Ordinary Differential Equations of First Order

CHAPTER 1 Ordinar Differential Equations of First Order 1.1 INTRODUCTION Differential equations pla an indispensable role in science technolog because man phsical laws relations can be described mathematicall in the fm of differential equations such as oscillations of mechanical electrical sstems, conduction of heat etc. It is also known that an tpe of motion can be modelled b a differential equation with appropriate initial boundar conditions, the solution gives the equation of the curve traced out during the motion. Therefe knowledge about various methods of solving differential equations is a must f ever student of engineering science. In this chapter we discuss the dinar differential equations (ODE) of first der in two parts (1) ODEs of first der of first degree () ODEs of first der of higher degree. 1. BASIC CONCEPTS An relation between one me independent variables, a dependent variable their differential coefficients differentials is called a differential equation. The following are eamples of differential equations: d sin...(1) d d d 5 cos...() d d 3 d d d 3 3 d d d...(3) 3 3 3 d d d 3 4 sin...(4) d d d d ed 0...(5) d d 3 0...(6) d d

A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. II] u u u k...(7) Ordinar Differential Equation u u u 0...(8) z An equation involving onl one independent variable, a dependent variable their differential coefficients differentials is called an dinar differential equation (ODE). Equations (1) to (6) are ODEs. In the present treatise onl dinar differential equations will be considered. Therefe the wd dinar will be omitted in most cases. Partial Differential Equation An equation involving two me independent variables, a dependent variable partial derivatives of the dependent variable with respect to the independent variables is called a partial differential equation (PDE). Equations (7) (8) are PDEs. Order of a Differential Equation The der of a differential equation is the der of the highest dered derivative appearing in the equation. F eamples, the der of each of the differential equations (1), (5), (6) is one, the der of () is two the der of each of (3) (4) is three. Degree of a Differential Equation The degree of a differential equation is the highest power of the highest dered derivative appearing in it after it has been epressed rational integral as far as derivatives are concerned. Thus in the previous eamples, (1) (5) are of first der first degree; () is of second der first degree; (3) (4) are of third der first degree; (6) is of first der second degree. We net consider the following two equations: 1 d d The equation (9) can be written as 3/ d d a 0, d d equation (10), when squared, becomes 3 d d a...(9) d d d k...(10) d 1 d d d k d Therefe the equation (9) is of first der second degree the equation (10) is of second der third degree.

ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER 3 Linear Differential Equation If a differential equation is of first degree in the dependent variable its derivatives (therefe there cannot be an term involving the product of the dependent variable its derivatives), then it is called a linear differential equation, otherwise it is known as non-linear differential equation. A general fm of a linear ODE of der n in the unknown function ( the dependent variable ) the independent variable is n n1 0 1... n n1 n1 d d d a a a an ƒ()...(11) d d d where a 0, a 1,..., a n are either given functions of constants. The ODEs that cannot be written in the fm (11) are known as non-linear ODEs. F eample, d d d 3 is a linear ODE while d d d, d d 0 are non-linear ODEs. Fmation of Differential Equations A differential equation is a concise fm of epressing the general properties of a famil of functions. It is well known that man general laws in Phsics, Chemistr, Biolog, Economics, Space Science in an branch of human knowledge can be epressed in concise fms b differential equations. We illustrate through few eamples Eample 1: Find the curves in the -plane the tangent at ever point of which is perpendicular to the line joining the igin to that point. Solution: Let P (, ) be a point on the required curve. The slope of the line perpendicular to the line joining the igin O to P is. B the given condition we have d d. This clearl gives the general propert of a famil of concentric circles with centre at the igin. Eample : A particle moves in a straight line such that its acceleration at a point is proptional to the square of its displacement measured from a fied point on the line. Describe the motion mathematicall. Solution: Let denotes the displacement of the particle measured from a fied point on the line of motion. Then b the given condition, we get the equation of motion of the particle as: d. Eample 3: Predat-pre population model: This model involves two species, one pre, sa, rabbits one predat, sa, foes we assume the following: 1. Rabbits have unlimited food suppl hence if there were no foes, the number of rabbits d1 1 (t) will grow eponentiall, i.e., a 1, a > 0.

