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Chapter 9 DIFFERENTIAL EQUATIONS 9.1 Overview (i) An equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation. (ii) A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation and a differential equation involving derivatives with respect to more than one independent variables is called a partial differential equation. (iii) Order of a differential equation is the order of the highest order derivative occurring in the differential equation. (iv) Degree of a differential equation is defined if it is a polynomial equation in its derivatives. (v) Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it. (vi) A relation between involved variables, which satisfy the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called the general solution and the solution free from arbitrary constants is called particular solution. (vii) To form a differential equation from a given function, we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants. (viii) The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves. (ix) ‘Variable separable method’ is used to solve such an equation in which variables can be separated completely, i.e., terms containing x should remain with dx and terms containing y should remain with dy. 180 MATHEMATICS (x) A function F (x, y) is said to be a homogeneous function of degree n if F (λx, λy )= λn F (x, y) for some non-zero constant λ. (xi) A differential equation which can be expressed in the form dy = F (x, y) or dx dx = G (x, y), where F (x, y) and G (x, y) are homogeneous functions of degree dy zero, is called a homogeneous differential equation. dy = F (x, y), we make dx substitution y = vx and to solve a homogeneous differential equation of the type (xii) To solve a homogeneous differential equation of the type dx = G (x, y), we make substitution x = vy. dy dy + Py = Q, where P and Q are constants or dx functions of x only is known as a first order linear differential equation. Solution (xiii) A differential equation of the form of such a differential equation is given by y (I.F.) = ∫ ( Q × I.F.) dx + C, where Pdx I.F. (Integrating Factor) = e ∫ . (xiv) Another form of first order linear differential equation is dx + P1x = Q1, where dy P1 and Q1 are constants or functions of y only. Solution of such a differential equation is given by x (I.F.) = ∫ ( Q1 × I.F.) dy + C, where I.F. = e∫ P dy . 1 9.2 Solved Examples Short Answer (S.A.) Example 1 Find the differential equation of the family of curves y = Ae2x + B.e–2x. Solution y = Ae2x + B.e–2x DIFFERENTIAL EQUATIONS 181 dy = 2Ae2x – 2 B.e–2x and dx d2y = 4Ae2x + 4Be–2x dx 2 d2y d2y = 4y i.e., – 4y = 0. dx 2 dx 2 Thus Example 2 Find the general solution of the differential equation dy y = dx x Solution ⇒ dy dx = y x ⇒ dy = y dy y = . dx x dx x ⇒ logy = logx + logc ⇒ y = cx Example 3 Given that Solution dy = yex ⇒ dx dy = yex and x = 0, y = e. Find the value of y when x = 1. dx dy y = ⇒ e x dx logy = ex + c Substituting x = 0 and y = e,we get loge = e0 + c, i.e., c = 0 (∵ loge = 1) Therefore, log y = ex. Now, substituting x = 1 in the above, we get log y = e ⇒ y = ee. Example 4 Solve the differential equation Solution The equation is of the type dy y + = x2. dx x dy + Py = Q , which is a linear differential dx equation. 