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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., , 20/04/2018
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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, , 20/04/2018
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182 MATHEMATICS, , x x 2 dx , i.e. yx =, , y.x =, , x4, +c, 4, , x3 c, + ., 4 x, , Hence y =, , 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, , ⇒, , y, ., x, , dy, y, = y+, dx, x, , = y 1+, , 1, x, , dy, 1, = 1+, dx, y, x, , Integrating both sides, we get, logy = x + logx + c ⇒, , log, , y, =x+c, x, , 20/04/2018
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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, dx, – y+x, dy, dy, , =0, , ...(1), , Therefore, the length of perpendicular from origin to (1) is, y+ x, , dx, dy, , dx, 1+, dy, , ...(2), , 2, , Also distance between P and x-axis is |y|. Thus, we get, y+ x, , dx, dy, , dx, 1+, dy, , 2, , = |y|, , 2, , , dx , dx, 2, ⇒ y + x = y 1+, dy , dy, , , or, , 2, , ⇒, , (, , ), , dx dx 2, dx, =0, x – y 2 + 2 xy = 0 ⇒, dy, dy dy, , 2 xy, dx, = 2 2, y –x, dy, , 20/04/2018
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186 MATHEMATICS, , Substituting x = 1, y =, , k=, , π, , we get, 2, , 1 3, y, 3, = − 2 + is the required solution., , therefore, tan, 2x, 2x 2, 2, , Example 11 State the type of the differential equation for the equation., xdy – ydx =, , x 2 + y 2 dx and solve it., , Solution Given equation can be written as xdy =, , (, , ), , x 2 + y 2 + y dx , i.e.,, , x2 + y 2 + y, dy, =, dx, x, , ... (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, , x 2 + v 2 x 2 + vx, dv, =, dx, x, , dv, = 1+ v2, dx, , ⇒, , dv, 1+ v, , 2, , i.e. v + x, , =, , dv, = 1+ v 2 + v, dx, , 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+, , x 2 + y 2 = cx2, , 20/04/2018
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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, log, x, 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, , 20/04/2018
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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, , 2, dy, y, Solution Correct answer is (B). Given equation can be written as dx + x log x = x ., 1, , Therefore,, , dx, I.F. = ∫ x log x, = elog, e, , (logx), , = log x., , Example 18 A solution of the differential equation, (A) y = 2, , (B) y = 2x, , dy, dx, , 2, , −x, , dy, + y = 0 is, dx, , (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, y, +, x, 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., , 20/04/2018
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DIFFERENTIAL EQUATIONS 189, , Example 21 The solution of the differential equation x, , (A) y =, , x2 + c, 4 x2, , x2, (B) y = + c, 4, , Solution Correct answer is (D). I.F. = e, is y . x2 =, , 2, ∫ x .xdx =, , 2, dx, x, , dy, + 2 y = x 2 is, dx, , (C) y =, , x4 + c, x4 + c, y, =, (D), x2, 4 x2, , = e2log x = elog x = x2 . Therefore, the solution, 2, , x4 + c, x4, + k , i.e., y =, ., 4x2, 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, − x2, y, , xy log, , 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, , 20/04/2018
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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, x, dx 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, , 1, − dx, x, , = e–logx =, , x2, + c ; I.F. =, e, 2, , dy, = Acosx – Bsinx, dx, ⇒, , (x), , 1, dx, x, , = elogx = x and the solution is y . x = x .1 dx =, , and, , x2, +C., 2, , d2y, = –Asinx – Bcosx, dx 2, , d2y, + y = 0 is the differential equation., dx 2, , 1, ; the given equation can be written as, x, , 20/04/2018
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DIFFERENTIAL EQUATIONS 191, , dy, e –2 x, y, −, =, dx, x, 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+, , d2y, dy, 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, − y = cos x is ex., dx, , (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, + y 2 + 1= 0, dx, , 20/04/2018
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192 MATHEMATICS, , (x), , y = x is a particular solution of the differential equation, , d 2 y 2 dy, −x, + xy = x ., dx, dx 2, , Solution, (i), , True, since the equation representing the given family is, , x2 y 2, + = 1 , which, a 2 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, −2 y, dx =, dy, 2, 1+ x, 1+ y 2, , (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 ) = secx + tanx, the solution is,, , y (secx + tanx) = (sec x + tan x) tan xdx =, , ∫ (sec x tan x + sec x − 1) dx =, 2, , secx + tanx – x +k, (ix), , True, x + y = tan–1y ⇒ 1+, , ⇒, , dy, dx, , dy, 1 dy, =, dx 1+ y 2 dx, , 1, , dy − (1 + y 2 ), =, ,, i.e.,, which satisfies the given equation., –, 1, =, 1, , , 2, dx, y2, 1+ y, , , 20/04/2018
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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, = 2 y− x ., dx, Find the differential equation of all non vertical lines in a plane., Find the solution of, , dy −2 y, =e, 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 = e mx, dx, dy, + 1= e x + y, dx, 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, of y, , 1, dy, + 2xy = 2, x −1 ., dx, , dy, = y., dx, , 2 + sin x dy, = – cosx and y (0) = 1, then find the value, 1+ y, dx, , π, ., 2, , 12. If y(t) is a solution of (1 + t), y (1) = –, , dy, – ty = 1 and y (0) = – 1, then show that, dt, , 1, ., 2, , 20/04/2018
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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, , dy, is:, dx, , (D) not defined, dy, dx, , 3, 2 2, , =, , d2y, is, dx 2, , (C) not defined, , (D) 2, , d2y, dy, +, 36. The order and degree of the differential equation, 2, dx, dx, , respectively, are, (A) 2 and not defined, , (B) 2 and 2, , (C) 2 and 3, , 1, 4, , 1, , + x5 = 0 ,, , (D) 3 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, , d2y, dy, −2 +2y = 0, (B), 2, dx, dx, (D), , d2y, + 2y =0, dx 2, , 20/04/2018
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DIFFERENTIAL EQUATIONS 197, , 45. The number of solutions of, (A) none, , dy y +1, =, dx x −1 when y (1) = 2 is :, , (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 :, (B) ex siny = k, (A) ex cosy = 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, (B) y = xe–x, (A) y = ex (x – 1), –x, (C) y = xe + 1, (D) y = (x + 1)e–x, , 52. The solution of, , 20/04/2018
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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 = ex + 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, = e 2 + xy is :, 63. The general solution of the differential equation, 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, , 2 x −1, (D) 2 y − 1 = k, , 20/04/2018
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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, (C) y = log {k (y + 1) (e + 1)}, , 74. The solution of the differential equation, , e x +1, +k, y +1, , (D) y = log, , x, , dy, = ex–y + x2 e–y is :, dx, x3, +c, 3, , (A) y = ex–y – x2 e–y + c, , (B) ey – ex =, , x3, (C) e + e =, +c, 3, , x3, (D) e – e =, +c, 3, , x, , y, , x, , y, , dy 2 xy, 1, =, 75. The solution of the differential equation dx +, 2, 1+ x (1+ x 2 ) 2 is :, 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, + e = 0 is _________., The degree of the differential equation, dx 2, , (ii), , The degree of the differential equation 1+, , dy, dx, , 2, , = x is _________., , 20/04/2018
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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., 1, 3, , (vii), , dy, y, The solution of, =, dx, x, , (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, , y, xdy, y, = cx, = y + x tan is sin, x, dx, x, , d2x, =0 ., dy 2, , 20/04/2018
