DIFFERENTIAL EQUATIONS

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1 DIFFERENTIAL EQUATIONS 379 Chapter 9 DIFFERENTIAL EQUATIONS He who seeks f methods without having a definite problem in mind seeks f the most part in vain. D. HILBERT 9. Introduction In Class XI and in Chapter 5 of the present book, we discussed how to differentiate a given function f with respect to an independent variable, i.e., how to find f () f a given function f at each in its domain of definition. Further, in the chapter on Integral Calculus, we discussed how to find a function f whose derivative is the function g, which ma also be fmulated as follows: F a given function g, find a function f such that = g(), where = f ()... () An equation of the fm () is known as a differential equation. A fmal definition will be given later. Henri Poincare (854-9 ) These equations arise in a variet of applications, ma it be in Phsics, Chemistr, Biolog, Anthropolog, Geolog, Economics etc. Hence, an indepth stu of differential equations has assumed prime imptance in all modern scientific investigations. In this chapter, we will stu some basic concepts related to differential equation, general and particular solutions of a differential equation, fmation of differential equations, some methods to solve a first der - first degree differential equation and some applications of differential equations in different areas. 9. Basic Concepts We are alrea familiar with the equations of the tpe: = 0... () sin + cos = 0... () + = 7... (3)

2 380 MATHEMATICS Let us consider the equation: + = 0... (4) We see that equations (), () and (3) involve independent and/ dependent variable (variables) onl but equation (4) involves variables as well as derivative of the dependent variable with respect to the independent variable. Such an equation is called a differential equation. In general, an equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation. A differential equation involving derivatives of the dependent variable with respect to onl one independent variable is called an dinar differential equation, e.g., 3 d + = 0 is an dinar differential equation... (5) Of course, there are differential equations involving derivatives with respect to me than one independent variables, called partial differential equations but at this stage we shall confine ourselves to the stu of dinar differential equations onl. Now onward, we will use the term differential equation f dinar differential equation. Note. We shall prefer to use the following notations f derivatives: 3 d d =, =, = 3. F derivatives of higher der, it will be inconvenient to use so man dashes as supersuffi therefe, we use the notation n f nth der derivative n d n Order of a differential equation Order of a differential equation is defined as the der of the highest der derivative of the dependent variable with respect to the independent variable involved in the given differential equation. Consider the following differential equations: = e... (6)

3 DIFFERENTIAL EQUATIONS 38 d + = 0... (7) 3 3 d d 3 + = 0... (8) The equations (6), (7) and (8) involve the highest derivative of first, second and third der respectivel. Therefe, the der of these equations are, and 3 respectivel. 9.. Degree of a differential equation To stu the degree of a differential equation, the ke point is that the differential equation must be a polnomial equation in derivatives, i.e.,,, etc. Consider the following differential equations: 3 d d sin = 0... (9) = 0... (0) + sin = 0... () We observe that equation (9) is a polnomial equation in, and, equation (0) is a polnomial equation in (not a polnomial in though). Degree of such differential equations can be defined. But equation () is not a polnomial equation in and degree of such a differential equation can not be defined. B the degree of a differential equation, when it is a polnomial equation in derivatives, we mean the highest power (positive integral inde) of the highest der derivative involved in the given differential equation. In view of the above definition, one ma observe that differential equations (6), (7), (8) and (9) each are of degree one, equation (0) is of degree two while the degree of differential equation () is not defined. Note Order and degree (if defined) of a differential equation are alwas positive integers.

4 38 MATHEMATICS Eample Find the der and degree, if defined, of each of the following differential equations: (i) cos 0 = (ii) (iii) + + e = 0 + = 0 d Solution (i) (ii) The highest der derivative present in the differential equation is, so its der is one. It is a polnomial equation in and the highest power raised to is one, so its degree is one. d The highest der derivative present in the given differential equation is, so its der is two. It is a polnomial equation in d and and the highest (iii) d power raised to is one, so its degree is one. The highest der derivative present in the differential equation is, so its der is three. The given differential equation is not a polnomial equation in its derivatives and so its degree is not defined. EXERCISE 9. Determine der and degree (if defined) of differential equations given in Eercises to d + sin( ) = = ds d s + 3s = 0 dt dt 4. d cos 0 + = 5. d cos3 sin 3 = + 6. ( ) + ( ) 3 + ( ) = = 0

