Mathematics-18.Unit-14 Differential Equation

DIFFERENTIAL EQUATION BASIC DEFINITION An equation containing an independent variable, dependent variable and differential coefficients of dependent variable with respect to independent variable is called a differential equation. An ordinary differential equation is one is which there is only one independent variable. For Example : .........(1) .........(2) ..........(3) .........(4) .........(5) .........(6) Order and Degree of a Differential Equation: The order of differential equation is the order of highest order derivative appearing in the equation. For Example : Orders of differential equations (1), (2), (3), (4), (5) and (6) are 1, 2, 1, 3, 2 and 2 respectively. The degree of differential equation is the degree of the highest order derivative involved in it, when the differential coefficients are free from radicals and fractions (i.e. write differential equations as polynomial in derivatives) For Example : Degrees of differential equations (1), (2), (3), (4), (5) and (6) are 1, 1, 1, 4, 3 and 2 respectively. Illustration 1: Find the order and degree (if defined) of the following differential equations : (i) (ii) Solution : (i) The given differential equation can be re-written as . Hence its order is 3 and degree 2. (ii) . Hence its order is 2 and degree 1. Formation of differential Equation: We know y2 = 4ax is a parabola whose vertex is origin and axis as the x-axis . If a is a parameter, it will represent a family of parabola with the vertex at (0, 0) and axis as y = 0 . Differentiating y2 = 4ax . . (1) . . (2) From (1) and (2), y2 = 2yx ⇒ y = 2x This is a differential equation for all the members of the family and it does not contain any parameter ( arbitrary constant). (i) The differential equation of a family of curves of one parameter is a differential equation of the first order, obtained by eliminating the parameter by differentiation. (ii) The differential equation of a family of curves of two parameter is a differential equation of the second order, obtained by eliminating the parameter by differentiating the algebraic equation twice. Similar procedure is used to find differential equation of a family of curves of three or more parameter. Example -1: Find the differential equation of the family of curves y = Aex + Be3x for different values of A and B. Solution: y = A ex + Be3x . . . . (1) y1 = Aex + 3Be3x . . . (2) y2 = Aex + 9B3x . . . (3) Eliminating A and B from the above three, we get = 0 ⇒ ex e3x = 0 ⇒ 3y + 4y1 – y2 = 0 ⇒ 3y + 4 SOLUTION OF DIFFERENTIAL EQUATION The general solution of a differential equation is the relation in the variables x, y obtained by integrating (removing derivatives ) where the relation contains as many arbitrary constants as the order of the equation. The general solution of a differential equation of the first order contains one arbitrary constant while that of the second order contains two arbitrary constants. In the general solution, if particular values of the arbitrary constant are put , we get a particular solution which will give one member of the family of curves. To solve differential equation of the first order and the first degree: Simple standard form of differential equation of the first order and first degree are as follows: (i) Variable Separable Form f(x) dx + φ(y) dy = 0 Method: Integrate it i.e., find ∫ f(x) dx + ∫ φ(y)dy = c Example -2: Solve . Solution: Given ⇒ ⇒ Integrating, we get ln y – ex =c (ii) Reducible into Variable Separable Method: Sometimes differential equation of the first order cannot be solved directly by variable separation but by some substitution we can reduce it to a differential equation with separable variables. A differential equation of the is solved by writing ax + by + c = t Example -3: Solve (x – y)2 . Solution: Put z = x –y ⇒ ⇒ Now z2 ⇒ ⇒ dx = , which is in the form of variable separable Now integrating, we get x = z + ⇒ Solution is x =(x – y) + (iii) Homogeneous Equation When is equal to a fraction whose numerator and the denominator both are homogeneous functions of x and y of the same degree then the differential equation is said to be homogeneous equation. i.e. when , where f(km, ky) = f(x, y) then this differential equation is said to be homogeneous differential equation . Method: Put y = vx Example -4: Solve . Solution: (homogeneous ) . Put y = vx ∴ ⇒ , Integrate C + lnx = - ln(1 –v2) ⇒ lnkx + ln(1 –v2) =0 ⇒ kx(1- v2) = 1 ⇒ k(x2 – y2) = x . (iv) Non-homogeneous Differential Equation Form Method: If , put x = X + h , y = Y + k such that a1h +b1k + c1 = 0, a2h + b2k +c2 = 0 . In this way the equation becomes homogeneous in X, Y. then use the method for homogeneous equation. If . Put a1x+b1y=v or a2x + b2y = v. The equation in the form of variable separable in x, v. Example -5: . Solution: Here Hence we put x- y = v ⇒ or, 1 – ⇒ = dx or, Integrate 2v + ln (v +2) = x + C, Put the value of v ∴ x – 2y + ln (x – y +2) = C (v) Linear Equation Form , where P(x) and Q(x) are functions of x Method: Multiplying the equation by e∫P(x)dx, called integrating factor. Then the equation becomes Integrating (vi) Reducible into Linear Equation Form R(y) + P(x) S(y) = Q(x) , such that Method: Put S(y) =z then ∴ The equation becomes , which is in the linear form Example -6: . Solution: Multiplying both sides by I.F. and integrating Put ⇒ ⇒ 2 y (vii) Exact Differential Equations Mdx + Ndy = 0, where M and N are functions of x and y. If , then the equation is exact and its solution is given by ∫ Mdx + ∫ N dy = c To find the solution of an exact differential equation Mdx + N dy = 0, integrate as if y were constant. Also integrate the terms of N that do not contain x w.r.t y. Equate the sum of these integrals to a constant. Example -7: (x2 –ay)dx + (y2 –ax)dy = 0. Solution: Here M = x2 –ay N = y2 –ax ⇒ ∴ equation is exact solution is = c – ayx + = c or x3 –3axy + y3 = 3c. Integrating Factor: A factor, which when multiplied to a non exact, differential equation makes it exact, is known as integrating factor e.g. the non exact equation y dx –x dy = 0 can be made exact on multiplying by the factor . Hence is the integrating factor for this equation. Notes: In general such a factor exist but except in certain special cases, it is likely to be difficult to determine. The number of integrating factor for equation M dx + N dy = 0 is infinite. Some Useful Results: d(xy) = xdy + ydx = d tan-1 = d(sin-1 xy) = Example -8: Solve x dy – y dx = . Solution: ⇒ ⇒ Integrating ⇒ ln ⇒ ln = 2 ln x + ln k y + = kx2. (viii) Linear Differential Equation with