Chapter 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. (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. dy(xi) A differential equation which can be expressed in the form = F (x, y) ordx dx dy = G (x, y), where F (x, y) and G (x, y) are homogeneous functions of degree zero, is called a homogeneous differential equation. dy(xii) To solve a homogeneous differential equation of the type = F (x, y), we makedxsubstitution y = vx and to solve a homogeneous differential equation of the type dx dy = G (x, y), we make substitution x = vy. dy(xiii) A differential equation of the form + Py = Q, where P and Q are constants or dxfunctions of x only is known as a first order linear differential equation. Solution of such a differential equation is given by y (I.F.) = ∫(×QI.F.) dx + C, where I.F. (Integrating Factor) = e∫Pdx . dx (xiv) Another form of first order linear differential equation is dy + P1x = Q1, where P1 and Q1 are constants or functions of y only. Solution of such a differential Q× I.F. dy P1dyequation is given by x (I.F.) = ∫( 1) + C, where I.F. = ∫ . e9.2 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 2dy dy= 2Ae2x – 2 B.e–2x and 2 = 4Ae2x + 4Be–2x dxdx22dy dyThus 2 = 4y i.e., 2 – 4y = 0.dxdx dy yExample 2 Find the general solution of the differential equation .dx = x dyy dydx dy dxSolution ⇒ = ⇒ = dx = x yx yx ⇒ logy = logx + logc ⇒ y = cx dyExample 3 Given that dx = yex and x = 0, y = e. Find the value of y when x = 1. dydy xSolution dx = yex ⇒ y = edx ⇒ 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. dy yExample 4 Solve the differential equation + = x2. dxx dySolution The equation is of the type +Py = Q , which is a linear differential dx equation. dx = elogxNow I.F. = ∫ 1 = x.x Therefore, solution of the given differential equation is 4x2y.x = xxdx , i.e. yx = c4 3xcHence y = .4 x Example 5 Find the differential equation of the family of lines through the origin. dySolution Let y = mx be the family of lines through origin. Therefore, dx = m dy dyEliminating m, we get y = dx . x or x – y = 0.dxExample 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. dx Therefore, ab = 0.dy Again, differentiating both sides w.r.t. y, we get 22dx dx a = 0 ⇒ = 0.dy2 dy2 Example 7 Find the equation of a curve whose tangent at any point on it, different yfrom origin, has slope y . x dy y 1 Solution Given y = y 1 dxx x dy 1⇒ 1 dx yx Integrating both sides, we get ylogy = x + logx + c ⇒ log = x + c x y y⇒ = ex + c = ex.ec ⇒ = k . ex x 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. Solution Let the equation of normal at P(x, y) be Y – y = –dx (X– x) ,i.e.,dy dx dxY + X – yx = 0 ...(1)dydyTherefore, the length of perpendicular from origin to (1) is dx yx dy ...(2) 1 dx 2 dy Also distance between P and x-axis is |y|. Thus, we get dx yx dy ⎛ dx ⎞22 dx 2 dx dx 22 dx⇒ ⎜ yx ⎟ = y ⇒ x – y 2xy 0+ 1 ⇒ 0 ⎝ dy ⎠ dydydy dy dx 2xy or = dy y2– x2 dxCase I: = 0 ⇒ dx = 0dyIntegrating 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). dx 2xy dy y2 x2 Case II: = 22 . Substituting y = vx, we getdyyx dx 2xy 222 2dvvx x dvv 1 vx 2 ⇒ x. vdx 2vx dx 2v 22v dx (1 v )−+ dv= ⇒ 21 vx2v 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, y = 1, we get c = 2. Therefore, x2 + y2 – 2x = 0 is the required equation. Example 9 Find the equation of a curve passing through 1, 4 if the slope of the y 2 ytangent to the curve at any point P (x, y) is −cos . xx Solution According to the given condition dy y 2 y=−cos ... (i)dxx x This is a homogeneous differential equation. Substituting y = vx, we get dv dv v + x dx = v – cos2v ⇒ x = – cos2vdxdx⇒ sec2v dv = − ⇒ tan v = – logx + c x y⇒ tan +log x =c ...