Higher Engineering Mathematics - BS Grewal

486 Higher Engineering Mathematics Now try the following exercise 6. In a galvanometer the deflection θ satisfies d2θ dθ the differential equation +4 + 4 θ = 8. Exercise 189 Further problems on dt 2 dt Solve the equation for θ given that when t = 0, differential equations of the form θ = dθ = 2. d2y dy dt [θ = 2(t e−2t + 1)] a d x2 + b d x + cy = f (x) where f (x) is a constant or polynomial. In Problems 1 and 2, find the general solutions of 51.4 Worked problems on differential the given differential equations. d2 y dy equations of the form 2 dx2 5 1. + − 3y = 6 d2y dy dx a + b + cy = f (x) where d x2 d 1 x f (x) is an exponential function y = Ae 2 x + B e−3x − 2 2. d2 y + dy − 2y = 3x − 2 Problem 4. Solve the equation 6 dx2 4 d2 y dy 3e4x dx dx2 2 1 − dx + y = given the boundary 3 y = Ae x + B e− x −2 − 3 x dy 2 dx conditions that when x = 0, y= − 2 and = 4 1 3 3 In Problems 3 and 4 find the particular solutions of Using the procedure of Section 51.2: the given differential equations. 3. d2 y + dy − 4y = 8; when x = 0, y=0 and d2 y dy 3 dx2 dx dx2 2 dy = 0. (i) − + y = 3e4x in D-operator form is dx (D2 − 2D + 1)y = 3e4x . dx 4 3 y = 2 (3e− x + 4ex ) −2 (ii) Substituting m for D gives the auxiliary 7 equation m2 − 2m + 1 =0. Factorizing gives: 4. d2 y − dy + 4y = 3x − 1; when x = 0, (m − 1)(m − 1) = 0, from which, m = 1 twice. 9 dx2 12 (iii) Since the roots are real and equal the C.F., dx u = (Ax + B)ex. y = 0 and d y = − 4 dx 3 (iv) Let the particular integral, v = ke4x (see Table 51.1(c)). y = − 2 + 3 x e 2 x + 2 + 3 x 4 3 4 5. The charge q in an electric circuit at time t sat- (v) Substituting v = ke4x into (D2 − 2D + 1)v = 3e4x gives: isfies the equation d2q + R dq + 1 q=E, L dt2 dt C where L, R, C and E are constants. Solve the (D2 − 2D + 1)ke4x = 3e4x i.e. D2(ke4x ) − 2D(ke4x ) + 1(ke4x ) = 3e4x equation given L = 2H , C = 200 ×10−6 F and i.e. 16ke4x − 8ke4x + ke4x = 3e4x E = 250 V, when (a) R = 200 and (b) R is negligible. Assume that when t = 0, q = 0 and dq =0 Hence 9ke4x = 3e4x , from which, k = 1 dt 3 ⎤ ke4x 1 e4x . ⎡ q = 1 − 5t + 1 e−50t ⎥⎥⎦ Hence the P.I., v = = 3 20 2 20 ⎢⎢⎣ (a) (vi) The general solution is given by y = u + v, i.e. (b) q = 1 (1 − cos 50t ) 20 y = (Ax + B)ex + 1 e4x . 3 (vii) When x = 0, y= − 2 thus 3

Second order differential equations of the form a d2 y + b dy + cy = f (x) 487 dx2 dx − 2 = (0 + B)e0 + 1 e0, from which, B = −1. =2 ke 3 x 9 x + 3 − ke 3 x 3 x + 1 3 3 2 4 2 2 dy dx = ( Ax + B)ex + ex (A) + 4 e4x . 