4 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. II]. In the presence of foes, 1 is decreasing because of the kill b foes at a rate proptional to d1 1 (sa), where (t) denotes the number of foes at time t. Therefe, a 1 b 1, a, b > 0. d 3. In the absence of rabbits, foes ( ) will be decreasing eponentiall to zero, i.e., d, d > 0. However is increasing at a rate proptional to 1, i.e., proptional to the d number of encounters between foes rabbits; together we get c 1 d, c, d > 0. This gives the following predat-pre sstem: d 1 a b 1 1 d c1 d. Eample 4: Elimination of arbitrar constants: Fm the differential equation b eliminating arbitrar constants from the equation c1cos csin. Solution: On differentiation d d c1sin ccos. Again differentiating d c 1 cos c sin ( c 1 cos c sin ) d d d, d d 0. This is a differential equation of second der first degree. Solutions of a Differential Equation An equation containing dependent independent variables free from derivatives which satisfies the underling differential equation identicall is called a solution ( an integral) of the differential equation. d Thus sin, cos are solutions of the differential equation. d Note that c 1 sin c cos is also a solution f ever values of the constants c 1, c. This solution is known as the complete the general solution sin, cos are called particular solutions which are obtained b imposing some conditions. The solution of a differential equation which contains the same number of independent arbitrar constants as the der of the differential equation is known its general solution ( complete integral complete primitive). An solution obtained from the general solution b giving particular values to the arbitrar constants is said to be a particular solution.

ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER 5 It can be shown that the general solution does not alwas include all possible solutions of a differential equation, i.e., there ma eist solutions which are not deducible from the general solution b giving particular values to the arbitrar constants. Such solutions are known as singular solutions. F eample, consider the differential equation p a p b...(1) d where a, b are fied constants p. d Here (1) is satisfied b c a c b...() where c is an arbitrar constant. This is the general solution of (1). But observe that the equation 1...(3) a b satisfies (1) it is not deducible from (). Therefe (3) is a singular solution of (1). 1.3 ODEs OF FIRST ORDER AND FIRST DEGREE An dinar differential equation (ODE) of first der first degree can be written as d ƒ(, )...(1) d It can also be written as M d + N d 0...() where M M(, ) N N(, ) are functions of constants. Unftunatel it is not possible to solve all kinds of equations of tpe (1) (). However, b using elementar processes, the general solutions of the following classes of equations of the above tpe can be found in terms of known functions. These classes are as follows: 1. Equations solvable b separation of variables. Homogeneous equations 3. Eact equations 4. Linear equations of first der Among these four classes the first two are familiar to the students. Here we discuss the last two classes onl. 1.4 EXACT EQUATIONS A differential equation of the fm () is said to be eact, if there eists a function j(, ) such that M d + N d dj...(3) Then () becomes dj 0, which on integration gives j(, ) c, where c is an arbitrar constant. F eample, d + d 0 is an eact differential equation, since d + d d( + ). Its general solution is + c, where c is an arbitrar constant.

6 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. II] Imptant Differentials 1. d() d + d. d( + ) d + d 3. d d d 4. d d d 5. d log d d d 6. d 1 d tan 1 d d 7. 1 log( ) d 1 ( ) d d d 8. 9. 1 d log 1 d{sec ( )} d d d( ) ( ) 1 d d 1 10. 1 d d d 11. d 1 dd 4 1. e d e d e d 13. d d d We state below an imptant theem without proof. Theem: The necessar sufficient condition that M d + N d to be eact is M N. Wking rules f solving M d + N d 0 when it is eact Step 1. Integrate Md considering as constant. Step. Integrate Nd considering as constant, omit those terms alread obtained in Step 1. Step 3. The sum of these integrals will give a function j(, ) such that M d + N d dj. Step 4. j(, ) c is the general solution of M d + N d 0. Note: The condition of the previous theem gives a test f eactness. This condition is often called the condition of integrabilit of the differential equation (). ILLUSTRATIVE EXAMPLES Eample 1: Solve: ( ) d + ( ) d 0. Solution: Method 1: The given equation is of the fm M d + N d 0, where M N.

ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER 7 M N Hence the given differential equation is eact. Md ( ) 1 3 1 3 (assuming to be constant) Nd 1 3 1 ( ) d 3 (assuming to be constant) 1 3 3 1 Therefe, the required solution is ( ) c, rejecting the repeated term where 3 c is an arbitrar constant. Method : ( ) d + ( ) d 0. Regrouping the terms, we get ( d d) ( d d) 0 3 3 d ( ) d ( ) 3 Integrating, we get 1 3 3 d( ) d( ) 3 0 1 3 3 1 ( ) c 3 (c is an arbitrar constant), which is the required general solution. N 3 Eample : Solve: ( e 4 ) d(e 3 ) d 0. 3 Solution: The given equation is of the fm M d + N d 0, where M e 4 e 3. M Here, e e e (1 ) N e e e (1 ) M N hence the given equation is eact. 3 Md ( e 4 ) d 4 e (assuming constant) Nd ( e 3 ) d e Therefe, the required solution is 3 (assuming constant) 4 3 c (c is an arbitrar constant), rejecting the repeated term. e

8 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. II] Eample 3: Solve: e sin d ( e 1)cos d 0. (W.B.U.T. 006) Solution: The given equation is of the fm M d + N d 0, where M e sin N (e + 1) cos. M N e cos hence the given equation is eact. Here Md e sin d e sin (assuming constant) Nd ( e 1)cosd ( e 1)sin e sin sin (assuming constant) Therefe, the required solution is e sin sin c (c is an arbitrar constant), rejecting the repeated term. 1 Eample 4: Solve: 1 cos d ( log sin ) d 0. 1 Solution: The given equation is of the fm M d + N d 0, where M 1 cos N log sin. M 1 N 1 sin hence the given equation is eact. Here Md 1 1 cosd ( log ) cos (assuming constant) Nd ( logsin ) d ( log ) cos (assuming constant) All the terms in Nd are alread obtained in Md hence the general solution of the given differential equation is ( log ) cos c, where c is an arbitrar constant. d cos sin Eample 5: Solve: 0. d sin cos Solution: The given equation can be written as ( cossin ) d(sincos ) d 0. This is of the fm M d + N d 0, where M cos sin N sin cos. M N cos cos 1 hence the given equation is eact. Here Md ( cossin ) d

ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER 9 sin (sin ) (assuming constant) Nd (sin cos ) d (sin ) sin (assuming constant) sin (sin ) All the terms in Nd are alread obtained in Md hence the general solution of the given differential equation is sin (sin ) c, where c is an arbitrar constant. Eample 6: Solve: ( 3 7) d(3 8) d 0. Solution: Let us write the given equation in the fm: d d 3 7 3 8 d d 5( 3) d d 1 d d d d 5 3 1 d d 3 5(d d) 1 d( 3) d( 1) 5 3 1 Integrating, we get d( 3) d( 1) 5 3 1 log( 3) 5log( 1) logc 3 5 c ( 1) This is the required general solution where c is an arbitrar constant. Eample 7: Show that the equation (b componendo dividendo) 3 3 3 ( 3 ) d( ) d 0 is eact find the solution if 1 when 1. Solution: The given equation is of the fm M d + N d 0, where M 3 3 N 3 3. M 3 4 N hence the given equation is eact.

10 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. II] Here Md 3 ( 3 ) d 4 4 3 (assuming constant) 3 3 Nd ( ) d 3 (assuming constant) 4 Therefe, the general solution is 1 4 3 1 4 c (c is an arbitrar constant), rejecting the repeated terms. 4 4 B question when 1, 1, therefe c 0. Hence the required solution is 1 4 3 1 4 4 3 4 0, 4 4 0. 4 4 4 1.5 INTEGRATING FACTORS Suppose M d + N d 0 is not eact. If there eists a function ƒ(, ) such that ƒ(, ) (M d + N d) dj, f some function j(, ), then ƒ(, ) is said to be an integrating fact (I.F.) of the equation M d + N d 0. 1 F eample, the equation d d 0 is not eact. But after multipling it b, this d d equation becomes 0, d 0, which is eact has the general solution c, c is an arbitrar constant. Thus, I.F. is an multipling fact b which the equation M d + N d 0 becomes eact. Note: It can be proved that if a differential equation has one integrating fact, then it has an infinite number of integrating facts. Rules f determining integrating facts M N We consider the differential equation M d + N d 0, which is not eact, i.e.,. Rule 1. Integrating fact b inspection In art. 1.4 we have seen differentials of some functions which are eact. Observe that some of them are made eact b multipling the differentials b some functions (I.F.s). F eample, d d is not d d eact, but is eact is equal to d. Therefe, 1 is an I.F. f the differential d d. 1 Rule. If M + N ¹ 0 M, N are both homogeneous functions in, of same degree, then M N is an integrating fact of the equation M d + N d 0. 1 Rule 3. If M N ¹ 0, then is an integrating fact of the equation M d + N d 0, where M N M, N can be written in the fm M ƒ(), N g().