5 DIFFERENTIAL EQUATIONS = e 9. + ( ) + = sin = 0. The degree of the differential equation 3 d sin = is (A) 3 (B) (C) (D) not defined. The der of the differential equation d + = is 3 0 (A) (B) (C) 0 (D) not defined 9.3. General and Particular Solutions of a Differential Equation In earlier Classes, we have solved the equations of the tpe: + = 0... () sin cos = 0... () Solution of equations () and () are numbers, real comple, that will satisf the given equation i.e., when that number is substituted f the unknown in the given equation, L.H.S. becomes equal to the R.H.S.. 0 d Now consider the differential equation + =... (3) In contrast to the first two equations, the solution of this differential equation is a function φ that will satisf it i.e., when the function φ is substituted f the unknown (dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S.. The curve = φ () is called the solution curve (integral curve) of the given differential equation. Consider the function given b = φ () = a sin ( + b),... (4) where a, b R. When this function and its derivative are substituted in equation (3), L.H.S. = R.H.S.. So it is a solution of the differential equation (3). π Let a and b be given some particular values sa a = and b =, then we get a 4 π function = φ () = sin +... (5) 4 When this function and its derivative are substituted in equation (3) again L.H.S. = R.H.S.. Therefe φ is also a solution of equation (3).

6 384 MATHEMATICS Function φ consists of two arbitrar constants (parameters) a, b and it is called general solution of the given differential equation. Whereas function φ contains no arbitrar constants but onl the particular values of the parameters a and b and hence is called a particular solution of the given differential equation. The solution which contains arbitrar constants is called the general solution (primitive) of the differential equation. The solution free from arbitrar constants i.e., the solution obtained from the general solution b giving particular values to the arbitrar constants is called a particular solution of the differential equation. Eample Verif that the function = e 3 is a solution of the differential equation d = Solution Given function is = e 3. Differentiating both sides of equation with respect to, we get Now, differentiating () with respect to, we have d =9e 3, 3 = 3e... () d Substituting the values of and in the given differential equation, we get L.H.S. = 9 e 3 + ( 3e 3 ) 6.e 3 = 9 e 3 9 e 3 = 0 = R.H.S.. Therefe, the given function is a solution of the given differential equation. Eample 3 Verif that the function = a cos + b sin, where, a, b R is a solution of the differential equation 0 d + = Solution The given function is = a cos + b sin... () Differentiating both sides of equation () with respect to, successivel, we get = a sin + b cos d = a cos b sin

7 DIFFERENTIAL EQUATIONS 385 d Substituting the values of and in the given differential equation, we get L.H.S. = ( a cos b sin ) + (a cos + b sin ) = 0 = R.H.S.. Therefe, the given function is a solution of the given differential equation. EXERCISE 9. In each of the Eercises to 0 verif that the given functions (eplicit implicit) is a solution of the cresponding differential equation:. = e + : = 0. = + + C : = 0 3. = cos + C : + sin = 0 4. = + : = + 5. = A : = ( 0) 6. = sin : = + ( 0 and > < ) 7. = log + C : = ( ) 8. cos = : ( sin + cos + ) = 9. + = tan : + + = 0 0. = a ( a, a) : + = 0 ( 0). The number of arbitrar constants in the general solution of a differential equation of fourth der are: (A) 0 (B) (C) 3 (D) 4. The number of arbitrar constants in the particular solution of a differential equation of third der are: (A) 3 (B) (C) (D) Fmation of a Differential Equation whose General Solution is given We know that the equation = 0... () represents a circle having centre at (, ) and radius unit.

8 386 MATHEMATICS Differentiating equation () with respect to, we get = + ( )... () which is a differential equation. You will find later on [See (eample 9 section 9.5..)] that this equation represents the famil of circles and one member of the famil is the circle given in equation (). Let us consider the equation + = r... (3) B giving different values to r, we get different members of the famil e.g. + =, + = 4, + = 9 etc. (see Fig 9.). Thus, equation (3) represents a famil of concentric circles centered at the igin and having different radii. We are interested in finding a differential equation that is satisfied b each member of the famil. The differential equation must be free from r because r is different f different members of the famil. This equation is obtained b differentiating equation (3) with respect to, i.e., + = 0 + = 0... (4) which represents the famil of concentric circles given b equation (3). Again, let us consider the equation = m + c... (5) B giving different values to the parameters m and c, we get different members of the famil, e.g., = (m =, c = 0) = 3 (m = 3, c = 0) Fig 9. = + (m =, c = ) = (m =, c = 0) = (m =, c = ) etc. ( see Fig 9.). Thus, equation (5) represents the famil of straight lines, where m, c are parameters. We are now interested in finding a differential equation that is satisfied b each member of the famil. Further, the equation must be free from m and c because m and

