D] = 3 D + 4⎥⎦ 3 D2 + 4 [Put D2 = − 22 = − 4] 1 x = x sin 2 x = − sin 2 x. 3(− 4) + 4 8 Case III. When X = xm. 1 Here P.I. = xm = [f (D)] 1 xm. f ( D) Expand [f (D)] 1 in ascending powers of D as far as the term in Dm and operate on xm term by term. Since the (m + 1)th and higher derivatives of xm are zero, we need not consider terms beyond Dm. d2 y dy Example 13.10. Find the P.I. of + = x2 + 2x + 4. 2 dx dx Solution. Given equation in symbolic form is (D2 + D)y = x2 + 2x + 4. 1 1 ∴ P.I. = ( x2 + 2 x + 4) = (1 + D)−1 ( x2 + 2 x + 4) D( D + 1) D 1 = (1 − D + D2 −...)( x2 + 2 x + 4) = 1 [ x2 + 2 x + 4 − (2 x + 2) + 2] D D x3 = ∫( x2 + 4)dx = + 4 x. 3 Case IV. When X = eax V, V being a function of x. If u is a function of x, then D(eaxu) = eaxDu + aeaxu + eax(D + a)u D2(eaxu) = aaxD2u + 2aeaxDu + a2eaxu = eax(D + a)2u and in general, Dn(eaxu) = eax(D + a)nu ∴ f (D)(eaxu) = eax f (D + a)u Operating both sides by 1/f (D), 1 1. f ( D)(eax u) = [ eax f ( D + a)u] f ( D) f ( D) 1 eaxu = [ eax f ( D + a)u] f ( D) 1 1 1 Now put f ( D + a)u = V , i.e., u = V , so that eax V = ( eax V ) f ( D + a) f ( D + a) f ( D) i.e., 1 1...(6) (eax V) = eax V. f(D) f(D + a) Example 13.11. Find P.I. of (D2 2D + 4)y = ex cos x. 1 Solution. P.I. = e x cos x [Replace D by D + 1] 2 D − 2D + 4 x 1 1 = e cos x = e x cos x [Put D2 = 12 = 1] 2 2 ( D + 1) − 2( D + 1) + 4 D +3 x 1 1 = e cos x = e x cos x. −1+3 2  LINEAR DIFFERENTIAL EQUATIONS 479 Case V. When X is any other function of x. 1 Here P.I. = X. f ( D) If f (D) = (D m1)(D m2)... (D mn), resolving into partial fractions, 1 A1 A2 An = + +... + f ( D) D − m1 D − m2 D − mn ⎡ A1 A2 An ⎤ P.I. = ⎢ +... + X mn ⎥⎦ ∴ + ⎣ D − m1 D − m2 D − 1 1 1 = A1 X + A2 X +... + An X D − m1 D − m2 D − mn ∫ Xe ∫ Xe ∫ Xe m1 x − m1 x m2 x − m2 x mn x − mn x = A1. e dx + A2. e dx +... + An. e dx [By §13.5...(3)] Obs. This method is a general one and can, therefore, be employed to obtain a particular integral in any given case. 13.7 WORKING PROCEDURE TO SOLVE THE EQUATION dn y dn − 1 y dy n + k1 n −1 +... + kn − 1 + kn y = X dx dx dx of which the symbolic form is n n −1 ( D + k1 D +... + kn − 1 D + kn ) y = X. Step I. To find the complementary function (i) Write the A.E. i.e., Dn + k1Dn 1 +... + kn 1D + kn = 0 and solve it for D. (ii) Write the C.F. as follows : Roots of A.E. C.F. 1. m1, m2, m3... (real and different roots) c1 em1 x + c2 em2 x + c3 em3 x +... 