3 3 3 dy 1 13 4 −3 k x e 2 x = 5e 2 x When x = 0, = 4 , thus = B + A + dx 3 3 3 from which, A = 4, since B = −1. i.e. 9 kx e 3 x + 3 x − 3 3 x − 3 x 2 2 2 Hence the particular solution is: 6ke 2 xke 2 ke 2 y = (4x − 1)ex + 1 e4x − 3kx e 3 x = 5e 3 x 3 2 2 Problem 5. Solve the differential equation Equating coefficients of e 3 x gives: 5k = 5, from 2 d2 y dy 3 2 dx2 − dx − 3y = 5e 2 x . which, k = 1. Hence the P.I., v = kxe 3 x = xe 3 x. 2 2 Using the procedure of Section 51.2: (v) The general solution is y = u + v, i.e. d2 y dy 3 x y = Ae 3 x + Be−x + xe 3 x. 2 dx2 dx 2 2 2 (i) − − 3 y = 5e in D-operator form is (2D2 − D − 3)y = 5e 3 x . d2 y −4dy 2 dx2 dx Problem 6. Solve + 4y = 3e2x . (ii) Substituting m for D gives the auxiliary equation 2m2 − m − 3 = 0. Factorizing gives: (2m − 3)(m + 1) = 0, from which, m = 3 or Using the procedure of Section 51.2: 2 m = −1. Since the roots are real and different then 3 d2 y dy the C.F., u = Ae 2 x + Be−x. (i) dx2 − 4 dx + 4 y = 3e2x in D-operator form is (iii) Since e 3 x appears in the C.F. and in the (D2 − 4D +4)y = 3e2x . 2 right hand side of the differential equation, let the (ii) Substituting m for D gives the auxiliary equation m2 − 4m + 4 = 0. Factorizing gives: P.I., v = k x e 3 x (see Table 51.1(c), snag case (i)). 2 (m − 2)(m − 2) = 0, from which, m = 2 twice. (iv) Substituting v = k x e 3 x into (2D2 − D − 3)v = (iii) Since the roots are real and equal, the C.F., 2 u =(Ax + B)e2x. 5e 3 x gives: (2D2 − D − 3)k x e 3 x = 5e 3 x . (iv) Since e2x and xe2x both appear in the C.F. let the 2 2 2 P.I., v = kx2e2x (see Table 51.1(c), snag case (ii)). D kx e 3 x = (kx) 3 e 3 x + e 3 x (k), (v) Substituting v = kx2e2x into (D2 − 4D + 4)v = 2 2 2 2 3e2x gives: (D2 − 4D + 4)(kx2 e2x ) = 3e2x by the product rule, = ke 3 x 3 x + 1 D(kx2e2x ) = (kx2 )(2e2x ) + (e2x )(2kx) 2 2 = 2ke2x (x2 + x) D2 kx e 3 x =D ke 3 x 3 x + 1 D2(kx2e2x ) = D[2ke2x (x2 + x)] 2 2 2 = (2ke2x )(2x + 1) + (x2 + x)(4ke2x ) = 2ke2x (4x + 1 + 2x2) = ke 3 x 3 2 2 Hence (D2 − 4D + 4)(kx2 e2x ) = [2ke2x (4x + 1 + 2x2)] + 3 x + 1 3 ke 3 x − 4[2ke2x (x2 + x)] + 4[kx2 e2x ] 2 2 2 = 3e2x = ke 3 x 9 x + 3 2 4 Hence (2D2 − D − 3) kx e 3 x 2

488 Higher Engineering Mathematics from which, 2ke2x = 3e2x and k= 3 2 51.5 Worked problems on Hence the P.I., v = kx2e2x = 3 x2e2x . 2 differential equations of the d2y dy (vi) The general solution, y = u + v, i.e. form a dx2 + b dx + cy= f (x) y = (Ax + B)e2x + 3 x2e2x where f (x) is a sine or cosine 2 function Now try the following exercise Exercise 190 Further problems on Problem 7. Solve the differential equation differential equations of the form 2 d2 y + dy − 5y = 6 sin 2x. dx2 3 d2y dy a dx2 + b +cy = f (x) where f (x) is an dx dx exponential function Using the procedure of Section 51.2: In Problems 1 to 4, find the general solutions of the (i) 2 d2 y + dy − 5y = 6 sin 2x in D-operator form given differential equations. dx2 3 dx is (2D2 + 3D − 5)y = 6 sin 2x 1. d2 y − dy − 6y = 2ex dx2 dx (ii) The auxiliary equation is 2m2 + 3m −5 = 0, from which, y = Ae3x + B e−2x − 1 ex 3 2. d2 y − 3 dy − 4y = 3e−x (m − 1)(2m + 5) = 0, dx2 dx i.e. m = 1 or m = −25 y = Ae4x + B e− x − 3 x e−x 5 (iii) Since the roots are real and different the C.F., d2 y 3. dx2 + 9y = 26e2x u = Aex + Be− 5 x. 2 [ y = A cos 3x + B sin 3x + 2e2x ] (iv) Let the P.I., v = A sin 2x + B cos 2x (see Table 51.1(d)). 