9 DIFFERENTIAL EQUATIONS 387 c are different f different members of the famil. This is obtained b differentiating equation (5) with respect to, successivel we get = Y =+ = m = d, and 0 =... (6) X The equation (6) represents the famil of straight lines given b equation (5). Note that equations (3) and (5) are the general solutions of equations (4) and (6) respectivel. = = Procedure to fm a differential equation that will represent a given famil of curves (a) If the given famil F of curves depends on onl one parameter then it is represented b an equation of the fm F (,, a) = 0... () F eample, the famil of parabolas = a can be represented b an equation of the fm f (,, a) : = a. Differentiating equation () with respect to, we get an equation involving,,, and a, i.e., g (,,, a) = 0... () The required differential equation is then obtained b eliminating a from equations () and () as F(,, ) = 0... (3) (b) If the given famil F of curves depends on the parameters a, b (sa) then it is represented b an equation of the from F (,, a, b) = 0... (4) Differentiating equation (4) with respect to, we get an equation involving,,, a, b, i.e., g (,,, a, b) = 0... (5) But it is not possible to eliminate two parameters a and b from the two equations and so, we need a third equation. This equation is obtained b differentiating equation (5), with respect to, to obtain a relation of the fm h (,,,, a, b) = 0... (6) Y O Fig 9. X

10 388 MATHEMATICS The required differential equation is then obtained b eliminating a and b from equations (4), (5) and (6) as F (,,, ) = 0... (7) Note The der of a differential equation representing a famil of curves is same as the number of arbitrar constants present in the equation cresponding to the famil of curves. Eample 4 Fm the differential equation representing the famil of curves = m, where, m is arbitrar constant. Solution We have = m... () Differentiating both sides of equation () with respect to, we get = m Substituting the value of m in equation () we get = =0 which is free from the parameter m and hence this is the required differential equation. Eample 5 Fm the differential equation representing the famil of curves = a sin ( + b), where a, b are arbitrar constants. Solution We have = a sin ( + b)... () Differentiating both sides of equation () with respect to, successivel we get = a cos ( + b)... () d = a sin ( + b)... (3) Eliminating a and b from equations (), () and (3), we get d + = 0... (4) which is free from the arbitrar constants a and b and hence this the required differential equation.

11 DIFFERENTIAL EQUATIONS 389 Eample 6 Fm the differential equation representing the famil of ellipses having foci on -ais and centre at the igin. Solution We know that the equation of said famil of ellipses (see Fig 9.3) is a + =... () b Fig 9.3 Differentiating equation () with respect to, we get 0 a + b = = b a Differentiating both sides of equation () with respect to, we get... () d + =0 + d = 0... (3) which is the required differential equation. Eample 7 Fm the differential equation of the famil of circles touching the -ais at igin. Solution Let C denote the famil of circles touching -ais at igin. Let (0, a) be the codinates of the centre of an member of the famil (see Fig 9.4). Therefe, equation of famil C is + ( a) = a + = a... () where, a is an arbitrar constant. Differentiating both sides of equation () with respect to,we get + = a X Y O Y Fig 9.4 X

12 390 MATHEMATICS + = a a = + Substituting the value of a from equation () in equation (), we get... () + + = ( + ) = + = This is the required differential equation of the given famil of circles. Eample 8 Fm the differential equation representing the famil of parabolas having verte at igin and ais along positive direction of -ais. Solution Let P denote the famil of above said parabolas (see Fig 9.5) and let (a, 0) be the focus of a member of the given famil, where a is an arbitrar constant. Therefe, equation of famil P is =4a... () Differentiating both sides of equation () with respect to, we get =4a... () Substituting the value of 4a from equation () in equation (), we get = ( ) =0 which is the differential equation of the given famil of parabolas. Fig 9.5

