DIFFERENTIAL EQUATION. Contents. Theory Exercise Exercise Exercise Exercise

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1 DIFFERENTIAL EQUATION Contents Topic Page No. Theor 0-0 Eercise Eercise - - Eercise Eercise Answer Ke 0 - Sllabus Formation of ordinar differential equations, solution of homogeneous differential equations, variables separable method, linear first order differential equations Name : Contact No. ARRIDE LEARNING ONLINE E-LEARNING ACADEMY A-479 indra Vihar, Kota Rajasthan 4005 Contact No

2 DEFINITIONS : KEY CONCEPT. An equation that involves independent and dependent variables and the derivatives of the dependent variables is called a DIFFERENTIAL EQUATION.. A differential equation is said to be ordinar, if the differential coefficients have reference to a single independent variable onl and it is said to be PARTIAL if there are two or more independent variables u u u. We are concerned with ordinar differential equations onl. eg. + + = 0 is a partial z differential equation.. Finding the unknown function is called SOLVING OR INTEGRATING the differential equation. The solution of the differential equation is also called its PRIMITIVE, because the differential equation can be regarded as a relation derived from it. 4. The order of a differential equation is the order of the highest differential coefficient occurring in it : 5. The degree of a differential equation which can be written as a polnomial in the derivatives is the degree of the derivative of the highest order occurring in it, after it has been epressed in a form free from radicals & fractions so far as derivatives are concerned, thus the differential equation : p ( ) q é m ù é m- d d ù f (, ) ê ú + f(,) ê ú + m m-... = 0 is order m & degree p. ë û ë û Note that in the differential equation e + = 0 order is three but degree doesn t appl. 6. FORMATION OF A DIFFERENTIAL EQUATION : If an equation in independent and dependent variables having some arbitrar constant is given, then a differential equation is obtained as follows : F Differentiate the given equation w.r.t the independent variable (sa ) as man times as the number of arbitrar constants in it. F Eliminate the arbitrar constants. The eliminant is the required differential equation. Consider forming a differential equation for = 4a( + b) where a and b are arbitrar constant. Note : A differential equation represents a famil of curves all satisfing some common properties. This can be considered as the geometrical interpretation of the differential equation. 7. GENERAL AND PARTICULAR SOLUTIONS : SOLUTION. DIFFERENTIAL EQUATIONS The solution of a differential equation which contains a number of independent arbitrar constants equal to the order of the differential equation is called the GENERAL SOLUTION (OR COMPLETE INTEGRAL OR COMPLETE PRIMITIVE). A solution obtainable from the general solution b giving particular values to the constants is called a PARTICULAR Note that the general solution of a differential equation of the n th order contains n & onl n independent arbitrar constants. The arbitrar constants in the solution of a differential equation are said to be independent, when it is impossible to deduce from the solution an equivalent relation containing fewer arbitrar constants. Thus the two arbitrar constants A, B in the equation = A e + b are not independent since the equation can be written as = Ae B. e = C e. Similarl the solution = A sin + B cos ( + C) appears to contain three arbitrar constants, but the are reall equivalent to two onl. A-479 Indra Vihar, Kota Rajasthan 4005 Page No. #

3 8. ELEMENTARY TYPES OF FIRST ORDER & FIRST DEGREE DIFFERENTIAL EQUATIONS. TYPE VARIABLES SEPARABLE : If the differential equation can be epressed as ; f() +g() = 0 then this is said to be variable separable tpe. A general solution of this is gives b f () + g () = c ; ò ò where c is the arbitrar constant. consider the eample (/) = e +.e. TYPE : = f (a + b + c), b ¹ 0. To solve this, substitute t = a + b + c. Then the equation reduces to separable tpe in the variable t and which can be solved. Consider the eample ( + ) = a Tpe. HOMOGENEOUS EQUATIONS : A differential equation of the form f (,) = f (,) where f(, ) & f (, ) are homogeneous functions of &, and of the same degree, is called æ HOMOGENEOUS. This equation ma also be reduced to the form = g ç & is solved b putting è ø = v so that the dependent variable is changed to another variable v, where v is some unknown function, the differential equation is transformed to an equation with variables separable. Consider ( + ) + = 0 TYPE 4. EQUATIONS REDUCIBLE TO THE HOMOGENEOUS FORM : If a + b + c = ; where a a + b + c b a b ¹ 0, i.e. a ¹ then the substitution = u + h, = v + k transform this equation to a homogeneous tpe in the new variables u and v where h and k are arbitrar constants to be chosen so as to make the given equation homogeneous which can be solved b the method as given in tpe. If (i) a b -a b =0, then a substitution u = a +b transforms the differential equation to an equation with variables separable. and (ii) b +a = 0, then a simple cross multiplication and substituting d () for + & integrating term b term ields the result easil. (iii) In an equation of the form : f () + g () = 0 the variables can be separated b the substitution = v. IMPORTANT NOTE : (a) The function f (, ) is said to be a homogeneous function of degree n if for an real number t (¹0), we have f (t, t) = t n f(, ). For e.g. f (, ) = a / + h /. / +b / is homogeneous function of degree /. (b) A differential equation of the form f(,) b a b = is homogeneous if f(, ) is a homogeneous function of degree zero i.e. f(t, t) = t 0 f(, ) = f(, ). The function f does not depend on & separatel but onl on their ratio or. A-479 Indra Vihar, Kota Rajasthan 4005 Page No. #