4. d2 y − dy + y = t (v) Substituting v = A sin 2x + B cos 2x into 9 dt2 6 (2D2 + 3D −5)v = 6 sin 2x gives: 12e 3 (2D2 + 3D−5)(A sin 2x + B cos 2x) = 6 sin 2x. dt y = ( At + B)e 1 t + 2 t 2e 1 t 3 3 3 D(A sin 2x + B cos 2x) In problems 5 and 6 find the particular solutions of = 2 A cos 2x − 2B sin 2x the given differential equations. D2(A sin 2x + B cos 2x) d2 y dy 1 = D(2 A cos 2x − 2B sin 2x) 5 dx2 dx 4 5. +9 − 2y = 3ex ; when x = 0, y = = −4 A sin 2x − 4B cos 2x Hence (2D2 + 3D −5)(A sin 2x + B cos 2x) and d y = 0. dx y= 5 e−2x − e 1 x + 1 ex = −8 A sin 2x − 8B cos 2x + 6 A cos 2x 5 44 4 − 6B sin 2x − 5 A sin 2x − 5B cos 2x 6. d2 y − 6 dy + 9y = 4e3t ; when t = 0, y =2 = 6 sin 2x dt dt 2 Equating coefficient of sin 2x gives: and d y = 0 [ y = 2e3t (1 − 3t + t 2)] dt −13 A − 6B = 6 (1)

Second order differential equations of the form a d2 y + b dy + cy = f (x) 489 dx2 dx Equating coefficients of cos 2x gives: D[x(C sin 4x + D cos 4x)] = x(4C cos 4x − 4D sin 4x) 6A − 13B = 0 (2) + (C sin 4x + D cos 4x)(1), by the product rule 6 × (1)gives : −78 A − 36B = 36 (3) D2[x(C sin 4x + D cos 4x)] 13 × (2)gives : 78 A − 169B = 0 (4) = x(−16C sin 4x − 16D cos 4x) + (4C cos 4x − 4D sin 4x) (3) + (4)gives : − 205B = 36 + (4C cos 4x − 4D sin 4x) from which, B = −36 205 Substituting B = −36 into equation (1) or (2) Hence (D2 + 16)[x(C sin 4x + D cos 4x)] 205 = −16Cx sin 4x −16Dx cos 4x + 4C cos 4x − 4D sin 4x + 4C cos 4x − 4D sin 4x gives A = −78 + 16Cx sin 4x + 16Dx cos 4x 205 = 10 cos4x, Hence the P.I., v = −78 sin 2x − 36 cos 2x. 205 205 (vi) The general solution, y = u +v, i.e. y = Aex + Be− 5 x 2 − 2 (39 sin 2x + 18 cos 2x) 205 Problem 8. Solve d2 y + 16y = 10 cos4x given i.e. −8D sin 4x + 8C cos 4x = 10 cos4x dx2 dy y = 3 and = 4 when x = 0. Equating coefficients of cos 4x gives: dx 8C = 10, from which, C = 10 = 5 84 Using the procedure of Section 51.2: Equating coefficients of sin 4x gives: −8D = 0, from which, D = 0. (i) d2 y + 16y = 10 cos 4x in D-operator form is dx2 Hence the P.I., v = x 5 4 sin 4x . (D2 + 16)y = 10 cos4x (vi) The general solution, y = u +v, i.e. (ii) The auxili√ary equation is m2 + 16 = 0, from y = A cos 4x + B sin 4x + 5 x sin 4x which m = −16 = ± j 4. 