13 DIFFERENTIAL EQUATIONS 39 EXERCISE 9.3 In each of the Eercises to 5, fm a differential equation representing the given famil of curves b eliminating arbitrar constants a and b.. + =. a b = a (b ) 3. = a e 3 + b e 4. = e (a + b) 5. = e (a cos + b sin ) 6. Fm the differential equation of the famil of circles touching the -ais at igin. 7. Fm the differential equation of the famil of parabolas having verte at igin and ais along positive -ais. 8. Fm the differential equation of the famil of ellipses having foci on -ais and centre at igin. 9. Fm the differential equation of the famil of hperbolas having foci on -ais and centre at igin. 0. Fm the differential equation of the famil of circles having centre on -ais and radius 3 units.. Which of the following differential equations has = c e + c e as the general solution? d (A) 0 + = d (B) 0 = d (C) 0 + = d (D) 0 =. Which of the following differential equations has = as one of its particular solution? d (A) + = d (B) + + = (C) 0 d + = (D) 0 d + + = 9.5. Methods of Solving First Order, First Degree Differential Equations In this section we shall discuss three methods of solving first der first degree differential equations Differential equations with variables separable A first der-first degree differential equation is of the fm =F(, )... ()

14 39 MATHEMATICS If F (, ) can be epressed as a product g () h(), where, g() is a function of and h() is a function of, then the differential equation () is said to be of variable separable tpe. The differential equation () then has the fm = h (). g()... () If h () 0, separating the variables, () can be rewritten as = g()... (3) h ( ) Integrating both sides of (3), we get h ( ) = g( )... (4) Thus, (4) provides the solutions of given differential equation in the fm H() = G() + C Here, H () and G () are the anti derivatives of C is the arbitrar constant. h ( ) Eample 9 Find the general solution of the differential equation Solution We have and g () respectivel and + =, ( ) = +... () Separating the variables in equation (), we get ( ) =( + )... () Integrating both sides of equation (), we get ( ) = ( + ) = C = C = 0, where C = C which is the general solution of equation (). C

15 DIFFERENTIAL EQUATIONS 393 Eample 0 Find the general solution of the differential equation + =. + Solution Since + 0, therefe separating the variables, the given differential equation can be written as = + + Integrating both sides of equation (), we get + + = tan = tan + C which is the general solution of equation (). Eample Find the particular solution of the differential equation that =, when = () = 4 given Solution If 0, the given differential equation can be written as = 4... () Integrating both sides of equation (), we get = 4 = + C = C Substituting = and = 0 in equation (), we get, C =.... () Now substituting the value of C in equation (), we get the particular solution of the given differential equation as = +. Eample Find the equation of the curve passing through the point (, ) whose differential equation is = ( + ) ( 0).

16 394 MATHEMATICS Solution The given differential equation can be epressed as * = + * = + Integrating both sides of equation (), we get... () = + = + log + C... () Equation () represents the famil of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the famil which passes through the point (, ). Therefe substituting =, = in equation (), we get C = 0. Now substituting the value of C in equation () we get the equation of the required curve as = + log. Eample 3 Find the equation of a curve passing through the point (, 3), given that the slope of the tangent to the curve at an point (, ) is. Solution We know that the slope of the tangent to a curve is given b. so, =... () Separating the variables, equation () can be written as =... () Integrating both sides of equation (), we get = 3 3 = + C... (3) * The notation due to Leibnitz is etremel fleible and useful in man calculation and fmal transfmations, where, we can deal with smbols and eactl as if the were dinar numbers. B treating and like separate entities, we can give neater epressions to man calculations. Refer: Introduction to Calculus and Analsis, volume-i page 7, B Richard Courant, Fritz John Spinger Verlog New Yk.