4 LINEAR DIFFERENTIAL EQUATIONS : A differential equation is said to be linear if the dependent variable & its differential coefficients occur in the first degree onl and are not multiplied together. The nth order linear differential equation is of the form ; n n- d d a0 () + a () an () n n-. = f (). Where a 0 (), a ()... a n () are called the coefficients of the differential equation. Note that a linear differential equation is alwas of the first degree but ever differential equation of d æ the first degree need not be linear. e.g. the differential equation + ç + = 0 is not linear, è ø though its degree is. TYPE - 5 LINEAR DIFFERENTIAL EQUATIONS OF FIRST ORDER : The most general form of a linear differential equations of first order is functions of.to solve such an equation multipl both sides b Note : ò p e. + P = Q, where P & Q are () The factor e ò p. on multipling b which the left hand side of the differential equation becomes the differential coefficient of some function of &, is called integrating factor of the differential equation popularl abbreviated as I.F. () It is ver important at remember that on multipling b the integrating factor, the left hand side becomes the derivative of the product of and the I.F. () Some times a given differential equation becomes linear if we take as the independent variable and as the dependent variable. e.g. the equation ; ( + + ) = + can be written as ( + ) = + + which is a linear differential equation. TYPE - 6 EQUATIONS REDUCIBLE TO LINEAR FORM : The equation n + p = Q. where P & Q function of, is reducible to the linear form b dividing it b n & then substituting n+ =Z. Its solution can be obtained as in Tpe 5. Consider the eample ( +)=. The equation + p = Q n is called BERNOULI S EQUATION. Note : Following eact differentials must be remembered : - æ (i) + = d() (ii) = d ç è ø (iii) - æ = d + ç è (iv) = d ( In) ø + - (v) = d ( l n ( + ) æ (vi) = d çln + è ø (vii) - æ = d çl n - (viii) = æ ç - d tan è ø + è ø (i) - = æ ç - æ d tan e e - e + è () d ç = ø è ø æ e e - e (i) d ç = è ø A-479 Indra Vihar, Kota Rajasthan 4005 Page No. #

5 PART - I : OBJECTIVE QUESTIONS * Marked Questions are having more than one correct option. Section (A) : Degree & Order, Differential equation formation A-. The order and degree of the differential equation é æ ù ê+ ç ú ê ë è ø úû r = d / are respectivel : (A), (B), (C), (D) none of these A- The order of the differential equation whose general solution is given b = (C + C ) sin ( + C ) C C5 4 e + is (A) 5 (B) 4 (C) (D) A-. The order and degree of differential equation of all tangent lines to parabola = 4 is : (A), (B), (C), (D) 4, A-4. If p and q are order and degree of differential equation æ d ç è ø æ + ç ø è / + = sin, then : (A) p > q (B) q p = (C) p = q (D) p < q A-5. Famil = A + A of curve represented b the differential equation of degree (A) three (B) two (C) one (D) none of these A-6*. If = e cos and n + k n = 0, where n = n n and k n, n Î N are constants. (A) k 4 = 4 (B) k 8 = 6 (C) k = 0 (D) k 6 = 4 Section (B): Variable separable, Homogeneous equation, polar substitution B-. If = e and = 0 when = 5, the value of for = is : B-. (A) e 5 (B) e 6 + (C) If f() = f () and f() =, then f() equals e (D) log e 6 (A) e (B) e (C) e (D) e B-. If = and ( ) = 0, then function is (A) ( ) / e - (B) (+ ) / e - (C) log e ( + ) (D) + A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 4