1 Now I.F. = ∫ x dx = e logx = x. Therefore, solution of the given differential equation is 182 MATHEMATICS x x 2 dx , i.e. yx = y.x = x3 4 Hence y = x4 4 c c . x Example 5 Find the differential equation of the family of lines through the origin. Solution Let y = mx be the family of lines through origin. Therefore, dy =m dx dy dy . x or x – y = 0. dx dx Example 6 Find the differential equation of all non-horizontal lines in a plane. Solution The general equation of all non-horizontal lines in a plane is ax + by = c, where a ≠ 0. Eliminating m, we get y = Therefore, a dx b = 0. dy Again, differentiating both sides w.r.t. y, we get a d2x d 2x = 0 ⇒ = 0. dy 2 dy 2 Example 7 Find the equation of a curve whose tangent at any point on it, different from origin, has slope y Solution Given ⇒ dy y dy dx 1 y y . x y x = y 1 1 x 1 dx x Integrating both sides, we get logy = x + logx + c ⇒ log y =x+c x DIFFERENTIAL EQUATIONS 183 ⇒ y = ex + c = ex.ec ⇒ x y = k . ex x ⇒ y = kx . ex. Long Answer (L.A.) Example 8 Find the equation of a curve passing through the point (1, 1) if the perpendicular distance of the origin from the normal at any point P(x, y) of the curve is equal to the distance of P from the x – axis. – dx Solution Let the equation of normal at P(x, y) be Y – y = dy ( X – x ) ,i.e., Y+ X dx – dy y x dx dy =0 ...(1) Therefore, the length of perpendicular from origin to (1) is y x 1 dx dy dx dy ...(2) 2 Also distance between P and x-axis is |y|. Thus, we get y x dx dy dx dy 1 2 2 = |y| ⎛ dx ⎞ 2 ⇒ ⎜ y+x ⎟ = y 1 dy ⎠ ⎝ or dx dy 2 xy dx = 2 2 y –x dy 2 ⇒ dx dx 2 x – y2 dy dy 2 xy 0 ⇒ dx 0 dy 184 MATHEMATICS dx = 0 ⇒ dx = 0 dy Case I: Integrating both sides, we get x = k, Substituting x = 1, we get k = 1. Therefore, x = 1 is the equation of curve (not possible, so rejected). 2x y dx = 2 2 y x dy Case II: v x = y 2 x2 . Substituting y = vx, we get 2 xy dy dx dv v 2 x 2 x 2 dx 2vx 2 −(1 + v 2 ) 2v ⇒ x. ⇒ dv v 2 1 v 2v dx 2v dv 1 v2 dx x Integrating both sides, we get log (1 + v2) = – logx + logc ⇒ ⇒ log (1 + v2) (x) = log c ⇒ (1 + v2) x = c x2 + y2 = cx. Substituting x = 1, Therefore, y = 1, we get c = 2. x2 + y2 – 2x = 0 is the required equation. Example 9 Find the equation of a curve passing through 1, tangent to the curve at any point P (x, y) is 4 if the slope of the y y − cos 2 . x x Solution According to the given condition dy y y = − cos 2 dx x x ... (i) This is a homogeneous differential equation. Substituting y = vx, we get v+x dv = v – cos2v dx ⇒ x dv = – cos2v dx DIFFERENTIAL EQUATIONS 185 ⇒ sec2v dv = − ⇒ tan dx x ⇒ tan v = – logx + c y + log x = c x Substituting x = 1, y = y x tan ...(ii) 4 , we get. c = 1. Thus, we get + log x = 1, which is the required equation. 2 Example 10 Solve x dy ⎛ y⎞ π xy = 1 + cos ⎜ ⎟ , x ≠ 0 and x = 1, y = dx 2 ⎝x⎠ Solution Given equation can be written as x2 dy ⎛ y ⎞ xy = 2cos2 ⎜ ⎟ , x ≠ 0. dx ⎝ 2x ⎠ dy xy dx y ⇒ 2cos 2 2x x2 1 ⇒ y 2x sec 2 2 x2 dy xy dx 1 Dividing both sides by x3 , we get ⎛ y ⎞ sec 2 ⎜ ⎟ ⎡ x dy − y ⎤ ⎥ 1 ⎝ 2 x ⎠ ⎢ dx ⎢ ⎥= 3 2 2 ⎢ x ⎥ x ⎣ ⎦ Integrating both sides, we get tan y 2x 1 2 x2 k. ⇒ d y tan dx 2x 1 x3 186 MATHEMATICS Substituting x = 1, y = k= 2 , we get y 3 , therefore, tan 2x 2 1 2 x2 3 is the required solution. 2 Example 11 State the type of the differential equation for the equation. xdy – ydx = x2 y 2 dx and solve it. x2 Solution Given equation can be written as xdy = x2 dy dx y2 x y2 y dx , i.e., y ... (1) Clearly RHS of (1) is a homogeneous function of degree zero. Therefore, the given equation is a homogeneous differential equation. Substituting y = vx, we get from (1) v x x dv dx dv dx x2 v2 x2 x 1 v2 ⇒ vx dv 1 v 2 i.e. v x dv dx 1 v2 v dx x ... (2) Integrating both sides of (2), we get log (v + 1 v 2 ) = logx + logc ⇒ v + 1 v 2 = cx ⇒ y y2 + 1 2 = cx x x ⇒ y+ x2 y 2 = cx2 DIFFERENTIAL EQUATIONS 187 Objective Type Questions Choose the correct answer from the given four options in each of the Examples 12 to 21. 