C1 Order and Degree of a Differential Equation : Order : Order is the highest differential appearing in a differential equation. MDE 1 DIFFERENTIAL EQUATION Degree : It is determined b the degree of the highest order derivative present in it after the differential equation is cleared of radicals and fractions so far as the derivatives are concerned. m d f (, ) m d n m d f(, ) d 1 1 n 1 nk d...fk (, ) d 1 m The above differential equation has the order m and degree n 1. Practice Problems : 1. The order of the differential equation whose general solution is given b = (c 1 + c )e 4 + c is 4 1. The degree of the differential equation, of which = 4a( + a) is a solution, is 1 of these. The order of the differential equation whose general solution is given b : c = c 1 cos( + c ) (c + c 4 ) a 5 c6 sin( c 7 ) is 4 5 [Answers : (1) d () b () c] 0 c e C Formation of Differential Equation : The differential equation corresponding to a famil of curve can be obtained b using the following steps : Identif the number of essential arbitrar constant in equation of curve. If arbitrar constants appear in addition, subtraction, multiplication or division, then we can club them to reduce into one new arbitrar constant. Differentiate the equation of curve till the required order. Eliminate the arbitrar constant from the equation of curve and additional equation obtained in step above. 1. The differential equation of the famil of curves = e (A cos + B sin ), where A and B are arbitrar constants, is d d d d 0 0 d d d d d d 0 of these d d [Answers : (1) c] C Solution of a Differentible Equation : The solution or the integral of a differential equation is, therefore, a relation between dependent and independent variables (free from derivatives) such that it satisfies the given differential equation. Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Etn., Outer Ring Road New Delhi 110 018, Ph. : 916905, 85711111
MDE Practice Problems : 1. The solution of the differential equation, d d + d = 0 is constant constant constant. The solution of the differential equation; d ( e + e / ) = d (e / e ) is / = ln(e / + ) = ln(e / + ) + = ln(e / + ) = ln(e / + ) [Answers : (1) a () b] C4 Solution of elementar tpes of first order and first degree differential equations : d Equations Reducible to the Variables Separable form : Its general form is = f(a + b + c) d a, b 0. To solve this, put a + b + c = t. Homogeneous Differential Equation : C5 d f(, ) A differential equation of the form where f and g are homogeneous function of and, and d g(, ) of the same degree, is called homogeneous differential equation and can be solved easil b putting = v. Equations Reduciable to the Homogeneous form Equations of the form d d a b c A B C can be made homogeneous (in new variables X and Y) be substituting = X + h and = Y + k, where h and k are constants. Now, h and k are chosen such that ah + bk + c = 0, and Ah + Bk + C = 0; the differential equation can now be solved b putting Y = vx. C6 Linear differential equations of first order : The differential equation d P Q, is linear in, where P and Q are functions of. d Integrating Factor (I.F.) : It is an epression which when multiplied to a differential equation converts it into an eact form. I.F. for linear differential equation = Pd e (constant of integration will not be considered) d After multipling above equation b I.F. both side, it becomes (.e Pd ) = d.e Pd Q.e Pd d + c. Pd Qe Some times differential equation becomes linear if is taken as the dependent variable and as independent variable. The differential equation has then the following form : Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Etn., Outer Ring Road New Delhi 110 018, Ph. : 916905, 85711111
MDE d P d 1 Q 1 Practice Problems :. where P 1 and Q 1 are functions of. pthe I.F. now is d 1. Solution of the differential equation ( ) is d P 1 d e = (c + ) = (c ) = (c ) = (c + ). The equation of the curve whose tangent at an point (, ) makes an angle tan 1 ( + ) with -ais and which passes through (1, ) is 6 + 9 + = 6e ( 1) ( 1) 6 9 + = 6e 6 + 9 = 6e ( 1) None of these [Answers : (1) d () a] C7 Time saving tips : 1. The equation of the form d n P Q where P and Q are functions of onl and n is constant d n d 1n {n 0, 1} can be reduced to linear form b. P. Q and put 1 n = v. d. Some important integers factors (I.F.) for the quick solutions 1 d d (i) d tan (ii) d d dlog (iii) d( m n ) = m 1. n 1.(md + nd) (iv) 1 d d d log Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Etn., Outer Ring Road New Delhi 110 018, Ph. : 916905, 85711111