17 DIFFERENTIAL EQUATIONS 395 Substituting =, = 3 in equation (3), we get C = 5. Substituting the value of C in equation (3), we get the equation of the required curve as 3 = + 5 = (3 + 5) 3 3 Eample 4 In a bank, principal increases continuousl at the rate of 5% per ear. In how man ears Rs 000 double itself? Solution Let P be the principal at an time t. Accding to the given problem, dp dt = 5 P 00 dp = P dt 0 separating the variables in equation (), we get dp dt = P 0 Integrating both sides of equation (), we get t log P = + C 0 0 P = e e t C... ()... () 0 C P = C e (where e = C )... (3) Now P = 000, when t = 0 Substituting the values of P and t in (3), we get C = 000. Therefe, equation (3), gives P = 000 Let t ears be the time required to double the principal. Then t t e 0 t 000 = 000 e 0 t = 0 log e EXERCISE 9.4 F each of the differential equations in Eercises to 0, find the general solution:. cos =. + cos 4 ( ) = < <

18 396 MATHEMATICS 3. ( ) + = 4. sec tan + sec tan = 0 5. (e + e ) (e e ) = log = 0 8. ( ) ( ) = + + = = sin 0. e tan + ( e ) sec = 0 F each of the differential equations in Eercises to 4, find a particular solution satisfing the given condition:.. 3 ( ) = + ; = when = 0 ( ) = ; = 0 when = 3. cos = a (a R); = when = 0 4. tan = ; = when = 0 5. Find the equation of a curve passing through the point (0, 0) and whose differential equation is = e sin. 6. F the differential equation = ( + )( + ), find the solution curve passing through the point (, ). 7. Find the equation of a curve passing through the point (0, ) given that at an point (, ) on the curve, the product of the slope of its tangent and codinate of the point is equal to the codinate of the point. 8. At an point (, ) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point ( 4, 3). Find the equation of the curve given that it passes through (, ). 9. The volume of spherical balloon being inflated changes at a constant rate. If initiall its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

19 DIFFERENTIAL EQUATIONS In a bank, principal increases continuousl at the rate of r% per ear. Find the value of r if Rs 00 double itself in 0 ears (log e = 0.693).. In a bank, principal increases continuousl at the rate of 5% per ear. An amount of Rs 000 is deposited with this bank, how much will it wth after 0 ears (e 0.5 =.648).. In a culture, the bacteria count is,00,000. The number is increased b 0% in hours. In how man hours will the count reach,00,000, if the rate of growth of bacteria is proptional to the number present? 3. The general solution of the differential equation (A) e + e = C (C) e + e = C 9.5. Homogeneous differential equations Consider the following functions in and F (, ) = +, F (, ) = 3, (B) e + e = C + = e is (D) e + e = C F 3 (, ) = cos, F (, ) = sin + cos 4 If we replace and b λ and λ respectivel in the above functions, f an nonzero constant λ, we get F (λ, λ) = λ ( + ) = λ F (, ) F (λ, λ) = λ ( 3) = λ F (, ) λ F 3 (λ, λ) = cos = cos λ = λ0 F 3 (, ) F 4 (λ, λ) = sin λ + cos λ λ n F 4 (, ), f an n N Here, we observe that the functions F, F, F 3 can be written in the fm F(λ, λ) = λ n F (, ) but F can not be written in this fm. This leads to the following 4 definition: A function F(, ) is said to be homogeneous function of degree n if F(λ, λ) = λ n F(, ) f an nonzero constant λ. We note that in the above eamples, F, F, F 3 are homogeneous functions of degree,, 0 respectivel but F 4 is not a homogeneous function.

20 398 MATHEMATICS We also observe that F (, ) = + h = F (, ) = + = h F (, ) = 3 = h3 F (, ) = 3 = h4 F 3 (, ) = 0 cos 0 = h5 F 4 (, ) n h6, f an n N F 4 (, ) n h7, f an n N Therefe, a function F (, ) is a homogeneous function of degree n if F(, ) = n n g h A differential equation of the fm = F (, ) is said to be homogenous if F(, ) is a homogenous function of degree zero. To solve a homogeneous differential equation of the tpe F (, ) = = g... () We make the substitution = v.... () Differentiating equation () with respect to, we get Substituting the value of = dv v+... (3) from equation (3) in equation (), we get

21 DIFFERENTIAL EQUATIONS 399 dv v+ = g (v) dv = g (v) v... (4) Separating the variables in equation (4), we get dv =... (5) g () v v Integrating both sides of equation (5), we get dv = C g () v v +... (6) Equation (6) gives general solution (primitive) of the differential equation () when we replace v b. Note If the homogeneous differential equation is in the fm F(, ) = where, F (, ) is homogenous function of degree zero, then we make substitution = v i.e., = v and we proceed further to find the general solution as discussed above b writing = F(, ) = h. Eample 5 Show that the differential equation ( ) = + is homogeneous and solve it. Solution The given differential equation can be epressed as = +... () Let F(, ) = Now F(λ, λ) = + λ ( + ) 0 =λ f (, ) λ( )