2. m1, m1, m3... (two real and equal roots) (c1 + c2 x)em1 x + c3 em3 x +... 3. m1, m1, m1, m4... (three real and equal roots) (c1 + c2 x + c3 x2 ) em1 x + c4 em4 x +... m3 x 4. α + iβ, α iβ, m3... (a pair of imaginary roots) eax(c1 cos βx + c2 sin βx) + c3 e +... m5 x 5. α ± iβ, α ± iβ, m5... (2 pairs of equal imaginary eax[(c1 + c2x) cos βx + (c3 + c4x) sin βx] + c5 e +... roots) Step II. To find the particular integral 1 1 1 From symbolic form P.I. = X= or X n n−1 D + k1 D +... + kn−1 D + kn f ( D) φ( D2 ) (i) When X = eax 1 P.I. = eax , put D = a, [f (a) ≠ 0] f ( D) 1 = x eax , put D = a, [ f (a) = 0, f ′(a) ≠ 0] f ′( D) 1 = x2 eax , put D = a, [ f ′(a) = 0, f ″(a) ≠ 0] f ( D) ′′ and so on. where f ′(D) = diff. coeff. of f (D) w.r.t. D f ″(D) = diff. coeff. of f ′(D) w.r.t. D, etc.  480 HIGHER ENGINEERING MATHEMATICS (ii) When X = sin (ax + b) or cos (ax + b). 1 P.I. = sin (ax + b) [or cos (ax + b)], put D2 = a2 [φ( a2) ≠ 0] φ( D2 ) 1 = x sin (ax + b) [or cos (ax + b)], put D2 = a2 [φ( a2) = 0, φ′( a2) ≠ 0] φ′( D2 ) 1 = x2 sin (ax + b) [or cos (ax + b)], put D2 = a2 [φ′( a2) ≠ 0, φ″( a2) ≠ 0] φ″( D2 ) and so on. where φ′(D2) = diff. coeff. of φ(D2) w.r.t. D, φ″(D2) = diff. coeff. of φ′(D2) w.r.t. D, etc. (iii) When X = xm, m being a positive integer. 1 P.I. = xm = [ f(D)] 1 xm f ( D) To evaluate it, expand [ f(D)] 1 in ascending powers of D by Binomial theorem as far as Dm and operate on m x term by term. (iv) When X = eaxV, where V is a function of x. 1 1 P.I. = eax V = eax V f ( D) f ( D + a) 1 and then evaluate V as in (i), (ii), and (iii). f ( D + a) (v) When X is any function of x. 1 P.I. = X f ( D) 1 Resolve into partial fractions and operate each partial fraction on X remembering that f ( D) 1 X = eax ∫ Xe− ax dx. D−a Step III. To find the complete solution Then the C.S. is y = C.F. + P.I. d2 y dy Example 13.12. Solve + + y = (1 e x) 2. dx 2 dx Solution. Given equation in symbolic form is (D2 + D + 1)y = (1 e x )2 (i) To find C.F. 1 Its A.E. is D2 + D + 1 = 0, ∴ D= (− 1 + 3i ) 2 x/2 ⎛ 3 3 ⎞ Thus C.F. = e ⎜⎜ c1 cos x + c2 sin x⎟ ⎟ ⎝ 2 2 ⎠ (ii) To find P.I. 1 1 P.I. = (1 − 2e x + e2 x ) = ( e0 x − 2e x + e2 x ) 2 2 D + D+1 D + D+1 1 1 1 2 x e2 x = e0 x − 2. ex + e2 x = 1 − e + 02 + 0 + 1 12 + 1 + 1 22 + 2 + 1 3 7 ⎛ 3 3 ⎞ 2 e2 x (iii) Hence the C.S. is y = e x/2 ⎜⎜ c1 cos x + c2 sin x ⎟ + 1 − ex +. 