6 + B-4. Integral curve satisfing = -, () =, has the slope at the point (, ) of the curve, equal to (A) 5 (B) (C) (D) 5 B-5. Solution of differential equation = 0 represents : (A) rectangular hperbola. (B) straight line passing through origin. (C) parabola whose verte is at origin. (D) circle whose centre is at origin. B-6*. The solution of + 6 = 0 are (A) = C (B) = C (C) log = C+ log (D) = C Section (C) : Linear upon linear, Linear diff. eq. & bernaullis diff. eq. C-. The solution of the differential equation k = 0,(0) =, approaches zero when, if (A) k = 0 (B) k > 0 (C) k < 0 (D) none of these C-. dv k The solution of + v = g is : dt m (A) v = k t m ce - mg k (B) v = c mg k k t e - m (C) v k t m e - = c mg k k t (D) ve m = c mg k Section (D) : Eact diff. eq., Higher degree differential equation and clairut's form D-. The differential equations of all conics whose centre lie at the origin is of order : (A) (B) (C) 4 (D) none of these D-. The differential equation for all the straight lines which are at a unit distance from the origin is : (A) æ ç - = è ø æ ç ø è (B) æ ç + = + è ø æ ç ø è (C) æ ç - = + è ø æ ç ø è (D) æ ç + = è ø æ ç ø è D-. The equation of the curve whose subnormal is constant a, is (A) = a + b (B) = a + b (C) a = a (D) none of these A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 5

7 PART - II : SUBJECTIVE QUESTIONS Section (A) : Degree & Order, Differential equation formation A-. Find the order and degree of the following differential equations- (i) æd æ ç + ç + = è ø è ø 4 0 (ii) 4 æ æ ç ç = è ø è ø (iii) - æ sin ç = + è ø (iv) æ ç + = è ø (v) e - + = 0 (vi) 5/ é æ ù ê+ ç ú = êë è ø úû A-. A-. (vii) d æ = sinç + è ø Identif the order of the following equations, (where a, b, c, d are parameters) (i) (sin a) + (cos a) = p (ii) = 4a e + b (iii) l n(a) = be + c (iv) = tan Form differential equations to the curves (i) = m (n ), where m, n are arbitrar constant. (ii) c( + c) =, where 'c' is an arbitrar constant. (iii) a + b =, where a & b are arbitrar constants. (iv) = ae + be æp æp ç + a tanç - a + c e è4 ø è4 ø b+ d Section (B): Variable separable, Homogeneous equation, polar substitution B-. Solve the following differential equations (i) ( + cos ) = ( cos ) (ii) - sin = log (iii) = 0 (iv) (ln + ) = sin+ cos B-. Solve: B-. (i) sin( ) cos ( ) = (ii) e + e = (iii) Solve: (i) + ( + ) = 0, given that =, when = (ii) (iii) æ = + sinç èø é æ æ ù é æ æ ù ê cosç sin sin cos 0 + ç - - = ú ê ç ç ú ë è ø è øû ë è ø è øû = - - B-4. Find the equation of the curve satisfing -- = + - and passing through (, ). A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 6

8 Section (C) : Linear upon linear, Linear diff. eq. & bernaullis diff. eq. C-. Solve: (i) tan sin = - (ii) (+ + ) + ( + ) = 0 (iii) ( + ), 0 = > (iv) (+ ) + = cos C-. (A) Find the integrating factor of the following equations (i) (log) log + = (ii) = tan sec C-. (B) If the integrating factor of ( ) + ( a p. ) = 0 is eò, then P is equal to Solve: (i) ( + ) + ( ) = 0 (ii) = (iii) ( + 5) + ( + 5) = 0 Section (D) : Eact diff. eq., Higher degree differential equation and clairut's form D-. Solve: (i) ( + e ) = e (ii) (iii) sin + ( cos + ) = 0 (iv) + = - = + 4 Section (E) : Geometrical and phsical problems E-. E-. Identif the conic whose differential equation is ( + ) = 0 and passing through (, 0). Also find its focii and eccentricit. If a curve passes through the point (, p/4) and its slope at an point (,) on it is given b / cos (/), then find the equation of the curve. E-. (i) The temperature T of a cooling object drops at a rate which is proportional to the difference T-S, where S is constant temperature of the surrounding medium. E-4. E-5. (ii) (iii) Thus, dt k (T S) dt = - -, where k > 0 is a constant and t is the time. Solve the differential equation if it is given that T(0) = 50. The surface area of a spherical balloon, being inflated changes at a rate proportional to time t. If initiall its radius is units and after seconds it is 5 units, find the radius after t seconds. The slope of the tangent at an point of a curve is l times the slope of the straight line joining the point of contact to the origin. Formulate the differential equation representing the problem and hence find the equation of the curve. Find the curves such that the distance between the origin and the tangent at an arbitrar point is equal to the distance between the origin and the normal at the same point. Find the curve such that the oridnate of an of its points is the geometric mean between the abscissa and the sum of the abscissa and subnormal at the point. A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 7