(ii)x Substituting x = 1, y = , we get. c = 1. Thus, we get 4 ytan + log x = 1, which is the required equation.x2 dy y π⎛⎞Example 10 Solve x xy = 1 + cos ⎜⎟, x ≠ 0 and x = 1, y = xdx⎝⎠2 Solution Given equation can be written as 2 dy y⎛⎞ x xy = 2cos2 ⎜⎟, x ≠ 0.dx⎝⎠2x 2 dyx xy 2 ydx sec 1 2x dy⇒ 2 y ⇒ x2 xy 12cos 2 dx2x Dividing both sides by x3 , we get ⎛⎞2 y ⎡dysec xx −y ⎤⎥⎜⎟⎝⎠ 2 ⎢dx 1 dy 1=⎢ 2 ⎥ 3 ⇒ tan 2 xx 3⎢⎥ dx 2xx ⎣⎦ Integrating both sides, we get y 1tan k .2x 2x2 Substituting x = 1, y = , we get2 3 y 13 tan k = 2 , therefore, 2 is the required solution.2x 2x2 Example 11 State the type of the differential equation for the equation. xdy – ydx = x2 y2 dx and solve it. Solution Given equation can be written as xdy = x2 y2 y dx , i.e., dy x2 y2 y ... (1)dx x 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) 2 22dvxvxvx dv 2i.e. vx 1 vvvxdx x dx dv dx dv x 1 v2 ⇒ ... (2)x1 v2dx Integrating both sides of (2), we get log (v + 1 v2 ) = logx + logc ⇒ v + 1 v2 = cx yy2 ⇒ + 1 2 = cx ⇒ y + x2 y2 = cx2 xxObjective Type Questions Choose the correct answer from the given four options in each of the Examples 12 to 21. 2⎛ dy ⎞3 ⎛ dy ⎞2 Example 12 The degree of the differential equation ⎜1 + ⎟=⎜ 2 ⎟ is ⎝ dx ⎠⎝ dx ⎠(A) 1 (B) 2 (C) 3 (D) 4 Solution The correct answer is (B). Example 13 The degree of the differential equation 22dy ⎛ dy ⎞22 ⎛ dy ⎞3 = x log +⎜⎟ ⎜⎟ isdx2 ⎝ dx ⎠⎝ dx 2 ⎠ (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. ⎡ 22 2⎛ dy ⎞⎤ dy 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. dyx – yExample 16 The solution of the differential equation 2. dx = 3 represents a family of (A) straight lines (B) circles (C) parabolas (D) ellipses Solution Correct answer is (C). Given equation can be written as 2dy dx ⇒ 2log (y + 3) = logx + logcy 3 x⇒ (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 isdx(A) ex (B) log x (C) log (log x) (D) x dy y 2 Solution Correct answer is (B). Given equation can be written as .dx x log xx ∫1 dx elog (logx)Therefore, I.F. = xlog x = = log x. e dy 2 dyExample 18 A solution of the differential equation xy 0 is dx dx (A) y = 2 (B) y = 2x (C) y = 2x – 4 (D) y = 2x2 – 4 Solution Correct answer is (C). Example 19 Which of the following is not a homogeneous function of x and y. 2 yy(A) x2 + 2xy (B) 2x – y (C) cos (D) sinx – cosyxxSolution Correct answer is (D). dx dy+=0Example 20 Solution of the differential equation isxy 11+= c(A) (B) logx . logy = c (C) xy = c (D) x + y = cxy Solution Correct answer is (C). From the given equation, we get logx + logy = logc giving xy = c. dy 2Example 21 The solution of the differential equation x dx2yx is 2 244x + c xx +cx +c(A) y = (B) y =+c (C) y = (D) y = 2 224x 4 x 4x 2 Solution Correct answer is (D). I.F. = x dx 2log x log x2 2 . Therefore, the solution e eex 4 4 x xc is y . x2 = ∫ x2.xdx =+ k , i.e., y = 2.4 4x Example 22 Fill in the blanks of the following: (i) Order of the differential equation representing the family of parabolas y2 = 4ax is __________ . 32⎛ dy ⎞⎛ dy ⎞2 (ii) 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 __________ . x2 + y2 +y(iv) F (x, y) = is a homogeneous function of degree__________ .x (v) An appropriate substitution to solve the differential equation x x2 log x2 ydx = is__________ .dy x xy log ydy(vi) Integrating factor of the differential equation x −y = sinx is __________ . dx dy x− y(vii) The general solution of the differential equation =e is __________ .dx dy y(viii) The general solution of the differential equation +=1 is __________ .dx x (ix) The differential equation representing the family of curves y = A sinx + B cosx is __________ . (i) One; a is the only arbitrary constant. (ii) Two; since the degree of the highest order derivative is two. (x) e 2 x x y x dx dy 1(x 0) when written in the form dy dx P Qy+ =, then P = __________ . Solution (iii) Zero; any particular solution of a differential equation has no arbitrary constant. (iv) Zero. (v) x = vy. 1 dy y sin x(vi) ; given differential equation can be written as −= and therefore x dxx x 11dx = e–logx =I.F. = ex . x (vii) ey = ex + c from given equation, we have eydy = exdx. 2 2x 1 x(viii) xy = c ; I.F. = exdx = elogx = x and the solution is y . x = x.1dx = +C.2 2 2dy y(ix) 2 +=0; Differentiating the given function w.r.t. x successively, we get dx 2dy dy= Acosx – Bsinx and 2 = –Asinx – Bcosxdx dx2dy⇒ dx2 + y = 0 is the differential equation. 