2 2 ⎟ 3 7 ⎝ ⎠ LINEAR DIFFERENTIAL EQUATIONS 481 Example 13.13. Solve y″ + 4y′ + 4y = 3 sin x + 4 cos x, y(0) = 1 and y′(0) = 0. (J.N.T.U., 2003) Solution. Given equation in symbolic form is (D2 + 4D + 4)y = 3 sin x + 4 cos x (i) To find C.F. Its A.E. is (D + 2)2 = 0 where D = 2, 2 ∴ C.F. = (c1 + c2x)e 2x. (ii) To find P.I. 1 1 P.I. = (3 sin x + 4 cos x) = (3 sin x + 4 cos x) 2 D + 4D + 4 − 1 + 4D + 4 4D − 3 (4 D − 3) = (3 sin x + 4 cos x) = (3 sin x + 4 cos x) 16 D − 9 2 − 16 − 9 −1 = {3(4 cos x 3 sin x) + 4( 4 sin x 3 cos x)} = sin x 25 (iii) C.S. is y = (c1 + c2) e 2x + sin x When x = 0, y = 1, ∴ 1 = c1 Also y′ = c2e 2x + (c1 + c2x)( 2)e 2x + cos x. When x = 0, y′ = 0, ∴ 0 = c2 2c1 + 1, i.e., c2 = 1. Hence the required solution is y = (1 + x) e 2x + sin x. Example 13.14. Solve (D 2)2 = 8(e2x + sin 2x + x2). (V.T.U., 2012) Solution. (i) To find C.F. Its A.E. is (D 2)2 = 0, ∴ D = 2, 2. Thus C.F. = (c1 + c2x) e2x. (ii) To find P.I. ⎡ 1 1 1 ⎤ P.I. = 8 ⎢ e2 x + sin 2 x + x2 ⎥ 2 ⎣⎢ ( D − 2) ( D − 2)2 ( D − 2)2 ⎦⎥ 1 1 2x Now 2 e2 x = x 2 e [∵ by putting D = 2, (D 2)2 = 0, 2(D 2) = 0] ( D − 2) 2(1) x 2 e2 x =. 2 1 1 1 2 sin 2 x = 2 sin 2 x = 2 sin 2 x ( D − 2) D − 4D + 4 ( − 2 ) − 4D + 4 1 1 ⎛ − cos 2 x ⎞ 1 = − 4 ∫ sin 2 x dx = − 4 ⎜⎝ 2 ⎟ = 8 cos 2 x ⎠ −2 2 1 1 ⎛ D⎞ 1 ⎡ ⎛ D ⎞ (− 2)( − 3) ⎛ D⎞ ⎤ 2 and x2 = ⎜⎝ 1 − 2 ⎟⎠ x2 = ⎢1 + (− 2) ⎜ ⎟ + ⎜⎝ 2 ⎟⎠ +...⎥ x − ( D − 2)2 4 4⎢ ⎝ 2⎠ 2! ⎥⎦ ⎣ 1 ⎛ 3 D2 ⎞ 1 ⎛ 2 3⎞ = ⎜ 1 D +...⎟ x2 = ⎜⎝ x + 2 x + 2 ⎟⎠ + + 4 ⎝ 4 ⎠ 4 Thus P.I. = 4x2e2x + cos 2x + 2x2 + 4x + 3. (iii) Hence the C.S. is y = (c1 + c2x) e2x + 4x2e2x + cos 2x + 2x2 + 4x + 3. Example 13.15. Find the complete solution of y″ 2y′ + 2y = x + ex cos x. (U.P.T.U., 2002) Solution. Given equation in symbolic form is (D2 2D + 2) y = x+ ex cos x (i) To find C.F. 2 ± (4 − 8) Its A.E. is D2 2D + 2 = 0 ∴ D= = 1 ± i. 2 Thus C.F. = ex (c1 cos x + c2 sin x) 482 HIGHER ENGINEERING MATHEMATICS (ii) To find P.I. 1 1 P.I. = 2 ( x) + 2 (e x cos x) D − 2D + 2 D − 2D + 2 −1 1⎡ ⎛ D2 ⎞ ⎤ 1 = ⎢1 − ⎜ D − ⎟ ⎥ ( x) + e x (cos x) 2⎢ ⎜ 2 ⎟⎥ 2 ( D + 1) − 2( D + 1) + 2 ⎣ ⎝ ⎠⎦ ⎛ 2⎞ = 1 1 + D − D x + ex 1 cos x [Case of failure] ⎜ ⎟ 2 2⎝ 2 ⎠ D +1 1 1 1 xe x 1 xe x = ( x + 1 − 0) + e x ⋅ x cos x = ( x + 1) + ∫ cos x dx = ( x + 1) + sin x 2 2D 2 2 2 2 1 xe x (iii) Hence the C.S. is y = ex(c1 cos x + c2 sin x) + ( x + 1) + sin x. 2 2 d2 y dy