9 PART - III : MISCELLANEOUS OBJECTIVE QUESTIONS MATCH THE COLUMN. Column - I Column - II (A) Solution of = + is : (p) = 5 + c (B) Solution of ( 0 ) + = 0 is : (q) sec = + + ce (C) Solution of sec + tan = is : (r) ( + ) ( ) = c (D) Solution of sin = cos ( cos ) is : (s) tan = + ce. Match the following Column - I (A) = ( + ), () = and ( 0 ) =, then 0 = (p) Column - II 4 (B) If (t) is solution of (t + ) dt t =, (0) =, then () = (q) 5 (C) ( + ) = and () = and ( 0 ) = e, then 0 = (r) - (D) + = 0, () =, then () = (s) e COMPREHENSION # COMPREHENSION Differential equations are solved b reducing them to the eact differential of an epression in & i.e., the are reduced to the form d(f(, )) = 0. e.g. : = Þ = æ Þ d ç + è æ = d ç ø è ø æ Þ d ç + + è ø = 0 \ solution is + + = c. Use the above method to answer the following question ( to 5) A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 8

10 . The general solution of ( ) + ( ) = 0 is : (A) = c (B) = c (C) 4 4 = c (D) = c 4. General solution of the differential equation + + æ ç - = 0 is : è + ø (A) + tan æ ç è ø æ = c (B) + tan = c (C) tan ç è ø = c (D) none of these 5. General solution of the differential equation e + (e ) = 0 is : (A) e = c (B) e = c (C) e + = c (D) e = c Comprehension # 6. Solve n n- d d In order to solve the differential equation of the form a0 + a n a n n = 0 -, where a 0, a, a are constants. We take the auiliar equation a 0 D n + a D n a n = 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 auiliar equation. One real roots a ce a Corresponding complementar function a. Two real and different roots a and a a ce + ce. Two real and equal roots a and a (c + c )e a 4. Three real and equal roots a, a, a (c+ c+ c)e a 5. One pair of imaginar roots a ± ib (c cos b + c sin b) e a 6. Two pair of equal imaginar roots a ± ib and a ± ib [(c + c ) cos b + (c + c ) sin b] e a Solution of the given differential equation will be = sum of all the corresponding parts of the complementar functions. d - + = 0 (A) = (c + c )e (B) = (c e + c e ) (C) = (c ) e (D) none of these 7. Solve a 0 + = (A) = (c cos a + c sin a) e a (C) = c e a + c e a (B) = c cos a + c sin a (D) none of these 8. Solve = (A) = (c + c + c ) e (B) = (c e + c e + c e ) (C) = c e + c e + c e (D) none of these A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 9

11 ASSERTION/REASONING 9. Statement - : The relation = A sin + B cos can be represented b the differential equation d + = 0. Statement - : Solution of sec + tan = is tan = / e c +. (A) Statement- is True, Statement- is True; Statement- is a correct eplanation for Statement-. (B) Statement- is True, Statement- is True; Statement- is NOT a correct eplanation for Statement- (C) Statement- is True, Statement- is False. (D) Statement- is False, Statement- is True. 0. Statement - : Solution of ( ) ( ) / = 0 is - + ( + ) + c = 0 Statement - : Solution of ( + ) + ( ) = 0 is ln - = c (A) Statement- is True, Statement- is True; Statement- is a correct eplanation for Statement-. (B) Statement- is True, Statement- is True; Statement- is NOT a correct eplanation for Statement- (C) Statement- is True, Statement- is False. (D) Statement- is False, Statement- is True.. Statement - : The equation of the curve passing through (, 9) which satisfies differential equation = + is 6 = æ Statement - : The solution of D.E. ç - è ø (e + e ) + = 0 is = c e + c e. (A) Statement- is True, Statement- is True; Statement- is a correct eplanation for Statement-. (B) Statement- is True, Statement- is True; Statement- is NOT a correct eplanation for Statement- (C) Statement- is True, Statement- is False. (D) Statement- is False, Statement- is True.. Statement- : The Solution of D.E. - - = m m is given b tan = + c Statement-: The solution of differential equation + = sin is ( + cos ) = sin + c (A) Statement- is True, Statement- is True; Statement- is a correct eplanation for Statement-. (B) Statement- is True, Statement- is True; Statement- is NOT a correct eplanation for Statement- (C) Statement- is True, Statement- is False. (D) Statement- is False, Statement- is True.. Statement-: Solution of the differential equation - = + is = c ( ) ( + ) Statement-: D.E. = f(). g() can be solved b seperating variables. f() g() = (A) Statement- is True, Statement- is True; Statement- is a correct eplanation for Statement-. (B) Statement- is True, Statement- is True; Statement- is NOT a correct eplanation for Statement- (C) Statement- is True, Statement- is False. (D) Statement- is False, Statement- is True. A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 0