constant coefficient Differential equation of the form , aI ∈ R for I = 0, 1 , 2, 3, . . . , n is called a linear differential equation with constant coefficients. In order to solve this differential equation, take the auxiliary equation as a0Dn + a1Dn-1 +…+ an =0 Find the roots of this equation and then solution of the given differential equation will be as given in the following table Roots of the auxiliary equation Corresponding complementary function 1 One real root α1 C1 2. Two real and differential root α1 and α2 C1 + C2 3. Two real and equal roots α1 and α2 (C1 + C2x) 4. Three real and equal roots α1, α2, α3 (C1 + C2x + C3x2 ) 5. One pair of imaginary roots α ± iβ (C1 cosβx + C2 sinβx) 6. Two Pair of equal imaginary roots α ± iβ and α ± iβ [ (C1 + C2x) cosβ + (C1 + C2x)sinβ] Example -9: Solve . Solution: Its auxiliary equation is D2 –3D + 2 = 0 ⇒ D = 1, D = 2 Hence its solution is y = C1ex + C2e2x So far only linear differential equation with constant coefficients of form a0 + a1 + ….+ an y = 0, aI ∈ R for i = 0, 1, 2, …., n were considered. Now we consider the following form a0 + a1 + ….+ an y = X where X is either constant or functions of x alone. Theorem: If y = f1(x) is the general solution of a0 + a1 + ….+ an y = 0 and y = f2(x) is a solution of a0 + a1 + ….+ an y = X Then y = f1(x) + f2(x) is the general solution of a0 + a1 + ….+ any = X Expression f1(x) is known as complementary function and the expression f2(x) is known as particular integral. These are denotes by C.F. and P.I. respectively. The nth derivative of y will be denoted Dny where D stands for and n denotes the order of derivative. If we take Differential Equation: + P1 + P2 + …. + Pny = X then we can write this differential equation in a symbolic form as Dny + P1Dn– 1y + P2Dn– 2y + ….+ Pny = X (Dn + P1Dn– 1 + P2Dn– 2 + ….+ Pn)y = X The operator Dn + P1Dn– 1 + P2Dn– 2 + ….+ Pn is denoted by f(D) so that the equation takes the form f(D)y = X y = Methods of finding P.I. : Notes: If both m1 and m2 are constants, the expressions (D – m1)(D – m2)y and (D– m2)(D – m1)y are equivalent i.e. the expression is independent of the order of operational factors. We will explain the method with the help of following Example -10: Solve . Solution: The equation can be written as (D2 – 5D + 6)y = e3x (D – 3) (D – 2)y = e3x C.F. = c1 e3x + c2e2x And P.I. = = = = e3x = x. e3x ∴ y = c1 e3x + c2e2x + x e3x P.I. can be found by resolving Into partial functions ∴ P.I. = = = = x e3x – e3x . Second term can be neglected as it is included as it inclined in the first term of a C.F. Short Method of Finding P.I. : In certain cases, the P.I. can be obtained by methods shorter than the general method. (i). To find P.I. when X = eax in f(D) y = X, where a is constant y = if f(a) ≠ 0 if f(a) = 0 , where f(D) = (D – a)r φ(D) Example -11: Solve (D3 – 5D2 + 7D – 3)y = e3x. Solution: (D – 1)2 (D – 3) y = e3x C.F. = aex + bx ex + ce3x And P.I. = e3x = ∴ y = aex + bx ex + ce3x + (ii). To find P.I. when X = cosax or sinax f(D) y = X y = sinax If f( – a2) ≠ 0 then = If f(– a2) = 0 then (D2 + a2) is atleast one factor of f(D2) Let f (D2) = (D2 + a2)r φ (D2) Where φ(– a2) ≠ 0 ∴ = = when r = 1 sin ax = – Similarly If f(– a)2 ≠ 0 then cos ax = cosax and Example -12: Solve (D2 – 5D + 6)y = sin3x. Solution: (D – 2) (D – 3)y = sin3x C.F. = ae2x + be3x P.I.