Example 13.16. Solve 2 −3 + 2y = xe3x + sin 2x. dx dx (V.T.U., 2008 ; Kottayam, 2005 ; U.P.T.U., 2003) Solution. Given equation in symbolic form is (D2 3D + 2)y = xe3x + sin 2x (i) To find C.F. Its A.E. is D2 3D + 2 = 0 or (D 2)(D 1) = 0 whence D = 1, 2. Thus C.F. = c1ex + c2 e2x (ii) To find P.I. 1 1 1 P.I. = 2 ( xe3 x + sin 2 x) = 2 (e3 x. x) + 2 (sin 2 x) D − 3D + 2 D − 3D + 2 D − 3D + 2 3x 1 1 = e. 2 ( x) + (sin 2 x) ( D + 3) − 3( D + 3) + 2 − 4 − 3D + 2 −1 3x 1 3D − 2 e3 x ⎡ ⎧⎪ 3 D + D2 ⎫⎪ ⎤ (3 D − 2) = e. ( x) − (sin 2 x) =. ⎢1 + ⎨ ⎬⎥ x− (sin 2 x) D2 + 3 D + 2 9 D2 − 4 2 ⎩⎪ 2 ⎭⎪ ⎥⎦ 9(− 4) − 4 ⎣⎢ e3 x ⎛ 3D ⎞ 1 e3 x ⎛ 3⎞ 1 = ⎜ 1−...⎟ x + (6 cos 2 x − 2 sin 2 x) = x− ⎟ + (3 cos 2 x − sin 2 x) 2 ⎝ 2 ⎠ 40 2 ⎜⎝ 2 ⎠ 20 ⎛ x 3⎞ 1 (iii) Hence the C.S. is y = c1ex + c2e2x + e3x ⎜ − ⎟ + (3 cos 2x sin 2x). ⎝ 2 4 ⎠ 20 d2 y Example 13.17. Solve − 4y = x sinh x. dx2 Solution. Given equation in symbolic form is (D2 4) y = x sinh x. (i) To find C.F. Its A.E. is D2 4 = 0, whence D = ± 2. Thus C.F. = c1e2x + c2e 2x (ii) To find P.I. 1 1 ⎛ e x − e− x ⎞ 1⎡ 1 1 ⎤ P.I. = x sinh x = x⎜ ⎟= ⎢ 2 ex ⋅ x − e− x ⋅ x ⎥ 2 D −4 D2 − 4 ⎜⎝ 2 ⎟ 2 ⎢D − 4 ⎠ ⎣ 2 D −4 ⎥⎦ 1⎡ x 1 1 ⎤ 1⎡ x 1 1 ⎤ = ⎢e x − e− x x ⎥ = ⎢e x − e− x x ⎥ 2 ⎢⎣ ( D + 1)2 − 4 ( D − 1)2 − 4 ⎥⎦ 2 ⎢⎣ D2 + 2 D − 3 D2 − 2 D − 3 ⎥⎦ LINEAR DIFFERENTIAL EQUATIONS 483 ⎡ x −1 −1 ⎤ ⎧⎪ ⎛ 2 D D2 ⎞ ⎫⎪ e− x ⎧⎪ ⎛ 2 D D2 ⎞ ⎫⎪ = 1⎢e ⎨1 − ⎜ ⎬.x− ⎨1 + ⋅ x⎥ 2 ⎢− 3 + 3 ⎟⎠ ⎪ ⎜ 3 − 3 ⎟⎬ ⎥ ⎪⎩ ⎝ 3 ⎭ −3 ⎪⎩ ⎝ ⎠ ⎪⎭ ⎣ ⎦ 1 ⎡ x ⎛ 2D ⎞ ⎛ 2D ⎞ ⎤ 1 ⎡ ⎛ 2⎞ ⎛ 2⎞ ⎤ = e 1+ +...⎟ x − e− x ⎜ 1 − +...⎟ x⎥ = − ⎢ e x ⎜ x + ⎟ − e− x ⎜ x − ⎟ ⎥ 6 ⎢⎣ ⎜⎝ 3 ⎠ ⎝ 3 ⎠ ⎦ 6 ⎣ ⎝ 3⎠ ⎝ 3⎠ ⎦ x ⎛ e x − e− x ⎞ 2 ⎛ e x + e− x ⎞ x 2 = − ⎜ ⎟− ⎜ ⎟ = − sinh x − cosh x. 3 ⎜ 2 ⎟ 9⎜ 2 ⎟ 3 9 ⎝ ⎠ ⎝ ⎠ x 2 (iii) Hence the C.S. is y = c1e2x + c2e 2x sinh x − cosh x. 3 9 Example 13.18. Solve (D2 1) y = x sin 3x + cos x. Solution. (i) To find C.F. Its A.E. is D2 1 = 0, whence D = ± 1. ∴ C.F. = c1ex + c2e x (ii) To find P.I. P.I. = 1 1 1 ( x sin 3 x + cos x) = x (I.P. of e3ix ) + cos x 2 2 2 D −1 D −1 D −1 1 1 ⎡ 1 ⎤ cos x = I.P. of 2 e3ix. x + 2 cos x = I.P. of ⎢ e3ix 2 x⎥ − D −1 (− 1) − 1 ⎣⎢ ( D + 3i) − 1 ⎥⎦ 2 [Replacing D by D + 3i] ⎡ 3ix 1 ⎤ cos x = I.P. of ⎢ e 2 x⎥ − ⎣⎢ D + 6iD − 10 ⎦⎥ 2 ⎡ −1 ⎤ 1 ⎛ 3iD D2 ⎞ cos x = I.P. of ⎢ e3ix. ⎜1 − − ⎟ x⎥ − [Expand by Binomial theorem] ⎢ − 10 ⎜ 5 10 ⎟ ⎥ 2 ⎢⎣ ⎝ ⎠ ⎥⎦ ⎡ 1 ⎛ 3iD ⎞ ⎤ cos x ⎡ e3ix ⎛ 3i ⎞ ⎤ cos x = I.P. of ⎢ e3ix. ⎜ 1 + 5 +...... ⎟ x⎥ − = I.P. of ⎢ − ⎜ x + ⎟⎥ − ⎣ − 10 ⎝ ⎠ ⎦ 2 ⎢⎣ 10 ⎝ 5 ⎠⎥ 2 ⎦ ⎡− 1 ⎛ 3i ⎞ ⎤ cos x = I.P. of ⎢ (cos 3 x + i sin 3 x) ⎜ x + ⎟ ⎥ − ⎣ 10 ⎝ 5 ⎠⎦ 2 1 ⎡⎛ 3 sin 3 x ⎞ ⎛ 3 ⎞ ⎤ cos x = − I.P. of ⎢⎜⎝ x cos 3 x − ⎟ + i ⎝⎜ x sin 3 x + 5 cos 3 x⎠⎟ ⎥ − 2 10 ⎣ 5 ⎠ ⎦ 1 ⎛ 3 ⎞ cos x = − 10 ⎜ x sin 3 x + 5 cos 3 x ⎟ − 2. ⎝ ⎠ 1 (iii) Hence the C.S. is y = c1ex + c2e x (5x sin 3x + 3 cos 3x + 25 cos x). 50 d2 y dy Example 13.19. Solve −2 + y = xex sin x. 2 dx dx (V.T.U., 2013 ; C.S.V.T.U., 2011 S ; J.N.T.U., 2006) Solution. Given equation in symbolic form is (D2 2D + 1) y = xex sin x (i) To find C.F. Its A.E. is D2 2D + 1 = 0, i.e., (D 1)2 = 0 ∴ D = 1, 1. Thus C.F. = (c1 + c2x)ex 484 HIGHER ENGINEERING MATHEMATICS (ii) To find P.I. 1 1 P.I. = e x. x sin x = e x. x sin x 2 ( D − 1) ( D + 1 − 1)2 x 1 1 = e 2 x sin x = e x ∫ x sin x dx [Integrate by parts] D D 1 = ex ⎡ x(− cos x) − 1. (− cos x) dx ⎤ = e x ∫ ∫ [− x cos x + sin x] dx D⎣ ⎦ x = e ⎡− x sin x − ∫ 1. sin x dx − cos x ⎤ = ex[ x sin x { } cos x cos x] ⎣ ⎦ x = e (x sin x + 2 cos x). (iii) Hence the C.S. is y = (c1 + c2x) ex ex (x sin x + 2 cos x). Example 13.20. Solve (D4 + 2D2 + 1) y = x2 cos x. (Nagarjuna, 2008 ; Rajasthan, 2005) Solution. (i) To find C.F. Its A.E. is (D2 + 1)2 = 0 whose roots are D = ± i, ± i ∴ C.F. = (c1 + c2x) cos x + (c3 + c4x) sin x (ii) To find P.I. 1 1 P.I. = 2 2 x 2 cos x = 2 2 x 2 (Re.P. of eix) ( D + 1) ( D + 1) ⎧⎪ 1 ⎫⎪ ⎧ 1 ⎪⎫ = Re.P. of ⎨ 2 eix. x2 ⎬ = Re.P. of ⎪⎨ eix x2 ⎬ 2 2 2 ⎩⎪ ( D + 1) ⎭⎪ ⎪⎩ [( D + i) + 1] ⎪⎭ ⎧⎪ 1 ⎫⎪ ⎡ ⎧⎪ 1 ⎛ i ⎞ −2 ⎫⎪ ⎤ = Re.P. of ⎨eix 2 2 x2 ⎬ = Re.P. of ⎢ eix ⎨− 2 ⎜⎝ 1 − 2 D⎟⎠ x2 ⎬ ⎥ ⎩⎪ ( D + 2iD) ⎪⎭ ⎢⎣ ⎩⎪ 4 D ⎭⎪ ⎥⎦ ⎡ 1 1 ⎧⎪ iD ⎛ iD ⎞ 2 ⎫⎪ ⎤ = Re.P. of ⎢ − eix. 2 ⎨ 1 + 2 + 3 ⎜ ⎟ +...⎬ x2 ⎥ ⎢ 4 D ⎩⎪ 2 ⎝ 2 ⎠ ⎭⎪ ⎥⎦ ⎣ ⎧ 1 ix 1 ⎛ 2 3 ⎞⎫ ⎧⎪ 1 ix 1 ⎛ x3 3 ⎞ ⎫⎪ = Re.P. of ⎨− 4 e. 2 ⎜ x + 2ix − 2 ⎟ ⎬ = Re.P. of ⎨− e. ⎜ + ix 2 − x⎟ ⎬ ⎩ D ⎝ ⎠⎭ 4 D⎝ 3 2 ⎠ ⎭⎪ ⎩⎪ 1 ⎧⎪ ⎛ x4 ix3 3 2 ⎞ ⎪⎫ − 1 = − Re.P. of ⎨ eix ⎜ + − x ⎟⎬ = Re.P. of {(cos x + i sin x)(x4 + 4ix3 9x2)} 4 ⎝ 12 3 4 ⎠ 48 ⎩⎪ ⎭⎪ 1 = − [(x4 9x2) cos x 4x3 sin x] 48 1 (iii) Hence the C.S. is y = (c1 + c2x) cos x + (c3 + c4x) sin x + [4x3 sin x x2 (x2 9) cos x}. 