12 PART - I : OBJECTIVE QUESTIONS. The differential equation of all conics whose aes coincide with the aes of coordinates is of order : (A) (B) (C) 4 (D). If () is a solution of the differential equation + f() = 0, then a solution of differential equation + f() = r () is : (A) () ò r() () (B) () ò () (C) òr() () (D) none of these. If () and () are two solutions of + f() = r(), then () + () is solution of : (A) + f() = 0 (B) + f() = r() (C) + f() = r()(d) + f () = r() 4. The value of lim () obtained from the differential equation =, where (0) = is (A) zero (B) (C) (D) none of these 5. The slope of a curve at an point is the reciprocal of twice the ordinate at the point and it passes through the point (4, ). The equation of the curve is (A) = + 5 (B) = 5 (C) = + 5 (D) = The equation of the curve which is such that the portion of the ais of cut off between the origin and tangent at an point is proportional to the ordinate of that point is : (A) = (b a log ) (B) log = b + a (C) = (a b log ) (D) none of these 7. The solution of + e = 0 is (A) + e = C (B) e = 0 (C) + e = C (D) none of these - 8. The solution of + - = 0 is (A) sin, sin = C (B) sin = C sin (C) sin sin = C (D) sin +sin = C 9. S : The differential equation of parabolas having their vertices at origin and foci on the -ais is a equation whose variables are separable. S : Straight lines which are at a fied distance p from origin is a differential equation of degree. S : All conics whose aes coincide with the aes of coordinates is a equation of order. (A) TTT (B) TFT (C) FFT (D) TTF A-479 Indra Vihar, Kota Rajasthan 4005 Page No. #

13 0. The solution of the differential equation ( sin cos ) + ( cos sin sin ) = 0 is (A) sin = sin + C (B) sin + sin = C (C) sin + sin = C (D) sin + sin = C More than one choice tpe. Solution of the differential equation (A) tan + sin = c (C) tan. sin = c + + = 0 is - (B) tan + sin = c - - (D) cot + cos - = c. The solution of ( + + ) = are (A) + + = Ce (C) log ( + + ) = C (B) = = C log (D) log ( + + ) = C +. Solution of differential equation f() = f () + f() + f'() is (A) = f() + ce (B) = f() + ce (C) = f() + ce f() (D) = cf() + e PART - II : SUBJECTIVE QUESTIONS. Solve the following differential equations. (i) + = + (ii) (iv) ' sin = cos (sin ) - = e (e - e) (iii) 4 - = cos. Solve: = + ò, given =, where = 0 0. Find the integral curve of the differential equation (- l n) + = 0 which passes through (, /e). 4. Let and are two different solutions of the equation ' + P(). = Q(). (i) (ii) Prove that = + C ( ) is the general solution of the same equation (C is a constant) Find the relationship between the constants a and b such that the linear combination a + b be a solution of the given equation. 5. Solve the following differential equations = (ii) ( + tan) ( ) + = 0 a ( a ) 0 (i) ( ) 6. Find the curve for which sum of the lengths of the tangent and subtangent at an of its point is proportional to the product of the co-ordinates of the point of tangenc, the proportionalit factor is equal to k. 7. Find the nature of the curve for which the length of the normal at the point P is equal to the radius vector of the point P. 8. The perpendicular from the origin to the tangent at an point on a curve is equal to the abscissa of the point of contact. Find the equation of the curve satisfing the above condition and which passes through (, ). 9. Find all the curves possessing the following propert; the segment of the tangent between the point tangenc & the -ais is bisected at the point of the intersection with the -ais. A-479 Indra Vihar, Kota Rajasthan 4005 Page No. #

14 PART-I IIT-JEE (PREVIOUS YEARS PROBLEMS) *Marked Questions ma have more than one correct option.. If (t) is a solution of ( + t) dt t = and (0) =, then () is equal to : (A) / (B) e + / (C) e / (D) / [JEE - 00, (Screening) ]. An right circular cone of height H and radius R is pointed at bottom. It is filled with a volatile liquid completel. If the rate of evaporation is directl proportional to the surface area of the liquid in contact with air (constant of proportionalit k > 0), find the time in which whole liquid evaporates. [IIT-JEE-00, Mains, (4, 0)]. If P() = 0 and dp() > P() for all ³ then prove that P() > 0 for all >. [IIT-JEE-00, Mains, (4, 0)] + sin æ 4. If = () and + ç è ø = cos, (0) =, then æp ç equals : [JEE - 004, (Screening) ] èø (A) (B) (C) (D) - 5. A curve passes through (, 0) and slope at point P(, ) is ( + ) + ( -). Find equation of curve ( + ) and area between curve and -ais in 4 th quadrant. [JEE (Mains) 4 out of 60] 6. The solution of primitive integral equation ( + ), is = (). If () = and ( 0 ) = e, then 0 is : (A) e (B) (e - ) (C) (e + ) (D) (e + ) 7. For the primitive integral equation + = ;, > 0, = (), () =, then ( ) is : (A) (B) (C) (D) 5 [JEE - 005, (Screening) + ] 8. If length of tangent at an point on the curve = f() intercepted between the point and the -ais is of length. Find the equation of the curve. [JEE (Mains) 4 out of 60] 9*. A tangent drawn to the curve = f() at P(, ) cuts the -ais and -ais at A and B respectivel such that BP:AP = :, given that f() =, then : [JEE (5, ) out of 84] (A) equation of curve is - = 0. (B) normal at (, ) is + = 4. (C) curve passes through (, /8). 0. Let fbg be differentiable on the interval b0, g such that f (D) equation of curve is + = 0 t f bg- ft bg bg=, and lim = t t- for each > 0. Then f() is : [JEE 007 (, )out of 8] (A) 4 + (B) - + (C) - + (D) A-479 Indra Vihar, Kota Rajasthan 4005 Page No. #