1 (x) ; the given equation can be written asx –2 x –2 xdy e y dyye = i.e. + = dxxx dxxx dyThis is a differential equation of the type + Py = Q.dxExample 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. 2dy dy(ii) Degree of the differential equation 1+ 2= x+ is not defined.dx dx dy dy(iii) y 5 is a differential equation of the type +Py =Q but it can be solveddx dx using variable separable method also. y⎛⎞y cos +x⎜⎟x⎝⎠ (iv) F(x, y) = ⎛⎞y is not a homogeneous function.x cos ⎜⎟x⎝⎠x2 y2 (v) F(x, y) = is a homogeneous function of degree 1.xydy(vi) Integrating factor of the differential equation y cos x is ex.dx(vii) The general solution of the differential equation x(1 + y2)dx + y (1 + x2)dy = 0 is (1 + x2) (1 + y2) = k. dy(viii) The general solution of the differential equation +y sec x = tanx isdx y (secx – tanx) = secx – tanx + x + k. dy 2(ix) x + y = tan–1y is a solution of the differential equation y2 dx y 10 2dy dy (x) y = x is a particular solution of the differential equation 2 x2 xy x .dx dx Solution x2 y2 (i) True, since the equation representing the given family is a2 b2 1, which 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). e e – x(vi) False, because I.F = 1dx . (vii) True, because given equation can be written as 2x 2y 2 dx 2 dy1 x 1 y ⇒ log (1 + x2) = – log (1 + y2) + log k ⇒ (1 + x2) (1 + y2) = k sec xdx(viii) False, since I.F. = elog(sec x tan x) = secx + tanx, the solution is, e 2 y (secx + tanx) = (sec x tan x) tan xdx = ∫(sec x tan x + sec x −1) dx = secx + tanx – x +k dy 1 dy(ix) True, x + y= tan–1y ⇒ 12dx 1 y dx dy ⎛ 1 ⎞ dy (1 y2)⇒ ⎜ –1⎟=1, i.e., which satisfies the given equation.dx 1+ y2 dx y2⎝⎠ (x) False, because y = x does not satisfy the given differential equation. 9.3 EXERCISE Short Answer (S.A.) dy 2yx1. Find the solution of .dx 2. Find the differential equation of all non vertical lines in a plane. dy 2 y3. Given that e and y = 0 when x = 5.dxFind the value of x when y = 3. dy 1 4. Solve the differential equation (x2 – 1) + 2xy = 2 .dxx 1 dy5. Solve the differential equation dx 2xyy dy6. Find the general solution of ay emx dx dy7. Solve the differential equation dx 1 ex y 8. Solve: ydx – xdy = x2ydx. dy9. Solve the differential equation = 1 + x + y2 + xy2, when y = 0, x = 0.dx dy10. Find the general solution of (x + 2y3) dx = y. 2sin x dy11. If y(x) is a solution of dx = – cosx and y (0) = 1, then find the value1 y of y .2 dy12. If y(t) is a solution of (1 + t) – ty = 1 and y (0) = – 1, then show thatdt 1 y (1) = – 2. 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 dyequation (1 x2) 2xy 4x2.dx dy16. Solve : x2 = x2 + xy + y2.dxdy17. Find the general solution of the differential equation (1 + y2) + (x – etan–1y) = 0.dx18. 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] dy20. Solve : 2 (y + 3) – xy = 0, given that y (1) = – 2.dx 21. Solve the differential equation dy = cosx (2 – y cosecx) dx given that y = 2 when π x = .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). Long Answer (L.A.) d () = x (sinx + logx)25. Solve : y +xy dx 26. Find the general solution of (1 + tany) (dx – dy) + 2xdy = 0. dy27. Solve : = cos(x + y) + sin (x + y).[Hint: Substitute x + y = z]dxdy28. Find the general solution of dx 3y sin 2 x . 