48 Example 13.21. Solve (D2 4D + 4)y = 8x2e2x sin 2x. (M.T.U., 2013 ; V.T.U., 2013 ; Calicut, 2012 ; J.N.T.U., 2006) Solution. (i) To find C.F. Its A.E. is D2 4D + 4 = 0 i.e., (D 2)2 = 0. ∴ D = 2, 2 ∴ C.F. = (c1 + c2x) e2x (ii) To find P.I. 1 1 P.I. = (8 x2 e2 x sin 2 x) = 8 e2 x ( x2 sin 2 x) 2 2 ( D − 2) ( D + 2 − 2) 1 1 = 8e2 x 2 ( x2 sin 2 x) = 8 e2 x. ∫x 2 sin 2 x dx D D LINEAR DIFFERENTIAL EQUATIONS 485 1 ⎧ 2 ⎛ − cos 2 x ⎞ ⎛ − cos 2 x ⎞ ⎫ = 8 e2 x. ⎨x ∫ 2 x ⎜⎝ ⎟ dx ⎬ D ⎩ ⎜⎝ 2 ⎟− ⎠ 2 ⎠ ⎭ 2x 1 ⎧⎪ x2 sin 2 x sin 2 x ⎫⎪ = 8e ⎨− cos 2 x + x − ∫ 1. dx ⎬ D 2 2 2 ⎩⎪ ⎪⎭ ⎧ 2 2x ⎪ x x cos 2 x ⎫⎪ = 8 e ∫ ⎨− cos 2 x + sin 2 x + ⎬ dx ⎪⎩ 2 2 4 ⎪ ⎭ ⎡ ⎧⎪ − x2 sin 2 x sin 2 x ⎫⎪ ⎧ x ⎫ sin 2 x ⎤ = 8 e2 x ⎢ ⎨ − ∫ ( − x) dx ⎬ + ⎨ ∫ sin 2 x dx ⎬ + ⎥ ⎢⎣ ⎩⎪ 2 2 2 ⎭⎪ ⎩ 2 ⎭ 8 ⎥ ⎦ ⎡⎛ − x 2 1 ⎞ ⎤ = 8 e2 x ⎢⎜ + ⎟ sin 2 x + ∫ x sin 2 x dx⎥⎥ ⎢⎣⎜⎝ 4 8⎟ ⎠ ⎦ ⎡⎛ 1 x 2 ⎞ ⎛ − cos 2 x ⎞ ⎛ − cos 2 x ⎞ ⎤ = 8 e2 x ⎢⎜ − ⎟ sin 2 x + x ⎜ ⎟ − ∫ 1.⎜ ⎟ dx ⎥ ⎢⎣⎝⎜ 8 4 ⎟ ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎡⎛ 1 x 2 ⎞ x cos 2 x sin 2 x ⎤ = 8 e2 x ⎢ ⎜ − ⎟ sin 2 x − + ⎥ ⎢⎣⎜⎝ 8 4 ⎟ ⎠ 2 4 ⎥ ⎦ = e2x[(3 2x2) sin 2x 4x cos 2x] (iii) Hence the C.S. is y = e2x [c1 + c2x + (3 2x2) sin 2x 4x cos 2x]. 2 Example 13.22. Solve d y + a 2 y = sec ax. dx 2 Solution. Given equation in symbolic form is (D2 + a2)y = sec ax. (i) To find C.F. Its A.E. is D2 + a2 = 0 ∴ D = ± ia. Thus C.F. = c1 cos ax + c2 sin ax. (ii) To find P.I. 1 1 P.I. = sec ax = sec ax [Resolving into partial fractions] 2 2 ( D + ia)( D − ia) D +a = 1 ⎡ 1 1 ⎤ sec ax = 1 ⎡ 1 sec ax − 1 ⎤ sec ax ⎥ ⎢ − ⎥ 2ia ⎣ D − ia D + ia ⎦ ⎢ 2ia ⎣ D − ia D + ia ⎦ 1 ⎡ 1 ⎤ Now sec ax = eiax ∫ sec ax. e − iax dx ⎢∵ D − a X = e ax ∫ Xe− ax dx⎥ D − ia ⎣ ⎦ cos ax − i sin ax ⎛ i ⎞ = eiax ∫ dx = eiax ∫ (1 − i tan ax) dx = eiax ⎜ x + log cos ax ⎟ cos ax ⎝ a ⎠ Changing i to i, we have 1 ⎧ i ⎫ sec ax = e− iax ⎨ x − log cos ax ⎬ D + ia ⎩ a ⎭ 1 ⎡ iax ⎧ i ⎫ − iax ⎧ i ⎫⎤ Thus P.I. = 2ia ⎢ e ⎨ x + a log cos ax ⎬ − e ⎨ x − log cos ax ⎬⎥ a ⎣ ⎩ ⎭ ⎩ ⎭⎦ x eiax − e−iax 1 eiax + e−iax x 1 = + 2 log cos ax. = sin ax + 2 log cos ax. cos ax. a 2i a 2 a a (iii) Hence the C.S. is y = c1 cos ax + c2 sin ax + (1/a)x sin ax + (1/a2) cos ax log cos ax.  486 HIGHER ENGINEERING MATHEMATICS PROBLEMS 13.2 Solve : d2 y dy 1. 2 −6 + 9 y = 6e3x + 7e 2x log 2 (V.T.U., 2005) dx dx d2 y dy dy 2. +4 + 5y = 2 cosh x. Also find y when y = 0, = 1 at x = 0. dx2 dx dx d2 y dy d2 x dx 3. + = cos x. (Rohtak, 2012) 4. 2 +2 + 3 x = sin t. 2 dx dt dx dt 2 d y dy 5. 2 +3 + 2 y = 4 cos2 x. (Bhopal, 2002 S) 6. (D2 D + 2)y = sin 2x. (V.T.U., 2013) dx dx d3 y d2 y dy x d2 y dy 7. 