= sin3x = = = – (5D – 3). sin3x = ∴ ae2x + be3x + (iii). To find the P.I. when X = xm where m ∈ N f(D) y = xm y = we will explain the method by taking an example Example -13: Find P.I. of (D3 + 3D2 + 2D)y = x2. Solution: P.I. = = = = = = = = (iv). To find the value of eaxV where ‘a’ is a constant and V is a function of x Example -14: Solve (D2 + 2)y = x2 e3x. Solution: C.F. = a cos x + b sin x P.I. = x2 e3x = e3x = = x2 = x2 = x2 = ∴ a cos x + b sin x + (v). To find where V is a function of x Example -15: Solve (D2 + 4) y = x sin2x. Solution: C.F. = c1 cos 2x + c2 sin2x P.I. = = = = – = – = – y = c1 cos2x + c2 sin2x – . Some Results on Tangents and Normals: (i) The equation of the tangent at P(x, y) to the curve y= f(x) is Y – y = (ii) The equation of the normal at point P(x, y) to the curve y = f(x) is Y – y = (X – x ) (iii) The length of the tangent = CP = (iv) The length of the normal = PD = (v) The length of the cartesian subtangent = CA = (vi) The length of the cartesian subnormal = AD = (viii) The initial ordinate of the tangent =OB = y - x PROBLEMS SUBJECTIVE 1. A curve that passes through (2, 4) and having subnormal of constant length of 8 units can be; (A) y2 = 16x – 8 (B) y2 = -16x + 24 (C) x2 = 16y – 60 (D) x2 = -16y + 68 Solution: Let the curve be y = f(x). Subnormal at any point = ⇒ y = ±8 ⇒ y dy = ±8dx ⇒ = ±8x + c ⇒ y2 = 16 x+2c1, ⇒ c1= -8 or y2 = -16x +2c2, ⇒ c2= 24 Hence (A), (B) are correct answers. 2. Equation of a curve that would cut x2 + y2 - 2x - 4y - 15 = 0 orthogonally can be ; (A) (y-2) = λ(x-1) (B) (y-1) = λ(x-2) (C) (y+2) = λ(x+1) (D) (y+1) = λ(x+2) where λ ∈ R. Solution: Any line passing through the centre of the given circle would meet the circle orthogonally. Hence (A) is the correct answer. 3. Let m and n be the order and the degree of the differential equation whose solution is y = cx +c2- 3c3/2 +2, where c is a parameter. Then, (A) m = 1, n = 4 (B) m =1 , n =3 (C) m = 1, n = 2 (D) None Solution: = c ⇒ the differential equation is ; y = x. + -3. +2 Clearly its order is one and degree 4. Hence (A) is the correct answer. 4. y = a sinx + b cosx is the solution of differential equation : (A) + y = 0 (B) + y = 0 (C) = y (D) = y Solution: = a cosx - b sinx ⇒ = -a sinx – bcosx = -y Hence + y = 0 . Hence (A) is the correct answer. 5. For any differential function y = f(x), the value of + is : (A) always zero (B) always non-zero (C) equal to 2y2 (D) equal to x2 Solution: = for a differential equation or = -1 = - = - or + = 0. Hence (A) is the correct answer. 6. The degree of differential equation is : (A) 1 (B) 2 (C) 3 (D) none of these Solution: Since the equation is not a polynomial in all the differential coefficient so the degree of equation is not defined. Hence (D) is the correct answer. 7. The degree and order of the differential equation of all parabolas whose axis is x-axis are : (A) 2, 1 (B) 1, 2 (C) 3, 2 (D) none of these Solution: Equation of required parabola is of the form y2 = 4a(x –h) Differentiating, we have 2y = 4a ⇒ y = 2a Required differential equation . Degree of the equation is 1 and order is 2. Hence (B) is the correct answer. 