22 400 MATHEMATICS Therefe, F(, ) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation. Alternativel, + = = g... () R.H.S. of differential equation () is of the fm g and so it is a homogeneous function of degree zero. Therefe, equation () is a homogeneous differential equation. To solve it we make the substitution = v... (3) Differentiating equation (3) with respect to, we get Substituting the value of and = dv v +... (4) in equation () we get dv v+ = + v v dv = + v v v dv = v + v+ v v dv = v + v+ Integrating both sides of equation (5), we get v v + v+ dv = v + 3 dv = log + C v + v+

23 DIFFERENTIAL EQUATIONS 40 v + 3 log C dv = dv v v v + v+ 3 log v + v+ log C dv= + 3 v v + log v + v+. tan = log + C 3 3 v + log v + v+ + log = 3 tan + C (Wh?) 3 Replacing v b, we get + 3 log log = 3 tan + C + log + + = 3 tan + C 3 + log ( + + ) = 3 tan + C 3 log ( + + ) = 3 tan + C 3 which is the general solution of the differential equation () + Eample 6 Show that the differential equation cos = cos + homogeneous and solve it. Solution The given differential equation can be written as = cos + cos is... ()

24 40 MATHEMATICS It is a differential equation of the fm F(, ) =. Here F(, ) = Replacing b λ and b λ, we get cos + cos λ [ cos + ] 0 F(λ, λ) = =λ [F(, )] λ cos Thus, F(, ) is a homogeneous function of degree zero. Therefe, the given differential equation is a homogeneous differential equation. To solve it we make the substitution = v... () Differentiating equation () with respect to, we get Substituting the value of and = dv v+... (3) in equation (), we get dv v v+ = cos v+ cosv dv v = cos v+ v cos v dv = cosv Therefe cos v dv = cosvdv =

25 sin v = log + log C sin v = log C Replacing v b, we get sin = log C which is the general solution of the differential equation (). DIFFERENTIAL EQUATIONS 403 Eample 7 Show that the differential equation e + e = 0is homogeneous and find its particular solution, given that, = 0 when =. Solution The given differential equation can be written as = e e... () Let F(, ) = e e λ e 0 Then F(λ, λ) = [F(, )] =λ λ e Thus, F(, ) is a homogeneous function of degree zero. Therefe, the given differential equation is a homogeneous differential equation. To solve it, we make the substitution = v... () Differentiating equation () with respect to, we get = v+ dv

26 404 MATHEMATICS Substituting the value of and in equation (), we get dv v+ = ve v e v dv = ve v v e dv = v e v e v dv = e v dv = e v = log + C and replacing v b, we get e + log = C... (3) Substituting = 0 and = in equation (3), we get e 0 + log = C C = Substituting the value of C in equation (3), we get e + log = which is the particular solution of the given differential equation. Eample 8 Show that the famil of curves f which the slope of the tangent at an point (, ) on it is +, is given b = c. Solution We know that the slope of the tangent at an point on a curve is. Therefe, = +

27 DIFFERENTIAL EQUATIONS 405 = +... () Clearl, () is a homogenous differential equation. To solve it we make substitution = v Differentiating = v with respect to, we get = dv v+ dv v+ = + v v dv = v v v v dv v dv v = = v Therefe dv = v log v = log + log C log (v ) () = log C (v ) =± C Replacing v b, we get =± C ( ) = ± C = C

28 406 MATHEMATICS EXERCISE 9.5 In each of the Eercises to 0, show that the given differential equation is homogeneous and solve each of them.. ( + ) = ( + ). + = 3. ( ) ( + ) = 0 4. ( ) + = 0 5. = + 6. = + 7. cos sin + = sin cos 8. + sin = log = e + e = 0 F each of the differential equations in Eercises from to 5, find the particular solution satisfing the given condition:. ( + ) + ( ) = 0; = when =. + ( + ) = 0; = when = 3. sin 0; π + = = 4 when = 4. + cosec = 0 ; = 0 when = 5. + = 0; = when = 6. A homogeneous differential equation of the from h = can be solved b making the substitution. (A) = v (B) v = (C) = v (D) = v