4 (iii) Since the roots are complex the C.F., (vii) When x = 0, y = 3, thus u =e0(A cos 4x + B sin 4x) 3 = A cos 0 + B sin 0 + 0, i.e. A = 3. i.e. u =Acos 4x + B sin4x d y = −4 A sin 4x + 4B cos 4x dx (iv) Since sin 4x occurs in the C.F. and in the right hand side of the given differential equa- + 5 x (4 cos 4x ) + 5 sin 4x tion, let the P.I., v = x(C sin 4x + D cos 4x) (see 4 4 Table 51.1(d), snag case—constants C and D are used since A and B have already been used in the When x = 0, d y = 4, thus C.F.). dx 4 = −4 A sin 0 + 4B cos 0 + 0 + 5 sin 0 4 (v) Substituting v = x(C sin 4x + D cos 4x) into i.e. 4 =4B, from which, B = 1 (D2 + 16)v = 10 cos 4x gives: Hence the particular solution is (D2 + 16)[x(C sin 4x + D cos 4x)] y = 3 cos 4x + sin 4x + 5 x sin 4x = 10 cos 4x 4

490 Higher Engineering Mathematics Now try the following exercise Exercise 191 Further problems on given by: differential equations of the form y = e−4t (A cos 2t + B sin 2t ) d2y dy + 15 (sin 4t − 8 cos4t ) a dx2 + b + cy = f (x) where f (x) is a sine or 13 dx cosine function d2q dq 1 7. L dt 2 + R dt + C q = V0 sin ωt represents the In Problems 1 to 3, find the general solutions of the given differential equations. variation of capacitor charge in an elec- tric circuit. Determine an expression for 1. d2 y − dy − 3y = 25 sin 2x q at time t seconds given that R = 40 , 2 dx2 dx L =0.02 H, C = 50 × 10−6 F, V0 = 540.8 V and ω = 200 rad/s and given the boundary y = 3 x + B e− x conditions that when t = 0, q = 0 and dq = 4.8 Ae 2 dt q = (10t + 0.01)e−1000t − 1 (11 sin 2x − 2 cos 2x ) 5 + 0.024 sin 200t − 0.010 cos 200t d2 y dy 2. dx2 − 4 dx + 4y = 5 cos x y = ( Ax + B)e2x − 4 sin x + 3 cos x 5 5 3. d2 y + y = 4 cos x 51.6 Worked problems on dx2 differential equations of the [ y = A cos x + B sin x + 2x sin x] d2y dy 4. Find the particular solution of the differen- form a d x2 + b d x + cy = f (x) where f (x) is a sum or a product tial equation d2 y − dy − 4y = 3 sin x; when dx2 3 Problem 9. Solve dx x = 0, y = 0 and d y = 0. dx d2 y + dy − 6y = 12x − 50 sin x. dx2 dx ⎡ =1 ⎤ ⎢⎣⎢y 170 (6e4x − 51e−x ) Using the procedure of Section 51.2: −1 (15 sin x − 9 cos ⎥⎥⎦ d2y dy 34 x ) (i) dx2 + dx − 6y = 12x − 50 sin x in D-operator 5. A differential equation representing the form is motion of a body is d2 y + n2 y = k sin pt , (D2 + D − 6)y = 12x − 50 sin x dt 2 where k, n and p are constants. Solve the equa- (ii) The auxiliary equation is (m2 + m − 6) = 0, from which, tion (given n =0 and p2 = n2) given that when (m − 2)(m + 3) = 0, t = 0, y = dy = 0. i.e. m = 2 or m = −3 dt (iii) Since the roots are real and different, the C.F., kp u = Ae2x + Be−3x. y = n2 − p2 sin pt − sin nt n (iv) Since the right hand side of the given differential equation is the sum of a polynomial and a sine 6. The motion of a vibrating mass is given by function let the P.I. v = ax + b + c sin x + d cos x d2 y dy (see Table 51.1(e)). dt 2 + 8 dt + 20y = 300 sin4t . Show that the general solution of the differential equation is