2 3 dy ⎞ ⎛ d 2 y ⎞ ⎛ Example 12 The degree of the differential equation ⎜1 + ⎟ is ⎟ =⎜ dx ⎠ ⎝ dx 2 ⎠ ⎝ (A) 1 (B) 2 (C) 3 (D) 4 Solution The correct answer is (B). Example 13 The degree of the differential equation 2 ⎛ d2y ⎞ d2y ⎛ dy ⎞ 2 + 3 = x log ⎜ 2 ⎟ is ⎜ ⎟ dx 2 ⎝ dx ⎠ ⎝ dx ⎠ (A) 1 (B) 2 (C) 3 (D) not defined Solution Correct answer is (D). The given differential equation is not a polynomial equation in terms of its derivatives, so its degree is not defined. 2 ⎡ ⎛ dy ⎞ 2 ⎤ d 2 y Example 14 The order and degree of the differential equation ⎢1+ ⎜ ⎟ ⎥ = 2 ⎢⎣ ⎝ dx ⎠ ⎥⎦ dx respectively, are (A) 1, 2 (B) 2, 2 (C) 2, 1 (D) 4, 2 Solution Correct answer is (C). Example 15 The order of the differential equation of all circles of given radius a is: (A) 1 (B) 2 (C) 3 (D) 4 Solution Correct answer is (B). Let the equation of given family be (x – h)2 + (y – k)2 = a2 . It has two orbitrary constants h and k. Threrefore, the order of the given differential equation will be 2. Example 16 The solution of the differential equation 2 x . (A) straight lines (B) circles dy – y = 3 represents a family of dx (C) parabolas (D) ellipses 188 MATHEMATICS Solution Correct answer is (C). Given equation can be written as 2dy dx y 3 x ⇒ 2log (y + 3) = logx + logc ⇒ (y + 3)2 = cx which represents the family of parabolas Example 17 The integrating factor of the differential equation dy (x log x) + y = 2logx is dx (A) ex (B) log x (C) log (log x) (D) x dy Solution Correct answer is (B). Given equation can be written as dx y 2 x log x x . 1 Therefore, dx I.F. = ∫ x log x = elog (logx) e = log x. Example 18 A solution of the differential equation (A) y = 2 (B) y = 2x dy dx 2 x dy dx y 0 is (D) y = 2x2 – 4 (C) y = 2x – 4 Solution Correct answer is (C). Example 19 Which of the following is not a homogeneous function of x and y. (A) x2 + 2xy (B) 2x – y 2 (C) cos y x y x (D) sinx – cosy Solution Correct answer is (D). dx dy Example 20 Solution of the differential equation x + y = 0 is 1 1 (A) x + y = c (B) logx . logy = c (C) xy = c (D) x + y = c Solution Correct answer is (C). From the given equation, we get logx + logy = logc giving xy = c. DIFFERENTIAL EQUATIONS 189 Example 21 The solution of the differential equation x (A) y = x2 + c 4 x2 (B) y = Solution Correct answer is (D). I.F. = e is y . x2 = 2 ∫ x .xdx = 2 dx x x2 +c 4 dy 2 y x 2 is dx (C) y = e2log x elog x 2 x4 + c x4 + c y = (D) x2 4 x2 x 2 . Therefore, the solution x4 c x4 + k , i.e., y = . 4 x2 4 Example 22 Fill in the blanks of the following: (i) Order of the differential equation representing the family of parabolas y2 = 4ax is __________ . 2 (ii) 3 2 ⎛ dy ⎞ ⎛ d y ⎞ + The degree of the differential equation ⎜ ⎟ ⎜ 2 ⎟ = 0 is ________ . ⎝ dx ⎠ ⎝ dx ⎠ (iii) The number of arbitrary constants in a particular solution of the differential equation tan x dx + tan y dy = 0 is __________ . (iv) F (x, y) = (v) x2 + y 2 + y is a homogeneous function of degree__________ . x An appropriate substitution to solve the differential equation x 2 log dx = dy x y xy log x2 x y is__________ . dy − y = sinx is __________ . dx (vi) Integrating factor of the differential equation x (vii) The general solution of the differential equation dy = e x − y is __________ . dx 190 MATHEMATICS (viii) (ix) dy y + =1 is __________ . dx x The differential equation