29 DIFFERENTIAL EQUATIONS Which of the following is a homogeneous differential equation? (A) ( ) ( ) = 0 (B) () ( ) = 0 (C) ( 3 + ) + = 0 (D) + ( ) = Linear differential equations A differential equation of the from P + =Q where, P and Q are constants functions of onl, is known as a first der linear differential equation. Some eamples of the first der linear differential equation are + = sin + = e + log = Another fm of first der linear differential equation is P + =Q where, P and Q are constants functions of onl. Some eamples of this tpe of differential equation are + = cos + = e To solve the first der linear differential equation of the tpe + P = Q... () Multipl both sides of the equation b a function of sa g () to get g() + P. (g()) =Q.g ()... ()

30 408 MATHEMATICS Choose g () in such a wa that R.H.S. becomes a derivative of. g (). i.e. g () + P. g() = d [. g ()] g () + P. g() = g() P.g () = g () P = g ( ) g ( ) Integrating both sides with respect to, we get + g () P = g ( ) g( ) P = log (g ()) g () = P e On multipling the equation () b g() = P e, the L.H.S. becomes the derivative P of some function of and. This function g() = e is called Integrating Fact (I.F.) of the given differential equation. Substituting the value of g () in equation (), we get P P e + Pe = Q e d P ( e ) P P = Qe Integrating both sides with respect to, we get P = ( Q e ) e P = P P ( Q ) e e + C which is the general solution of the differential equation.

31 DIFFERENTIAL EQUATIONS 409 Steps involved to solve first der linear differential equation: (i) Write the given differential equation in the fm P Q + = where P, Q are constants functions of onl. (ii) P Find the Integrating Fact (I.F) = e. (iii) Write the solution of the given differential equation as (I.F) = ( Q I.F) + C In case, the first der linear differential equation is in the fm P Q + =, P where, P and Q are constants functions of onl. Then I.F = e and the solution of the differential equation is given b. (I.F) = ( Q I.F) + C Eample 9 Find the general solution of the differential equation Solution Given differential equation is of the fm Therefe I.F = P Q + =, where P = and Q = cos e = e Multipling both sides of equation b I.F, we get cos =. e e = e cos ( e ) = e cos On integrating both sides with respect to, we get e = e cos + C... () Let I = e cos e = cos ( sin ) ( e )

32 40 MATHEMATICS = cos e sin e = cos e sin ( e ) cos ( e ) = cos e + sin e cos e I = e cos + sin e I I = (sin cos ) e (sin cos e ) I = Substituting the value of I in equation (), we get e = sin cos e + C = sin cos + Ce which is the general solution of the given differential equation. Eample 0 Find the general solution of the differential equation ( 0) + =. Solution The given differential equation is + =... () Dividing both sides of equation () b, we get + = which is a linear differential equation of the tpe P Q + =, where So I.F = e = e log = e = [ as e = f ( )] Therefe, solution of the given equation is given b. = ( )( ) + C = log 3 + C log f ( ) = + C 4 which is the general solution of the given differential equation. P = and Q =.

33 DIFFERENTIAL EQUATIONS 4 Eample Find the general solution of the differential equation ( + ) = 0. Solution The given differential equation can be written as = This is a linear differential equation of the tpe P Q + =, where P = and log log( ) Q =. Therefe I.F = e = e = e = Hence, the solution of the given differential equation is = ( ) + C = ( ) + C = + C = + C which is a general solution of the given differential equation. Eample Find the particular solution of the differential equation given that = 0 when cot + = + cot ( 0) π =. Solution The given equation is a linear differential equation of the tpe P Q + =, where P = cot and Q = + cot. Therefe I.F = cot log sin = = e e sin Hence, the solution of the differential equation is given b. sin = ( + cot ) sin + C

34 4 MATHEMATICS sin = sin = sin + cos + C sin cos + cos + C sin = sin cos + cos + C sin = sin + C... () Substituting = 0 and π = in equation (), we get 0 = π π sin + C π C = 4 Substituting the value of C in equation (), we get sin = = sin π 4 π (sin 0) 4sin which is the particular solution of the given differential equation. Eample 3 Find the equation of a curve passing through the point (0, ). If the slope of the tangent to the curve at an point (, ) is equal to the sum of the codinate (abscissa) and the product of the codinate and codinate (dinate) of that point. Solution We know that the slope of the tangent to the curve is. Therefe, = + =... () This is a linear differential equation of the tpe P Q + =, where P = and Q =. Therefe, I. F = = e e