8. The differential equation of all ellipses centred at origin is : (A) y2 + xy12 –yy1 = 0 (B) xyy2 + xy12 –yy1 = 0 (C) yy2 + xy12 –xy1 = 0 (D) none of these Solution: Ellipse centred at origin are given by = 1 ……(1) where a and b are unknown constants ⇒ y1 = 0 ……(2) Differentiating again, we get (y12 + yy2) = 0 ……(3) Multiplying (3) with x and then subtracting from (2) we get ( yy1 – xy12 –xyy2) = 0 ⇒ xyy2 + xy12 –yy1 = 0. Hence (B) is the correct answer. 9. Particular solution of y1 + 3xy = x which passes through (0, 4) is : (A) 3y = 1 + 11 (B) y = + 11 (C) y = 1 + (D) y = + 11 Solution: + (3x)y = x I.F = ∴ Solution of given equation is y + c = + c If curve passes through (0, 4), then 4 – = c ⇒ c = y = ⇒ 3y = 1 + 11 . Hence (A) is the correct answer. 10. Solution of equation is : (A) (x –y)2 + c = log (3x –4y + 1) (B) x –y + c = log (3x –4y + 1) (C) x –y + c = = log (3x –4y –3) (D) x –y + c = log (3x –4y + 1) Solution: Let 3x –4y = z 3 –4 ⇒ Therefore the given equation ⇒ – ⇒ – dz = dx ⇒ –z + 4 log (z + 1) = x + c ⇒ log (3x –4y + 1) = x –y + c. Hence (B) is the correct answer. 11. The order of the differential equation, whose general solution is y = C1 ex + C2 e2x + C3 e3x + C4 , where C1, C2, C3, C4, C5 are arbitrary constants, is : (A) 5 (B) 4 (C) 3 (D) none of these Solution: y = (c1 + c4) ex + c2 e2x + c3 e3x + c4 y = c1 ex + c2 e2x + c3 e3x + c4 y = k1 ex + k2 e2x + k3 e3x + k4 Therefore 4 obituary constants Hence (B) is the correct answer. 12. I.F. for y ln is : (A) ln x (B) ln y (C) ln xy (D) none of these Solution: I.F. = = ln y Hence (B) is the correct answer. 13. Which one of the following is a differential equation of the family of curves y =Ae2x + Be-2x (A) (B) (C) (D) Solution: y = A e2x + b e−2x ⇒ = 2 (A e2x − b e−2x) = 4 (A e2x + b e−2x) = ln y Hence (C) is the correct answer. 14. Solution of is : (A) log tan = c – 2 sin (B) log cot = c – 2 sin (C) log tan = c – 2 cos (D) none of these Solution: = − 2 cos − ⇒ c − 2 sin = log tan Hence (A) is the correct answer. 15. Solution of is : (A) sin = kx (B) cos = kx (C) tan = kx (D) none of these Solution: = put y = vx ⇒ v + x = v + tan v cot v dv = Integrating, we get ln sin v = ln x + ln k ⇒ sin = kx Hence (A) is the correct answer. 16. Solution of = 0 is : (A) sin–1 x – sin–1 y = c (B) sin–1 y + sin–1 x = c (C) sin–1 x = c sin–1 y (D) (sin–1 x) (sin–1 y) = c Solution: ⇒ sin−1 y + sin−1 x = c Hence (B) is the correct answer. 17. General solution of = e–2x is : (A) y = e–2x + c (B) y = e–2x + cx + d (C) y = e–2x + cx + d (D) y = e–2x + cx2 + d Solution: = e−2x, + k1 Integrating, y = + k1 x + k2 ⇒ y = + cx + d Hence (C) is the correct answer. 18. Solution of = x2 is : (A) x + y = + c (B) x – y = + c (C) xy = x4 + c (D) y – x = x4 + c Solution: = x2 I.F. = = x Therefore solution is xy = + c Hence (C) is the correct answer. 19. Differential equation associated with primitive y = Ae3x + Be5x : (A) + 15y = 0 (B) – 15y = 0 (C) + y = 0 (D) none of these Solution: y = A e3x + B e5x y′ = 3 A e3x + 5 B e5x y″ = 9 A e3x + 25 B e5x therefore y″ − 8y + 15y = 0 Hence (A) is the correct answer. 20. The curve satisfying y = 2x is a : (A) family of parabola (B) family of circle (C) pair of straight line (D) none of these Solution: ⇒ ln y = ln y2 + ln c ⇒ y2 = kx it represents a family of parabola Hence (A) is the correct answer. 