representing the family of curves y = A sinx + B cosx is __________ . The general solution of the differential equation e 2 x (x) x y dx 1( x 0) when written in the form dy + Py = Q , then x dy dx P = __________ . Solution (i) (ii) (iii) (iv) (v) One; a is the only arbitrary constant. Two; since the degree of the highest order derivative is two. Zero; any particular solution of a differential equation has no arbitrary constant. Zero. x = vy. (vi) 1 dy y sin x − = ; given differential equation can be written as and therefore dx x x x (vii) 1 . x ey = ex + c from given equation, we have eydy = exdx. (viii) xy = (ix) d2y + y = 0; Differentiating the given function w.r.t. x successively, we get dx 2 I.F. = e x2 2 1 dx x = e–logx = c ; I.F. = e dy = Acosx – Bsinx dx = elogx = x and the solution is y . x = x .1 dx = and d2y = –Asinx – Bcosx dx 2 d2y + y = 0 is the differential equation. dx 2 ⇒ (x) 1 dx x 1 x ; the given equation can be written as x2 +C. 2 DIFFERENTIAL EQUATIONS 191 dy e –2 x = dx x y x i.e. y dy e –2 x + = x dx x dy + Py = Q. dx Example 23 State whether the following statements are True or False. (i) Order of the differential equation representing the family of ellipses having centre at origin and foci on x-axis is two. This is a differential equation of the type (ii) (iii) Degree of the differential equation 1+ dy d2y is not defined. 2 =x+ dx dx dy dy y 5 is a differential equation of the type + Py = Q but it can be solved dx dx using variable separable method also. ⎛ y cos ⎜ ⎝ y⎞ ⎟+ x x⎠ ⎛ y ⎞ is not a homogeneous function. x cos ⎜ ⎟ ⎝x⎠ (iv) F(x, y) = (v) x2 y 2 F(x, y) = is a homogeneous function of degree 1. x y dy dx y cos x is ex. (vi) Integrating factor of the differential equation (vii) The general solution of the differential equation x(1 + y2)dx + y (1 + x2)dy = 0 is (1 + x2) (1 + y2) = k. (viii) The general solution of the differential equation dy + y sec x = tanx is dx y (secx – tanx) = secx – tanx + x + k. (ix) x + y = tan–1y is a solution of the differential equation y2 dy y2 1 0 dx 192 MATHEMATICS (x) y = x is a particular solution of the differential equation d 2 y 2 dy x dx dx 2 xy x . Solution (i) True, since the equation representing the given family is x2 a2 y2 1 , which b2 has two arbitrary constants. (ii) True, because it is not a polynomial equation in its derivatives. (iii) True (iv) True, because f ( λx, λy) = λ° f (x, y). (v) True, because f ( λx, λy) = λ1 f (x, y). (vi) False, because I.F = e (vii) True, because given equation can be written as 1dx e– x . 2x 2y dx dy 2 1 x 1 y2 (viii) ⇒ log (1 + x2) = – log (1 + y2) + log k ⇒ (1 + x2) (1 + y2) = k False, since I.F. = e sec xdx elog(sec x tan x ) y (secx + tanx) = (sec x tan x) tan xdx = = secx + tanx, the solution is, ∫ ( sec x tan x + sec x − 1) dx 2 = secx + tanx – x +k (ix) True, x + y = tan–1y ⇒ 1 ⇒ dy dx 1 dy dy dx 1 y 2 dx ⎛ 1 ⎞ dy – 1⎟ =1 , i.e., ⎜ 2 dx ⎝ 1+ y ⎠ (1 y 2 ) which satisfies the given equation. y2 DIFFERENTIAL EQUATIONS 193 (x) False, because y = x does not satisfy the given differential equation. 9.3 EXERCISE Short Answer (S.A.) 1. 2. 3. dy 2y x . dx Find the differential equation of all non vertical lines in a plane. Find the solution of dy e 2 y and y = 0 when x = 5. dx Find the value of x when y = 3. Given that 4. Solve the differential equation (x2 – 1) 5. Solve the differential equation 6. Find the general solution of 7. Solve the differential equation 8. Solve: ydx – xdy = x2ydx. 9. Solve the differential equation dy 2 xy y dx dy ay emx dx dy 1 ex dx of y 2 dy = y. dx 2 sin x dy = – cosx and y (0) = 1, then find the value 1 y dx . 