3 +2 2 + =e + sin 2x. (C.S.V.T.U., 2012) 8. +2 + y = e2x cos2 x. (Delhi, 2002) dx dx dx dx2 dx d2 y 9. (D3 5D2 + 7D 3)y = e2x cosh x. (Nagarjuna, 2008) 10. − y = ex + x2ex. (Nagpur, 2009) dx2 11. (D3 D)y = 2x + 1 + 4 cos x + 2ex. (Mumbai, 2006) 12. (D3 6D2 + 11D 6)y = 2x + cos 2x. (V.T.U., 2013) d2 y dy 13. (D2 + 1)2 y = x4 + 2 sin x cos 3x. (Madras, 2006) 14. 2 +5 + 6y = e 2x sin 2x. (Bhopal, 2008) dx dx d2 y dy 15. (D2 3D + 2)y = sin (ex) (U.P.T.U., 2012) 16. +2 + 3 y = ex cos x. (V.T.U., 2010) dx2 dx 17. (D2 + 4D + 3)y = e x sin x + xe3x. (Raipur, 2005 ; Anna, 2002 S) d2 y d4 y 18. 2 + 2 y = x2 e3x + ex cos 2x. 19. − y = cos x cosh x. dx dx4 d2 y 20. (D3 + 2D2 + D)y = x2e2x + sin2 x. (P.T.U., 2003) 21. + 16 y = x sin 3x. (V.T.U., 2010 S) dx2 22. (D2 + 2D + 1)y = x cos x. (Rajasthan, 2006) 23. (D2 1)y = x sin x + (1 + x2)ex. d2 y dy x 24. 2 +3 + 2 y = ee. (C.S.V.T.U., 2011) 25. (D2 + a2)y = tan ax. (V.T.U., 2005) dx dx 13.8 TWO OTHER METHODS OF FINDING P.I. I. Method of variation of parameters. This method is quite general and applies to equations of the form y″ + py′ + qy = X...(1) y2 X y1 X where p, q, and X are functions of x. It gives P.I. = − y1 ∫ W dx + y2 ∫ W dx...(2) where y1 and y2 are the solutions of y″ + py′ + qy = 0...(3) y1 y2 and W = is called the Wronskian* of y1, y2. y1′ y2′ Proof. Let the C.F. of (1) be y = c1y1 + c2y2 Replacing c1, c2 (regarded as parameters) by unknown functions u(x) and v(x), let the P.I. be y = uy1 + vy2...(4) Differentiating (4) w.r.t. x, we get y′ = uy′1 + vy′2 + u′y1 + v′y2 *Named after the Polish mathematician and philosopher Hoene Wronsky (1778 1853).  LINEAR DIFFERENTIAL EQUATIONS 487 = uy1′ + vy2′...(5) on assuming that u′y1 + v′y2 = 0...(6) Differentiate (4) and substitute in (1). Then noting that y1 and y2, satisfy (3), we obtain u′y1′ + v′y2′ = X...(7) Solving (6) and (7), we get y2 X y X u′ = − , v′ = 1 where W = y1 y2′ y2 y1 ′ W W y2 X y X Integrating u = ∫ dx, v = ∫ 1 dx. Substituting these in (4), we get (2). W W Example 13.23. Using the method of variation of parameters, solve d2 y + 4y = tan 2x. (Delhi, 2012 ; Rohtak, 2012 ; V.T.U., 2012 ; Bhopal, 2007) dx 2 Solution. Given equation in symbolic form is (D2 + 4)y = tan 2x. (i) To find C.F. Its A.E. is D2 + 4 = 0, ∴ D = ± 2i Thus C.F. is y = c1 cos 2x + c2 sin 2x. (ii) To find P.I. Here y1 = cos 2x, y2 = sin 2x and X = tan 2x y1 y2 cos 2 x sin 2 x ∴ W= = =2 y1′ y2′ − 2 sin 2 x 2 cos 2 x y2 X y X Thus, P.I. = − y1 dx + y2 ∫ 1 dx ∫ W W sin 2 x tan 2 x cos 2 x tan 2 x = cos 2x ∫ dx + sin 2 x ∫ dx 2 2 1 1 = − cos 2 x ∫ (sec 2 x − cos 2 x)dx + sin 2 x ∫ sin 2 x dx 2 2 1 1 = − cos 2x[ log (sec 2x + tan 2x) sin 2x] sin 2x cos 2x 4 4 1 = − cos 2x log (sec 2x + tan 2x) 4 1 Hence the C.S. is y = c1 cos 2x + c2 sin 2x cos 2x log (sec 2x + tan 2x). 4 Example 13.24. Solve, by the method of variation of parameters, d2y/dx2 y = 2/(1 + ex). (V.T.U., 2005 ; Hissar, 2005) Solution. Given equation is D2 1 = 2/(1 + ex) A.E. is D2 1 = 0, D = ± 1, ∴ C.F. = c1ex + c2e x Here y1 = ex, y2 = e x and X = 2/(1 + ex) y1 y2 ex e− x ∴ W= = = e xe x ex e x = 2. y1′ y2′ ex − e− x y2 X y1 X e− x 2 ex 2 Thus P.I. = − y1 ∫ dx + y2 ∫ dx = − e x ∫ ⋅ dx + e− x ∫ − 2 ⋅ 1 + e x dx W W − 2 1 + ex ⎛ 1 1 ⎞ x ⎡ −x e− x ⎤ = ex ∫ ⎜ ex 1 + ex ⎟ ⎜ − ⎟ dx − e −x log (1 + e x ) = e ⎢ e − ∫ dx ⎥ − e− x log (1 + e x ) −x ⎝ ⎠ ⎣⎢ e +1 ⎦⎥ = e [ e + log (e + 1)] e log (1 + e ) = 1 + e log (e + 1) e x log (ex + 1) x x x x x x x Hence C.S. is y = c1ex + c2e x 1 + ex log (e x + 1) e x log (ex + 1).  488 HIGHER ENGINEERING MATHEMATICS Example 13.25. Solve by the method of variation of parameters y″ 6y′ + 9y = e3x/x2. (D.T.U., 2013 ; Nagpur, 2009 ; C.S.V.T.U., 2009) Solution. Given equation is (D2 6D + 9)y = e3x/x2 A.E. is D2 6D + 9 = 0 i.e. (D 3)2 = 0 ∴ C.F. = (c1 + c2x) e3x Here y1 = e3x, y2 = xe3x and X = e3x/x2 y1 y2 e3 x xe3 x ∴ W= = = e6x. y1′ y2′ 3e3 x e3 x + 3 xe3 x y2 X y1 X xe3 x e3 x e3 x e3 x Thus P.I. = − y1 ∫ W dx + y2 ∫ W dx = e3x ∫ e6 x x2 dx + xe3 x ∫e 6x ⋅ x2 dx dx = e3x x∫+ xe3 x x −2 dx = ∫ e3x (log x + 1) Hence C.S. is y = (c1 + c2x)e3x e3x (log x + 1). Example 13.26. Solve, by the method of variation of parameters, y″ 2y′ + y = ex log x. (Rohtak, 2011 ; V.T.U., 2006 ; Madras, 2003) Solution. Given equation in symbolic form is (D2 2D + 1) y = ex log x (i) To find C.F. Its A.E. is (D 1)2 = 0, ∴ D = 1, 1 Thus C.F. is y = (c1 + c2x)ex (ii) To find P.I. Here y1 = ex, y2 = xex and X = ex log x y1 y2 ex xe x ∴ W= = = e2x y1′ y2′ ex (1 + x) e x y2 X y X Thus P.I. = y1 ∫ W W ∫ dx + y2 1 dx xe x. e x log x e x. e x log x ∫ x log x dx + xe ∫ log x dx x = ex ∫ e 2x dx + xe x ∫ e 2x dx = ex ⎛x 2 ⎞2 1 x ⎛ 1 ⎞ = ex ⎜⎜ ⎝ 2 log x − ∫x.2 dx ⎟⎟ + x. e x ⎜ x log x − ⎠ ⎝ ∫ x. x dx ⎟⎠ ⎛ x2 x2 ⎞ x