15 . The differential equation = - determines a famil of circles with : (A) variable radii and a fied centre at (0, ). (B) variable radii and a fied centre at (0, ). (C) fied radius and variable centres along the -ais. (D) fied radius and variable centres along the -ais. [JEE - 007(, )out of 8]. Let a solution = () of the differential equation = 0 satisf () = Statement- : () = and æ p sec ç sec è 6 ø. Statement- : () is given b = - -. (A) Statement - is true, Statement - is true ; Statement - is correct eplanation for Statement -. (B) Statement- is true, Statement- is true;statement - is NOT correct eplanation for Statement-. (C) Statement - is true, Statement - is false. (D) Statement - is false, Statement - is true. [JEE 008(, )out of 80]. Let f be a non-negative function defined on the interval [0, ]. If ò -(f'(t) ) dt = òf(t) dt, 0 and f(0) = 0, then : [JEE 009(4, )out of 80] 0 0 æ (A) f ç < è ø æ and f ç > è ø æ æ (C) f ç < and fç < è ø è ø æ æ (C) f ç > and fç > è ø è ø æ æ (D) f ç > and fç < è ø è ø 4*. Match the statements/epression in Column I with the open intervals in Column II. Column I [JEE 009(8,0)out of 80] Column II æ p p (A) Interval contained in the domain of definition of non-zero (p) ç -, è ø æ p solutions of the differential equation ( -) + = 0 : (q) ç0, è ø æ p 5p (B) Interval containing the value of the integral (r) ç, è 8 4 ø 5 ò ( -)( - )( - )( - 4)( - 5) : æ p (C) Interval in which at least one of the points of local (s) ç0, è 8 ø maimum of cos + sin lies : (D) Interval in which tan (sin + cos) is increasing : (t) (-p, p) A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 4

16 5*. Match the statements/epression given in Column I with the values given in Column II. [JEE 009 (8,0)out of 80] Column I Column II (A) The number of solution of the equation (p) e sin cos = 0 in the interval (0, p/) : (B) Value(s) of k for which of the planes k z = 0, (q) 4 + k + z = 0 and + + z = 0 intersect in a straight line : (r) (C) Value(s) of k for which = 4k has integer solution(s) : (s) 4 (D) If = + and (0) =, then value(s) of (ln ) : (t) 5 6. Let f be a real-valued differentiable function on R (the set of all real numbers) such that f() =. If the - intercept of the tangent at an point P(, ) on the curve = f() is equal to the cube of the abscissa of P, then the value of f(-) is equal to. [JEE 00, (, 0) Out of 84] 7. Let : [, ) [, ) be a differentiable function such that () =. If, 6 ò (t)dt = () for all ³, then the value of () is. [JEE 0, (4, 0) Out of 8] 8. Let () + () g () = g() g (), (0) = 0, Î R, where f '() denotes df() and g() is a given nonconstant differentiable function on R with g(0) = g() = 0. Then the value of ( ) is. [JEE 0, (4, 0) Out of 8] 9*. If () satisfies the differential equation tan = sec and (0) = 0, then : [JEE 0] æp (A) ç è 4 ø = p 8 æp (C) ç è ø = p 9 æp (B) ç è4 ø = p 8 æp (D) ç è ø = 4 p p + æ p 0. A curve passes through the point ç, 6 è ø. Let the slope of the curve at each point (, ) be sec æ + ç, > è ø 0. Then the equation of the curve is [JEE-Advanced-0, Paper-I] (A) sin æ ç = log + è ø (B) cosec æ ç = log + è ø (C) sec æ æ ç è ø = log + (D) cos ç è ø = log + A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 5