12. If y(t) is a solution of (1 + t) y (1) = – y dy = 1 + x + y2 + xy2, when y = 0, x = 0. dx 10. Find the general solution of (x + 2y3) 11. If y(x) is a solution of 1 dy + 2xy = 2 . x 1 dx 1 . 2 dy – ty = 1 and y (0) = – 1, then show that dt 194 MATHEMATICS 13. Form the differential equation having y = (sin–1x)2 + Acos–1x + B, where A and B are arbitrary constants, as its general solution. 14. Form the differential equation of all circles which pass through origin and whose centres lie on y-axis. 15. Find the equation of a curve passing through origin and satisfying the differential 2 equation (1 x ) 16. Solve : x2 dy 2 xy 4 x 2 . dx dy = x2 + xy + y2. dx 17. Find the general solution of the differential equation (1 + y2) + (x – etan–1y) dy = 0. dx 18. Find the general solution of y2dx + (x2 – xy + y2) dy = 0. 19. Solve : (x + y) (dx – dy) = dx + dy.[Hint: Substitute x + y = z after seperating dx and dy] dy = 0, given that y (1) = – 2. dx 21. Solve the differential equation dy = cosx (2 – y cosecx) dx given that y = 2 when 20. Solve : 2 (y + 3) – xy π . 2 22. Form the differential equation by eliminating A and B in Ax2 + By2 = 1. 23. Solve the differential equation (1 + y2) tan–1x dx + 2y (1 + x2) dy = 0. 24. Find the differential equation of system of concentric circles with centre (1, 2). x= Long Answer (L.A.) d ( xy ) = x (sinx + logx) dx 26. Find the general solution of (1 + tany) (dx – dy) + 2xdy = 0. 25. Solve : y + 27. Solve : dy = cos(x + y) + sin (x + y).[Hint: Substitute x + y = z] dx dy 3 y sin 2 x . dx 29. Find the equation of a curve passing through (2, 1) if the slope of the tangent to 28. Find the general solution of x2 y2 the curve at any point (x, y) is . 2 xy DIFFERENTIAL EQUATIONS 195 30. Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve at any point (x, y) is y 1 . x2 x 31. Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abcissa and ordinate of the point. 32. Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P (x, y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB. dy y (log y – log x + 1) dx Objective Type Choose the correct answer from the given four options in each of the Exercises from 34 to 75 (M.C.Q) 33. Solve : x d2y dx 2 34. The degree of the differential equation (A) 1 (B) 2 (C) 3 35. The degree of the differential equation 1 (A) 4 2 (B) 3 2 dy dx 2 x sin (D) not defined dy dx 3 2 2 d2y is dx 2 (C) not defined d2y 36. The order and degree of the differential equation dx 2 respectively, are (A) 2 and not defined (B) 2 and 2 (C) 2 and 3 37. If y = e–x (Acosx + Bsinx), then y is a solution of d2y dy 2 0 (A) 2 dx dx (C) d2y dy 2 2y 0 2 dx dx dy is: dx d2y dy −2 + 2y = 0 (B) 2 dx dx (D) d2y + 2y =0 dx 2 (D) 2 dy dx 1 4 1 + x5 (D) 3 and 3 0, 196 MATHEMATICS 38. The differential equation for y = Acos αx + Bsin αx, where A and B are arbitrary constants is d2y (A) dx 2 2 d2y (B) dx 2 y 0 2 y 0 d2y d2y y 0 y 0 (C) (D) dx 2 dx 2 39. Solution of differential equation xdy – ydx = 0 represents : (A) a rectangular hyperbola (B) parabola whose vertex is at origin (C) straight line passing through origin (D) a circle whose centre is at origin 40. Integrating factor of the differential equation cosx (A) cosx (B) tanx (C) secx dy + ysinx = 1 is : dx (D) sinx 41. Solution of the differential equation tany sec2x dx + tanx sec2ydy = 0 is : (A) tanx + tany = k (B) tanx – tany = k tan x k tan y (D) tanx . tany = k (C) 42. Family y = Ax + A3 of