MDE 4 SINGLE CORRECT CHOICE TYPE 1. The solution of the differential equation, d d + d = 0 is constant constant constant. The solution of the differential equation; d ( e + e / ) = d (e / e ) is / = ln(e / + ) = ln(e / + ) + = ln(e / + ) = ln(e / + ). The differential equation of all the conics whose aes coincide with the coordinate aes is given b d d d.. 0 d d d d d d.. 0 d d d d d d. 0 d d d d d d.. 0 d d d 4. For an differential function = f(); the value of d d d. d is d d 0 1 5. The curve is satisfing the following differential equation d d ( ) 1 and d d = 0 for = 1, then the value of at = is 1 6. Tangent is drawn at an point P of a curve which passes through (1, 1) cutting -ais and -ais at A and B respectivel. If AP : BP = : 1, then the curve is passing through the point 1, 8 1, 8 1 1 1, 8,1 8 7. A curve is satisfing the following differential d d equation 0. The curve is d d passing through the point (, 1) (1, ) ( 1, 1) (, ) 8. Let f : R R and g : R R be twice differentiable functions satisfing f () g(),f (1) g(1) 4 and f() = g() = 9. The value of f(4) g(4) is equal to 6 16 10 8 9. The order of the differential equation whose c e general solution is given b = (c 1 + c )e + c 4 is 4 1 10. The solution of the differential equation d 1 cos is d 1 sin c 1 cos c 1 tan c Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Etn., Outer Ring Road New Delhi 110 018, Ph. : 916905, 85711111
MDE 5 EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE Comprehension-1 Consider the curve cos 1 1 sin t dt cos t sin dt (0 /) 1/ 8 1/ 8 and the curve satisfing the differential equation ( + )d = ( )d passing through (4, ) 1. The curve is parallel to -ais parallel to -ais is a straight line passing through the origin. The solution of the differential equation is + = 0 + = 0 + = 0. The area bounded b the curve and satisfing the differential equation is 4 1 8 16 1 8 16 Comprension- 1 8 16 For a certain curve = f() satisfing d 6 4 ; f() has a local minimum value 5 d when = 1. 4. The equation of the curve is = + + 5 = + + + 5 = + 5 5. The global maimum of f() is 5 7 9 6. The global minimum of f() is 5 7 9 MATRIX-MATCH TYPE Matching-1 Column - A Column - B (A) The order of the differe- (P) 1 ntial equation d d 1 is d d (B) The degree of the (Q) two differential equation d 1 is d (C) The degree of the (R) differential equation of the famil of all parabolas whose ais is the -ais, are respectivel (D) The order of the differen- (S) tial equation of the famil of the circles touching the -ais at the origin MULTIPLE CORRECT CHOICE TYPE d 1. The solution of ( + ) = 1 is d 1/ = + Ce / the solution of an equation which is reducible to linear equation. / = 1 + e / 1 / Ce. The differential equation of the curve for which the initial ordinate ( intercept) of an tangent is equal to the corresponding subnormal is linear is homogeneous has separable variables of these Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Etn., Outer Ring Road New Delhi 110 018, Ph. : 916905, 85711111
MDE 6. Consider the following famil of curves = e (Acos + Bsin) where A, B are constant. Chooose the correct statments its differential equation has second order the differential equation of the curve is given b d d. d d its differential equation has third order of these 4. The equation of the curve passing through (1, ) whose differential equation is ( + )d = ( )d, is given b a + b + c = 0, then a = 1 b = c = 5 all the above 5. A student studing a foreign language has 50 verbs to memorize. The rate at which the student can memorize these verbs is proportional to the number of verbs remaining to be momorized; that is, if the student memorizes verbs in t minutes. d k(50 ) dt Assume that initiall no verbs are momorized, and suppose that 0 verbs are memorized in the first 0 minutes. Then The number of verbs memorized b the students in one hour is The number of verbs memorized b the students in one hour is 4.5 The number of hours after which the student have onl one verb left to memorize is.8 (Answers) EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE 1. a. b. a 4. a 5. c 6. b MATRIX-MATCH TYPE 1. [A-Q; B-Q; C-P; D-P] MULTIPLE CORRECT CHOICE TYPE 1. a, b, d. a, b. a, b 4. a, b, c, d 5. a, b, c Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Etn., Outer Ring Road New Delhi 110 018, Ph. : 916905, 85711111