17 PART-II AIEEE (PREVIOUS YEARS PROBLEMS). The degree and order of the differential equation of the famil of all parabolas whose ais is -ais, are respectivel: [AIEEE 00] (), (), (), (4),. The solution of the differential equation ( + ) + ( e tan ) = 0, is : [AIEEE 00] () ( ) = k e tan () tan e = tan e + k tan () e = tan + k (4) e = e + k. The differential equation for the famil of curves + a = 0, where a is an arbitrar contant, is: () ( ) = () ( + ) = () ( ) = (4) ( + ) = tan tan [AIEEE 004] 4. The solution of the differential equation + ( + ) = 0 is : [AIEEE 004] () = c () + log = c () + log = c (4) log = c 5. The differential equation representing the famil of curves = c( + c ), where c > 0, is a parameter, is of order and degree as follows : [AIEEE 005] () order, degree () order, degree () order, degree (4) order, degree 6. If = (log log + ), then the solution of the equation is : [AIEEE 005] æ () log æ ç è = c () log ç ø è ø æ = c () log ç è ø æ = c (4) log ç è ø = c 7. The differential equation whose solution is A + B =, where A and B are arbitar constant, is of : () first order and second degree. () first order and first degree. [AIEEE 006] () second order and first degree. (4) second order and second degree. 8. The differential equation of all circles passing through the origin and having their centres on the -ais is : [AIEEE 007] () = + () = + () = + (4) = 9. The normal to a curve at P(, ) meets the -ais at G. If the distance of G from the origin is twice the abscissa of P, then the curve is a : () ellipse () parabola () circle (4) hperbola [AIEEE 007] A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 6

18 0. The solution of the differential equation = + satisfing the condition () = is : [AIEEE 008] () = log + () = log + () = e ( ) (4) = log + c. The differential equation which represents the famil of curves = c e is : () = () = (). = (4). = ( ), where c and c are arbitar constants [AIEEE 009]. Solution of the differential equation cos = (sin ), 0 < < p is: [AIEEE 00] () sec = tan + c () tan = sec + c () tan = (sec + c) (4) sec = (tan + c). The value of 8log (+ ) ò is : 0 + p () p log () log 8 p () log (4) log 4. Let l be the purchase value of an equipment and V(t) be the value after it has been used for t ears. The value V(t) depreciates at a rate given b differential equation dv(t) dt = k(t t), where k > 0 is a constant and T is the total life in ears of the equipment. Then the scrap value V(T) of the equipment is: [AIEEE 0] () l kt () l k(t t) () e kt (4) T k 5. If = + > 0 and (0) =, then (In ) is equal to : [AIEEE 0] () 7 () 5 () (4) 6. The curve that passes through the point (, ), and has the propert that the segment of an tangent to it ling between the coordinate aes is bisected b the point of contact is given b : [AIEEE 0] 6 æ æ () = 0 () = () + = (4) ç + = ç è ø è ø æ 7. Consider the differential equation + ç - = 0. If () =, then is given b : [AIEEE-0] è ø () e 4- - () e e e e - + () + - (4) - + e e e 8. The population p(t) at time t of a certain mouse species satisfies the differential equation dp(t) dt = 0.5p(t) 450. If p(0) = 850, then the time at which the population becomes zero is: [AIEEE 0] () ln 8 () ln 9 () ln 8 (4) ln 8 A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 7

19 NCERT BOARD QUESTIONS Short Answer. Find the solution of = -. Find the differential equation of all non vertical lines in a plane.. Given that - = e and = 0 when = 5. Find the value of when =. 4. Solve the differential equation ( ) + = - 5. Solve the differential equation + = 6. Find the general solution of a e + = m 7. Solve the differential equation e + + = 8. Solve : =. 9. Solve the differential equation = + + +, when = 0, = Find the general solution of ( + ) =.. If () is a solution of æ + sin ç è + ø = cos and (0) =, then find the value of æp ç è ø.. If (t) is a solution of ( + t) t dt - = and (0) =, then show that () =.. Form the differential equation having = (sin ) + Acos + B, where A and B are arbitrar constants, as its general solution. 4. Form the differential equation of all circles which pass through origin and whose centres lie on - ais. 5. Find the equation of a curve passing through origin and satisfing the differential equation ( + ) 4 + =. A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 8

20 6. Solve : = Find the general solution of the differential equation ( + ) + ( e tan ) 0 =. 8. Find the general solution of + ( + ) = Solve : ( + ) ( ) = Solve : ( + ) 0 =, given that () =.. Solve the differential equation = cos ( cosec ) given that = when = p.. Form the differential equation b eliminating A and B in A + B =.. Solve the differential equation ( + ) tan + (+ ) = Find the differential equation of sstem of concentric circles circles with centre (, ). Long Answer (L.A.) 5. Solve : d + () = (sin + log ) 6. Find the general solution of (+ tan) ( ) + = Solve : cos( ) sin( ) = Find the general solution of = sin. 9. Find the equation of a curve passing through (, ) if the slope of the tangent to the curve at an point (, ) is Find the equation of the curve through the point (, 0) if the slope of the tangent to the curve at an - point (, ) is. +. Find the equation of a curve passing through origin if the slope of the tangent to the curve at an point (,) is equal to the square of the difference of the abcissa and ordinate of the point.. Find the equation of a curve passing through the point (, ). If the tangent drawn at an point P(, ) on the curve meets the co-ordinate aes at A and B such that P is the mid-point of AB.. Solve : (log log ) = - + A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 9