21. The equation of the curve through the origin satisfying the equation dy = (sec x + y tanx) dx is : (A) y sinx =x (B) y cosx = x (C) y tanx = x (D) none of these Solution: − y tan x = sec x I.F. = = cos x y cos x = ∫ sec x cos x dx = x + c y cos x = x + c At (0, θ), y cos x = x Since c = 0 Hence (B) is the correct answer. 22. The differential equation y + x = a (where ‘a’ is a constant) represents : (A) a set of circles having centre on y-axis (B) a set of circles with centre on x-axis (C) a set of ellipses (D) none of these Solution: y dy = (a − x) dx = ax − ⇒ x2 + y2 = 2ax ⇒ x2 + 2ax + y2 = 0 It represents a set of sides with centre on x−axis Hence (B) is the correct answer. 23. If = e–2y and y = 0 when x = 5, then value of x where y = 3 is given by : (A) e5 (B) (C) e6 + 1 (D) loge 6 Solution: e2y dy = dx ⇒ + c = x ⇒ c = 5 − At. y = 3 = x ⇒ x = Hence (B) is the correct answer. 24. The equation of curve passing through and having slope 1 – at (x, y) is : (A) y = x2 + x + 1 (B) xy = x + 1 (C) xy = x2 + x + 1 (D) xy = y + 1 Solution: ⇒ y = x + + c At (2, 7/2) = 2 + + c ⇒ c = 1 Therefore y = x + + 1 ⇒ xy = x2 + x + 1 Hence (C) is the correct answer. 25. Given that = yex such that x = 0, y = e. The value of y (y > 0) when x = 1 will be : (A) e (B) (C) e (D) e2 Solution: = ex dx ⇒ ln y = ex + c At x = 0, y = 1; c = 0 ln y = ex Therefore At x = 1, y = ec Hence (C) is the correct answer. 26. The differential equation of all conics with centre at origin is of order (A) 2 (B) 3 (C) 4 (D) None of these Solutions : The general equation of all conics with centre at origin is ax2 + 2hxy + by2 = 1 Since it has three arbitrary constants, the differential equation is of order 3. Hence (B) is the correct answer. 27. The differential equation of all conics with the coordinate axes, is of order (A) 1 (B) 2 (C) 3 (D) 4 Solutions : The general equation of all conics with axes as coordinate axes is ax2 + by2 = 1. Since it has two arbitrary constants, the differential equation is of order 2. Hence (B) is the correct answer. 28. The differential equation of all non-vertical lines in a plane is : (A) (B) (C) (D) Solutions : The general equation of non-vertical lines in a plane is ax + by = 1, b ≠ 0 Hence (C) is the correct answer. 29. The differential equation of all non-horizontal lines in a plane is : (A) (B) (C) (D) Solutions : ax = by = 0, a ≠ 0 . Hence (D) is the correct answer. 30. The degree and order of the differential equation of all tangent lines to the parabola x2 = 4y is : (A) 2, 1 (B) 2, 2 (C) 1, 3 (D) 1, 4 Solutions : Equation of any tangent to x2 = 4y is , where m is arbitrary constant. ∴ Putting this value of m in , we get Which is a differential equation of order 1 and degree 2. Hence (A) is the correct answer. 31. If f(x) = f'(x) and f(1) = 2, then f(3) = (A) e2 (B) 2e2 (C) 3e2 (D) 2e3 Solutions : ⇒ log f(x) = x + c Since f(1) = 2 ∴ log 2 = 1 + c ∴ log f(x) = x + log 2 – 1 ∴ log f(3) = 3 + log 2 – 1 = 2 + log 2 ⇒ f(3) = e2+log2 = e2 . elog2 = 2e2. Hence (B) is the correct answer. 32. Equation of the curve passing through (3, 9) which satisfies the differential equation is : (A) 6xy = 3x2 – 6x + 29 (B) 6xy = 3x2 – 29x + 6 (C) 6xy = 3x3 – 29x – 6 (D) None of these Solutions : It passes through (3, 9) Hence (C) is the correct answer. 