MDE 7 INITIAL STEP EXERCISE (SUBJECTIVE) 1. Solve d d ( ) ( )( ). d. Solve. d 1 d d. Solve ( ) 0. d d 4. Determine all curve for which the ratios of the length of the segment intercept b an tangent on the - ais to the length of the radius vector is a constant. 5. Solve the equation (t)dt ( 1) t(t)dt, 0. 0 0 6. Given two curves : = f() passing through (0, 1) and f (t)dt passing through (0, 1/n). The tangent drawn to both the curves at the points with equal abscissas intersect on the -ais find the curve = f(). 7. A normal is drawn at a point P(, ) of a curve. It meets the -ais and the -ais in point A and B, 1 1 respectivel, such that 1, where O is OA OB the origin, find the equation of such a curve passing through (5, 4). 8. Solve the differential equation 9. Solve 10. Solve : ( + 4 + 4)d = ( + 4 + 1)d. d d d d 4. 1. Solve : cos (d d) sin (d d) 0 1. Reduce the equation d + ( + )d = 0 to a homogeneous equation and then solve it. d 14. If () (sin + log ), find (). d 15. Solve the following differential equation. 16. Solve : 17. Solve : (1 )d + ( 5 )d = 0. d (1 ) ( 1) a d f d d. f 18. The tangent at a point P of a curve meets the ais of in N; the line through P parallel to the ais of meets the ais of at M; O is the origin. If the area of MON is constant. Show that the curve is a hperbola. 19. Prove the orthogonal of curves 1, a b where is a parameter, is self orthogonal. 0. A and B are two separate reservoirs of water. The capacit of A is double that of B. Both the reservoirs are filled completel with water. Water is released simultaneousl from both the reservoirs. For each of the reservoirs, the rate of flow out at an instant is proportional to the quantit of water left in the reservoir. After one hour, the quantit of water in A is 1.5 times the quantit of water in B. After how man hours from the time of release of water, do both A and B have the same quantit of water? d e d 11. Solve : 1 e d e 1 d 0 Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Etn., Outer Ring Road New Delhi 110 018, Ph. : 916905, 85711111
MDE 8 FINAL STEP EXERCISE (SUBJECTIVE) 1. Let u() and v() satisf the differential equations du dv p()u f () and p()v g(), d d where p(), f() and g() are continuous functions. If u( 1 ) > v ( 1 ) for some 1 and f() > g(), for all > 1, prove that an point (, ), where > 1, does not satisf the equation = u() and = v().. Find the pair of curves such that : the tangents drawn at points with equal abscissas intersect on the -ais. the normal drawn at points with equal abscissas intersect on -ais. one curve passes through (1, 1) and other passes through (, ). Determine the equation of the curve passing through the points (a, a) (a > 0) in the form = f() which satisf the differential equation; a d. d 4. Find the famil of curves, the subtangent at an point of which is the arithmetic mean of the co-ordinates of the point of tangenc. 5. There are 100 million litres of fluoridated water in the reservoir containing a cit s water suppl, and the water contains 700 kg of fluoride. To decrease the fluoride content, fresh water runs into the reservoir at the rate of million litres per da, and the miture of water and fluoride, kept uniform, runs out of the reservoir at the same rate. How man kilograms of fluoride are in the reservoir 60 das after the pure water started to flow into the reservoir? 6. Find the time required for a clindrical tank of radius r and height H to empt through a round of area a at the bottom. The flow through a hole is according to the law v(t) = u gh(t) and v(t) and h(t) are respectivel the velocit of flow through the hole and the height of the water level above the hole at time t and g is the acceleration due to gravit (u is a constant quantit). 7. Let f(, ) = 0 be a curve passing through the point (0, ) such that all the ras coming from a fied point source when reflected from it are parallel. Determine all such possible curves. 8. If the area of a triangle formed b tangent at an point on the curve and co-ordinates aes is constant, then form the differential equations of all such curves and hence prove that the curve is either a straight line or a hperbola. 9. A curve is such that the length of perpendicular from origin on the tangent at an point P of the curve is equal to the absciassa of P. Prove that the differen- d tial equation of the curve is 0 d and hence find the curve. 10. A curve passing through (1, ) has its slope at an point (, ) equal to. Find the area of the region bounded b the curve and the line 4 = 0. 11. The normal at an point P(, ) of a curve meets the -ais at G. If the distance of G from the origin be twice the abscissa of P, prove that the curve is a rectangular hperbola. ANSWERS (SINGLE CORRECT CHOICE TYPE) 1. a. b. d 4. a 5. d 6. a 7. b 8. c 9. d 10. a Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Etn., Outer Ring Road New Delhi 110 018, Ph. : 916905, 85711111
MDE 9 ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE) 1. X Y X 4 v 1 Ke Here, X = +, Y = and v = Y/X. = (1 ) + c (1 ) 1/4. ( + c 1 ) ( + ln c ) = 0 4. c k ln 5. c e 1/ 6. f() = e n 7. ( 1) + ( 1) = 5 8. log 4 4 4 8 log c 1 9. k 10. log c e 11. c e 1. sin = c 1 1 1. tan log[ ( ) ] c 14. cos sin cos log 9 c 15. c 1 5 16. c 1 a 17. cf 0. log 4/ ANSWERS SUBJECTIVE (FINAL STEP EXERCISE).,. 1 e a 1 e k k 4. ( ) = c 5. 115.7 kg [approimatel] 6. r ua H g 7. = 4( + 1), = 4( 1) 9. + = c 10. 9 Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Etn., Outer Ring Road New Delhi 110 018, Ph. : 916905, 85711111