29. Find the equation of a curve passing through (2, 1) if the slope of the tangent to x2 y2 the curve at any point (x, y) is .2xy 30. Find the equation of the curve through the point (1, 0) if the slope of the tangent y 1 to the curve at any point (x, y) is 2 .xx 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. dy33. Solve : xy (log y – log x + 1)dxObjective Type Choose the correct answer from the given four options in each of the Exercises from 34 to 75 (M.C.Q) 222dy dy dy34. The degree of the differential equation 2 xsin is: dxdx dx (A) 1 (B) 2 (C) 3 (D) not defined 3 2dy 22 dy 35. The degree of the differential equation 1 2 is dx dx 3(A) 4 (B) (C) not defined (D) 2 2 1 21dy dy 436. The order and degree of the differential equation + x5 0,dx2 dx respectively, are (A) 2 and not defined (B) 2 and 2 (C) 2 and 3 (D) 3 and 3 37. If y = e –x (Acosx + Bsinx), then y is a solution of 2 2dydy dy dy (A) 220 (B) 2 −2 + 2y = 0 dxdx dx dx 2 2dydy dy(C) 222y 0 (D) 2 + 2y =0 dxdx dx 38. The differential equation for y = Acos αx + Bsin αx, where A and B are arbitrary constants is 2 2dy 2 dy 2(A) 2 y 0 (B) 2 y 0 dx dx 2 2dy dy(C) 2 y 0 (D) 2 y 0 dx dx 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 dy40. Integrating factor of the differential equation cosx + ysinx = 1 is :dx (A) cosx (B) tanx (C) secx (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 (C) k (D) tanx . tany = ktan y 42. Family y = Ax + A3 of curves is represented by the differential equation of degree : (A) 1 (B) 2 (C) 3 (D) 4 xdy43. Integrating factor of – y = x4 – 3x is :dx 1(A) x (B) logx (C) (D) – x x dy44. Solution of dx y 1, y (0) = 1 is given by (A) xy = – ex (B) xy = – e –x (C) xy = – 1 (D) y = 2 ex – 1 dy y +1 =45. The number of solutions of −when y (1) = 2 is :dx x 1 (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 dy47. Integrating factor of the differential equation (1 – x2) −xy =1 isdx x 21(A) – x (B) 2 (C) 1 x (D) log (1 – x2)1 x 2 48. tan–1 x + tan–1 y = c is the general solution of the differential equation: dy 1+y2 dy 1+x2 ==(A) (B)dx 1+x2 dx 1+y2 (C) (1 + x2) dy + (1 + y2) dx = 0 (D) (1 + x2) dx + (1 + y2) dy = 0 dy49. The differential equation y + 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 (C) ex = k cosy (D) ex = k siny dy dy 35 2 ⎛⎞ 51. The degree of the differential equation 2 +⎜ ⎟+6y =0 is :dx dx⎝⎠(A) 1 (B) 2 (C) 3 (D) 5 dy ye y52. The solution of += – x ,(0) =0 is :dx (A) y = ex (x – 1) (B) y = xe–x (C) y = xe–x + 1 (D) y =(x + 1)e–x dy53. Integrating factor of the differential equation y tan x –sec x 0 is:dx (A)cosx (B) secx (C) ecosx (D) esecx dy 1 y2 54. The solution of the differential equation is:dx 1 x2 (A) y = tan–1x (B) y – x = k (1 + xy) (C) x = tan–1y (D) tan (xy)= k dy 1+y+=55. The integrating factor of the differential equation y is:dx x x ex (A) x (B)e x (C) xex (D) ex 56. y = aemx + be–mx satisfies which of the following differential equation? dy dy(A) my 0 (B) my 0dx dx 2 2dy 2 dy 2(C) my 0 (D) my 0 dx2 dx2 57. The solution of the differential equation cosx siny dx + sinx cosy dy = 0 is : sin x c(A) (B) sinx siny = csin y (C) sinx + siny = c (D) cosx cosy = c dy58. The solution of x + y = ex is:dx ex k(A) y = (B) y = xex + cx xx ye k (C) y = xex + k (D) x = yy 59. The differential equation of the family of curves x2 + y2 – 2ay = 0, where a is arbitrary constant, is: dy dy(A) (x2 – y2) = 2xy (B) 2 (x2 + y2) dx = xydx dy dy(C) 2 (x2 – y2) dx = xy (D) (x2 + y2) = 2xydx60. Family y = Ax + A3 of curves will correspond to a differential equation of order (A) 3 (B) 2 (C) 1 (D) not defined dy61. The general solution of dx = 2x ex2 −y is : x2 −y x2(A) e = c (B) e –y + e = c x x + y(C) ey = e 2+ c (D) e 2 = 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 dy x2 63. The general solution of the differential equation e 2+ xy is :dx 2 2x x (A) 2 (B) 2yce yce 2 2x x (C) 2 (D) 2= (x + ) ycy ce )( xe 64. The solution of the equation (2y – 1) dx – (2x + 3)dy = 0 is : 2x 1 2y +1k = k(A) (B)2y 3 2x −3 2x 3 2x 1k k(C) (D)2y 1 2y 1 65. The differential equation for which y = acosx + bsinx is a solution, is : 22dy dy(A) + y = 0 (B) – y = 0dx2 dx2 22dy dy(C) + (a + b) y = 0 (D) + (a – b) y = 0dx2 dx2 dy66. The solution of + y = e –x, y (0) = 0 is :dx (A) y = e –x (x – 1) (B) y = xex (C) y = xe –x + 1 (D) y = xe–x 67. The order and degree of the differential equation 232dy dy dy 44 3 322 y are :dx dx dx (A) 1, 4 (B) 3, 4 (C) 2, 4 (D) 3, 2 2⎡⎛ dy ⎞2 ⎤ dy 1+=68. The order and degree of the differential equation ⎢⎜ ⎟⎥ 2 are :⎢⎝ dx ⎠⎥ dx⎣⎦ 3(A) 2, (B) 2, 3 (C) 2, 1 (D) 3, 42 69. The differential equation of the family of curves y2 = 4a (x + a) is : dy ⎛ dy ⎞ dy(A) y2 = 4 ⎜ x +⎟ (B) 2y 4adx ⎝ dx ⎠ dx 22 2dydy dy ⎛ dy ⎞(C) y 2 0 (D) 2x + y ⎜⎟ – ydxdx dx ⎝ dx ⎠ 2dy dy 70. Which of the following is the general solution of 2 −2 + y = 0?dx dx (A) y = (Ax + B)ex (B) y = (Ax + B)e –x (C) y = Aex + Be –x (D) y = Acosx + Bsinx dy71. General solution of + y tan x =sec x is :dx (A) y secx = tanx + c (B) y tanx = secx + c (C)tanx = y tanx + c (D) x secx = tany + c dy y72. Solution of the differential equation dx xsin x is : (A) x (y + cosx) = sinx + c (B) x (y – cosx) = sinx + c (C) xy cosx = sinx + c (D) x (y + cosx) = cosx + 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 (C) y = log {k (y + 1) (ex + 1)} (D) 1log 1 xe y k y 74. The solution of the differential equation dy dx = ex–y + x2 e–y is : (A) y = ex–y – x2 e –y + c (B) ey – ex = 3 3 x + c (C) ex + ey = 3 3 x + c (D) ex – ey = 3 3 x + c 75. The solution of the differential equation 2 2 2 2 1 1 (1 ) dy xy dx x x is : (A) y (1 + x2) = c + tan–1x (C) y log (1 + x2) = c + tan–1x (B) 21 y x = c + tan–1x (D) y (1 + x2) = c + sin–1x 76. Fill in the blanks of the following (i to xi) 2 dydy dx(i) The degree of the differential equation 2 e 0 is _________.dxdy 2 (ii) The degree of the differential equation 1 x is _________.dx(iii) The number of arbitrary constants in the general solution of a differential equation of order three is _________. dy y 1 (iv) is an equation of the type _________.dx xlog xxdx + P x =Q(v) General solution of the differential equation of the type 1 1dy is given by _________. xdy 2(vi) The solution of the differential equation dx2yx is _________. dy(vii) The solution of (1 + x2) +2xy – 4x2 = 0 is _________.dx (viii) The solution of the differential equation ydx + (x + xy)dy = 0 is ______. dy(ix) General solution of y = sinx is _________.dx (x) The solution of differential equation coty dx = xdy is _________. dy 1 y(xi) The integrating factor of y is _________.dx x77. State True or False for the following: + px =(i) Integrating factor of the differential of the form dx 1 Q1 is givendy ∫ 1by pdy .edx + px =Q1(ii) Solution of the differential equation of the type 1 is givendy by x.I.F. = (I.F) Q1dy . (iii) Correct substitution for the solution of the differential equation of the dytype dx f (, ) xy , where f (x, y) is a homogeneous function of zero degree is y = vx. (iv) Correct substitution for the solution of the differential equation of the type dx (, ) g xy where g (x, y) is a homogeneous function of thedy 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. 1 dy y 3 22(vii) The solution of is y3– x3= c.dx x(viii) Differential equation representing the family of curves 2dy dy y = ex (Acosx + Bsinx) is –22y 0 dx2 dx dy x + 2y(ix) The solution of the differential equation = is x + y = kx2.dx x xdy yy cx(x) Solution of yx tan is sin dx xx (xi) The differential equation of all non horizontal lines in a plane is 2dx 2 =0 .dy

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