33. Solution of , y = 1 is given by (A) y2 = sinx (B) y = sin2x (C) y2 = cosx + 1 (D) None of these Solutions : On dividing by sin x, Put y2 = v I.F. = e∫cotxdx = elog sinx = sin x ∴ Solution is v. sin x ∫sinx . (2 cos x) dx + c ⇒ y2 sin x = sin2x + c When , y = 1, then c = 0 ∴ y2 = sin x. Hence (A) is the correct answer. 34. The general solution of the equation is : (A) (B) (C) (D) None of these Solutions : ⇒ P = –x, Q = 1 I.F. = ∴ Solution is Hence (D) is the correct answer. 35. Solution of is : (A) ex(x + 1) = y (B) ex(x + 1) + 1 = y (C) ex(x – 1) + 1 = y (D) None of these Solutions : ⇒ e–y dy = (ex + e–x)dx ⇒ e–y = ex – e–x + c Hence (D) is the correct answer. 36. If y = e–x (A cosx + B sin x), then y satisfies (A) (B) (C) (D) Solutions : y = e–x (A cos x + B sin x) ....(1) (–A sin x + B cos x) – e–x (A cos x + B sin x) (–A sin x + B cos x) – y .....(2) (–A cos x – B sin x) – e–x (–A sin x + B cos x) Using (1) and (2), we get Hence (C) is the correct answer. 37. The solution of is : (A) (B) (C) (D) Solutions : .....(1) Put y = vx ∴ (1) becomes . Hence (A) is the correct answer. 38. Order and degree of differential equation are : (A) 4, 2 (B) 1, 2 (C) 1, 4 (D) 2, 4 Solutions : On simplification the equation becomes Hence (D) is the correct answer. 39. The order of differential equation is : (A) 2 (B) 3 (C) (D) 6 Solutions : Clearly the equation is of order 2. Hence (A) is the correct answer. 40. If m and n are order and degree of the equation , then (A) m = 3, n = 3 (B) m = 3, n = 2 (C) m = 3, n = 5 (D) m = 3, n = 1 Solutions : The given differential equation can be written as ⇒ m = 3, n = 2 Hence (B) is the correct answer. 41. The solution of the differential equation represent (A) straight lines (B) circles (C) parabolas (D) ellipses Solutions : ⇒ 2log(y + 3) = logx + log c ⇒ (y + 3)2 = ex, which represent parabolas. Hence (C) is the correct answer. 42. Solution of differential equation dy – sinx siny dx = 0 is : (A) (B) (C) cos x tan y = c (D) cos x sin y = c Solutions : Given equation can be written as : Hence (A) is the correct answer. 43. Solution of differential equation is : (A) (a + m) y = emx + c (B) yeax = memx + c (C) y = emx + ce–ax (D) (a + m) y = emx + ce–ax Solutions : I.F. = e∫adx = eax ∴ Solution is ⇒ (a + m)y = emx + c1 (a + m) e–ax ⇒ (a + m) y = emx + ce–ax where c = c1 (a + m) Hence (D) is the correct answer. 44. Integrating factor of the differential equation is : (A) cos x (B) tan x (C) sec x (D) sin x Solutions : Here P = tan x, Q = sec x ∴ I. F. = e∫tan x dx = elog secx = sec x. Hence (C) is the correct answer. 45. Solution of differential equation xdy – ydx = 0 represents (A) a rectangular hyperbola (B) straight line passing through origin (C) parabola whose vertex is at origin (D) circle whose centre is at origin Solutions : x dy – y dx = 0 ⇒ log y – log x = log c or y = cx which is a straight line. Hence (B) is the correct answer. 46. The integrating factor of the differential equation is given by (A) ex (B) logx (C) log(log x) (D) x Solutions : Hence (B) is the correct answer. 47. The solution of is : (A) (x + y)ex+y = 0 (B) (x + c)ex+y = 0 (C) (x – y)ex+y = 1 (D) (x – c)ex+y + 1 = 0 Solutions : Put e–y = z ∴ P = –1, Q = –ex. ∴ I.F. = e∫–1dx = e–x Solution is z . e–x = ∫ –ex . e–x dx + c = –x + c ⇒ e–y . e–x = c – x ⇒ e–(x+y) = c – x ⇒ (x – c) ex+y + 1 = 0 Hence (D) is the correct answer. 