curves is represented by the differential equation of degree : (A) 1 (B) 2 43. Integrating factor of (A) x 44. Solution of (A) xy (C) 3 (D) 4 xdy – y = x4 – 3x is : dx (B) logx (C) 1 x dy y 1 , y (0) = 1 is given by dx = – ex (B) xy = – e–x (C) xy (D) – x = – 1 (D) y = 2 ex – 1 DIFFERENTIAL EQUATIONS 197 dy y +1 45. The number of solutions of dx = x −1 when y (1) = 2 is : (A) none (B) one (C) two (D) infinite 46. Which of the following is a second order differential equation? (A) (y′)2 + x = y2 (B) y′y′′ + y = sinx (C) y′′′ + (y′′)2 + y = 0 (D) y′ = y2 47. Integrating factor of the differential equation (1 – x2) x 1 x2 dy − xy =1 is dx 1 log (1 – x2) 2 48. tan–1 x + tan–1 y = c is the general solution of the differential equation: (A) – x (B) (C) 1 x 2 (D) dy 1 + y 2 = (A) dx 1 + x 2 dy 1 + x 2 = (B) dx 1 + y 2 (C) (1 + x2) dy + (1 + y2) dx = 0 (D) (1 + x2) dx + (1 + y2) dy = 0 dy + x = c represents : dx (A) Family of hyperbolas (B) Family of parabolas (C) Family of ellipses (D) Family of circles 50. The general solution of ex cosy dx – ex siny dy = 0 is : (A) ex cosy = k (B) ex siny = k 49. The differential equation y (C) ex = k cosy (D) ex = k siny 3 d 2 y ⎛ dy ⎞ + ⎜ ⎟ + 6 y 5 = 0 is : 51. The degree of the differential equation 2 dx ⎝ dx ⎠ (A) 1 (B) 2 (C) 3 (D) 5 dy + y = e – x , y (0) = 0 is : dx (A) y = ex (x – 1) (B) y = xe–x –x (C) y = xe + 1 (D) y = (x + 1)e–x 52. The solution of 198 MATHEMATICS dy y tan x – sec x 0 is: dx (B) secx (D) esecx 53. Integrating factor of the differential equation (A) cosx (C) ecosx 54. The solution of the differential equation (A) y = tan–1x (C) x = tan–1y dy 1 y 2 is: dx 1 x 2 (B) y – x = k (1 + xy) (D) tan (xy) = k 55. The integrating factor of the differential equation (A) x ex (B) dy 1+ y +y = is: dx x ex x (C) xex (D) ex 56. y = aemx + be–mx satisfies which of the following differential equation? (A) dy my 0 dx (B) dy my 0 dx (C) d2y m2 y 0 dx 2 (D) d2y m2 y 0 dx 2 57. The solution of the differential equation cosx siny dx + sinx cosy dy = 0 is : (A) sin x c sin y (B) sinx siny = c (C) sinx + siny = c 58. The solution of x ex (A) y = x x k x (C) y = xe + k (D) cosx cosy = c dy + y = ex is: dx (B) y = xex + cx ey (D) x = y k y DIFFERENTIAL EQUATIONS 199 59. The differential equation of the family of curves x2 + y2 – 2ay = 0, where a is arbitrary constant, is: (A) (x2 – y2) dy = 2xy dx (C) 2 (x2 – y2) (B) 2 (x2 + y2) dy = xy dx (D) (x2 + y2) dy = xy dx dy = 2xy dx 60. Family y = Ax + A3 of curves will correspond to a differential equation of order (A) 3 (B) 2 61. The general solution of (A) e x 2 −y (C) 1 (D) not defined dy 2 = 2x e x − y is : dx 2 (B) e–y + e x = c =c 2 (C) ey = e x + c (D) e x 2 +y =c 62. The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the point is : (A) an ellipse (B) parabola (C) circle (D) rectangular hyperbola x2 dy 63. The general solution of the differential equation e 2 + xy is : dx (A) y ce x2 2 x2 (B) y ce 2 x2 x2 (C) y = ( x + c) e 2 (D) y (c x)e 2 64. The solution of the equation (2y – 1) dx – (2x + 3)dy = 0 is : 2x 1 (A) 2 y 3 k 2 y +1 (B) 2 x − 3 = k 2x 3 (C) 2 y 1 k 2x 1 (D) 2 y 1 k 200 MATHEMATICS 65. The differential equation for which y = acosx + bsinx is a solution, is : d2y (A) +y=0 dx 2 d2y (B) –y=0 dx 2 d2y + (a + b) y = 0 (C) dx 2 d2y (D) + (a – b) y = 0 dx 2 dy + y = e–x, y (0) = 0 is : dx (B) y = xex (A) y = e–x (x – 1) 66. The solution of (C) y = xe–x + 1 (D) y = xe–x 67. The order and degree of the differential equation d3y dx3 2 d2y dy 3 2 2 dx dx (A) 1, 4 4 (B) 3, 4 y 4 are : (C) 2, 4 (D) 3, 2 ⎡ ⎛ dy ⎞ 2 ⎤ d 2 y 68. The order and degree of the differential equation ⎢1 + ⎜ dx ⎟ ⎥ = dx 2 are : ⎢⎣ ⎝ ⎠ ⎥⎦ 3 (B) 2, 3 (C) 2, 1 (D) 3, 4 2 69. The differential equation of the family of curves y2 = 4a (x + a) is : (A) 2, 2 (A) y = 4 d2y (C) y 2 dx dy ⎛ dy ⎞ ⎜ x+ ⎟ dx ⎝ dx ⎠ dy dx 2 (B) 2 y dy 4a dx 2 0 dy ⎛ dy ⎞ + y⎜ ⎟ – y (D) 2 x dx ⎝ dx ⎠ 70. Which of the following is the general solution of (A) y = (Ax + B)ex (C) y = Aex + Be–x d2y dy −2 + y = 0? 