21 EXERCISE # PART # I A-. (A) A- (D) A-. (A) A-4. (D) A-5. (A) A-6*. (A, B) B-. (C) B-. (B) B-. (B) B-4. (A) B-5. (B) B-6*. (A,C,D) C-. (C) C-. (A) D-. (B) D-. (C) D-. (B) PART # II A-. (i) (, ) (ii) (, ) (iii), (iv), (v), degree is not applicable (vi), (vii), degree is not applicable A-. (i) (ii) (iii) (iv) A-. (i) + ( ) = 0 (ii) (') = [8 (') 7] (iii) d æ + 0 ç - = è ø (iv) d + = B-. (i) = tan / + c (ii) = - sin - cos + log log + c (iii) log + + = c + + (iv) sin = l n + c B- (i) æ+ log tanç + = + c è ø (ii) tan (e ) + = c (iii) = c( + ) - B-. (i) = + (ii) tan c = (iii) æ cosç = c è ø B-4. + = 0 C-. (i) = cos + c sec (ii) = c arc tan (iii) c = + (iv) ( + ) = c + sin C-. (A) (i) l n (ii) sec (B) ( -) (- ) C-. (i) + + = c (ii) - + ln ( ) = c 8 (iii) 4 + ( + ) 0 ( + ) = c D-. (i) e = - + c (ii) c = + (iii) sin = + c (iv) = ( + ) ln + + c(+ ) - A-479 Indra Vihar, Kota Rajasthan 4005 Page No. # 0

22 E-. Conic : = (hperbola) focii : ( ±,0 ),e = E-. tan / = log. E-. (i) T- S = e 50 - S -kt (ii) r = 4t + 9 units (iii) = kl where, k is some constant E-4. - ± tan + = ce E-5. = c or + l n = c PART # III. (A-r), (B-p) (C-s) (D-q). (A-q) (B-r) (C-s) (D-p). (B) 4. (A) 5. (A) 6. (A) 7. (B) 8. (C) 9. (C) 0. (B). (C). (D). (A) EXERCISE # PART # I. (A). (B). (C) 4. (B) 5. (C) 6. (A)\ 7. (A) 8. (D) 9. (A) 0. (A). (A, D). (A, D). (C) PART # II. (i) ( + ) = c (ii) e = c. ep ( e ) + e (iii) = sin + c (iv) c = sin+ sin. = (e - e + ) -e. (e + ln + ) = 4. (ii) a + b = 5. (i) ( + ) + a ( ) = c (ii) e (cos + sin ) = e sin + C 6. = ± l - 7. rectangular hperbola or circle 8. + = 0 9. = c k n c(k ) EXERCISE # PART # I. (A). t = H/k 4. (A) (A) 7. (C) log c + - = ± + 9*. (B, C, D) 0. (A). (C). (C). (C) 4*. (A-pqs), (B-pt) (C-pqrt) (D-s) 5*. (A-p) (B-qs) (C-qrst) (D-r) 6. 9 A-479 Indra Vihar, Kota Rajasthan 4005 Page No. #

23 7. Bonus (Taking =, the integral becomes zero, whereas the right side of the equation gives 5. Therefore, the function f does not eist) *. (A, D) 0. (A) PART # II. (). (). () 4. () 5. () 6. () 7. () 8. () 9. () 0. (4). (4). (4). () 4. () 5. () 6. () 7. () 8. () EXERCISE # 4 NCERT BOARD QUESTIONS.. 0 =. 6 e æ - ( - ) = logç + k è + ø 5. = c.e - m a 6. (a + m) = e + ce - 7. ( c) e + + = = ke 9. æ = tanç + è ø 0. = ( + c).. d ( ) = 4. ( ) - - = 0 5. = 4 + ( ) 6. - æ tan ç = log + c è ø 7. tan tan e - = e - + c 8. - æ tan ç + log = c è ø 9. + = ke 0. ( + ) = e +. -cos sin = +. " + (') ' = 0. (tan - ) log( ) c + + = 4. ( - ) + ( - ) = 0 5. sin cos log cos cs 9 - = (sin + cos ) = sin + ce 7. æ+ log + tanç = + c è ø 8. ésin + cosù = - ê + ce ú ë û 9. ( - ) = 0. ( ) ( + ) + = 0. ke ( + ) = +. =. æ logç = c è ø A-479 Indra Vihar, Kota Rajasthan 4005 Page No. #