48. Family y = Ax + A3 of curves is represented by the differential equation of degree (A) 3 (B) 2 (C) 1 (D) None of these Solutions : There is only one arbitrary constant. So, degree of the differential equation is one. Hence (C) is the correct answer. 49. Integrating factor of ; (x > 0) is : (A) x (B) log x (C) –x (D) ex Solutions : I.F. Hence (A) is the correct answer. 50. If integrating factor of x(1 – x2) dy + (2x2y – y – ax3) dx = 0 is e∫Pdx, then P is : (A) (B) 2x3 – 1 (C) (D) Solutions : Hence (D) is the correct answer. 51. For solving , suitable substitution is : (A) y = vx (B) y = 4x + v (C) y = 4x (D) y + 4x + 1 = v Solutions : y + 4x + 1 = v Hence (D) is the correct answer. 52. Integral curve satisfying has the slope at the point (1, 0) of the curve, equal to : (A) (B) –1 (C) 1 (D) Solutions : ⇒ slope at (1, 0) = Hence (C) is the correct answer. 53. A solution of the differential equation is : (A) y = 2 (B) y = 2x (C) y = 2x – 4 (D) y = 2x2 – 4 Solutions : Ans is (C) 54. The solution of is : (A) x + y = ce2x (B) y2 = 2x3 + c (C) xy2 = 2y5 + c (D) x(y2 + xy) = 0 Solutions : ⇒ –2x + 10y3 ⇒ ∴ I.F. = ∴ Solution is . Hence (C) is the correct answer. 55. A particle moves in a straight line with velocity given by (x being the distance described). The time taken by the particle to describe 99 metres is : (A) log10e (B) 2 loge 10 (C) 2 log10 e (D) Solutions : ⇒ log(x + 1) = t + c ....(1) Initially, when t = 0, x = 0 ∴ c = 0 ∴ log (x + 1) = t When x = 99, then t = loge (100) = 2 loge 10. Hence (B) is the correct answer. 56. The curve for which slope of the tangent at any point equals the ratio of the abscissa to the ordinate of the point is a/an (A) ellipse (B) rectangular hyperbola (C) circle (D) parabola Solutions : ⇒ ydy = xdx ⇒ y2 – x2 = 2c which is a rectangular hyperbola Hence (B) is the correct answer. 57. The solution of the differential equation (1 – xy – x5y5) dx – x2(x4y4 + 1)dy = 0 is given by (A) x = (B) x = (C) x = (D) None of these Solution: The given equation is dx – x (ydx+ xdy )= x5y4(ydx + xdy) ⇒ lnx = xy + ⇒ x = . Hence (A) is the correct answer. 58. If satisfies the relation then value of A and B respectively are: (A) –13, 14 (B) –13, –12 (C) –13, 12 (D) 12, –13 Solution: On differentiating , w.r.t x we get Putting these values in , we get . On solving we get A= –13 and B = –12 Hence (B) is the correct answer. 59. The differential equation of all circles which pass through the origin and whose centres lie on y-axis is : (A) (B) (C) (D) Solutions : If (0, a) is centre on y-axis, then its radius is a because it passes through origin. ∴ Equation of circle is x2 + (y – a)2 = a2 ⇒ x2 + y2 – 2ay = 0 ....(1) .....(2) Using (1) in (2), Hence (A) is the correct answer. 60. If (logy – logx + 1) then the solution of the equation is : (A) (B) (C) (D) Solutions : Put y = vx ⇒ ⇒ log(log v) = logx + logc = logcx ⇒ log v = cx . Hence (D) is the correct answer. 61. The degree of the differential equation satisfying (A) 2 (B) 3 (C) 4 (D) None of these Solution: Putting x = tanθ and y = tanφ. Then equation becomes secθ + secφ = A (tanθ secφ – tanφsecθ). ⇒ ⇒ ⇒ ⇒ ⇒tan –1x – tan –1y = 2cot –1A Differentiating, we get Which is a differential equation of degree 1. Hence (D) is the correct answer.