2 dx dx (B) y = (Ax + B)e–x (D) y = Acosx + Bsinx DIFFERENTIAL EQUATIONS 201 dy + y tan x = sec x is : dx (A) y secx = tanx + c (B) y tanx = secx + c 71. General solution of (C) tanx = y tanx + c (D) x secx = tany + c dy y sin x is : dx x (B) x (y – cosx) = sinx + c (D) x (y + cosx) = cosx + c 72. Solution of the differential equation (A) x (y + cosx) = sinx + c (C) xy cosx = sinx + c 73. The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is: (A) (y + 1) = k (ex + 1) (B) y + 1 = ex + 1 + k ex 1 y log (D) y 1 x (C) y = log {k (y + 1) (e + 1)} 74. The solution of the differential equation dy = ex–y + x2 e–y is : dx (A) y = ex–y – x2 e–y + c (C) ex + ey = k x3 +c 3 (B) ey – ex = x3 +c 3 (D) ex – ey = x3 +c 3 dy 2 xy 75. The solution of the differential equation dx 1 x2 1 is : (1 x 2 )2 y = c + tan–1x 1 x2 (A) y (1 + x2) = c + tan–1x (B) (C) y log (1 + x2) = c + tan–1x (D) y (1 + x2) = c + sin–1x 76. Fill in the blanks of the following (i to xi) dy (i) d 2 y dx 0 is _________. e The degree of the differential equation dx 2 (ii) The degree of the differential equation 1 dy dx 2 x is _________. 202 MATHEMATICS (iii) The number of arbitrary constants in the general solution of a differential equation of order three is _________. (iv) dy dx y x log x 1 is an equation of the type _________. x (v) dx General solution of the differential equation of the type dy + P1 x = Q1 is given by _________. 77. xdy 2 y x 2 is _________. dx (vi) The solution of the differential equation (vii) The solution of (1 + x2) (viii) The solution of the differential equation ydx + (x + xy)dy = 0 is ______. (ix) General solution of (x) The solution of differential equation coty dx = xdy is _________. (xi) The integrating factor of dy +2xy – 4x2 = 0 is _________. dx dy dx y = sinx is _________. dy 1 y y is _________. dx x State True or False for the following: (i) dx Integrating factor of the differential of the form dy + p1 x = Q1 is given by e ∫ p1dy . (ii) dx Solution of the differential equation of the type dy + p1 x = Q1 is given by x.I.F. = (I.F) Q1dy . (iii) Correct substitution for the solution of the differential equation of the dy f ( x, y ) , where f (x, y) is a homogeneous function of zero dx degree is y = vx. type DIFFERENTIAL EQUATIONS 203 (iv) Correct substitution for the solution of the differential equation of the type dx g ( x, y ) where g (x, y) is a homogeneous function of the dy degree zero is x = vy. (v) Number of arbitrary constants in the particular solution of a differential equation of order two is two. (vi) The differential equation representing the family of circles x2 + (y – a)2 = a2 will be of order two. y x 1 3 (vii) dy The solution of dx (viii) Differential equation representing the family of curves y = ex (Acosx + Bsinx) is 2 2 is y 3 – x 3 = c. d2y dy –2 2y 0 2 dx dx dy x + 2 y = is x + y = kx2. dx x (ix) The solution of the differential equation (x) Solution of (xi) The differential equation of all non horizontal lines in a plane is d2x =0 . dy 2 xdy dx y x tan y y is sin x x cx