35 Hence, the solution of equation is given b = ( ) e DIFFERENTIAL EQUATIONS 43 ( ) e + C... () Let I = ( ) e Let = t, then = dt = dt. Therefe, I = edt= e = e t t Substituting the value of I in equation (), we get e = e +C = + Ce... (3) Now (3) represents the equation of famil of curves. But we are interested in finding a particular member of the famil passing through (0, ). Substituting = 0 and = in equation (3) we get = + C. e 0 C = Substituting the value of C in equation (3), we get = + e which is the equation of the required curve. EXERCISE 9.6 F each of the differential equations given in Eercises to, find the general solution:. sin + =. π 4. + sec = tan 0 < + 3= e log + = ( + ) + = cot ( 0) 5. cos + = tan + = π 0 < log + = log

36 44 MATHEMATICS 9. cot 0 ( 0) + + = 0. ( + ) =. + ( ) = 0. ( + 3 ) = ( > 0). F each of the differential equations given in Eercises 3 to 5, find a particular solution satisfing the given condition: 3. π + tan= sin ; = 0 when = 3 4. ( ) ; 0 when + 5. π 3cot= sin ; = when = 6. Find the equation of a curve passing through the igin given that the slope of the tangent to the curve at an point (, ) is equal to the sum of the codinates of the point. 7. Find the equation of a curve passing through the point (0, ) given that the sum of the codinates of an point on the curve eceeds the magnitude of the slope of the tangent to the curve at that point b The Integrating Fact of the differential equation = is (A) e (B) e (C) (D) 9. The Integrating Fact of the differential equation + = a( < < ) is (A) (B) ( ) (C) Miscellaneous Eamples (D) Eample 4 Verif that the function = c e a cos b + c e a sin b, where c, c are arbitrar constants is a solution of the differential equation d ( a a b ) =

37 DIFFERENTIAL EQUATIONS 45 Solution The given function is = e a [c cosb + c sinb]... () Differentiating both sides of equation () with respect to, we get = a [ sin cos ] [ cos sin ] e bc b+ bc b + c b+ c b e a = a e [( bc + ac)cos b+ ( ac bc)sin b]... () Differentiating both sides of equation () with respect to, we get a d a = + + e [( bc ac ) ( bsin b) ( ac bc ) ( bcos b)] [( bc ac ) cos b ( ac bc ) sin b] e. a a = + + e [( a c abc b c )sin b ( a c abc b c )cos b] a Substituting the values of d, and in the given differential equation, we get a L.H.S. = e [ a c abc b c )sin b + ( a c + abc b c )cos b] a ae [( bc + ac )cos b + ( ac bc )sin b] = a + ( a + b ) e [ c cosb+ c sin b] ( ) a c abc b c a c abc a c b c sinb a e + ( ac+ abc bc abc ac+ ac+ bc)cosb a = e [0 sinb+ 0cos b] = e a 0 = 0 = R.H.S. Hence, the given function is a solution of the given differential equation. Eample 5 Fm the differential equation of the famil of circles in the second quadrant and touching the codinate aes. Solution Let C denote the famil of circles in the second quadrant and touching the codinate aes. Let ( a, a) be the codinate of the centre of an member of this famil (see Fig 9.6).

38 46 MATHEMATICS Equation representing the famil C is ( + a) + ( a) = a... () + + a a + a = 0... () Differentiating equation () with respect to, we get ( aa, ) Y + a a + =0 X O X + = a + a = Substituting the value of a in equation (), we get Y Fig = [ + + ] + [ ] = [ + ] ( + ) + [ + ] = [ + ] ( + ) [( ) + ] = [ + ] which is the differential equation representing the given famil of circles. Eample 6 Find the particular solution of the differential equation log = given that = 0 when = 0. Solution The given differential equation can be written as = e(3 + 4) = e3. e 4... () Separating the variables, we get = e 3 4 e Therefe 4 e = 3 e

39 DIFFERENTIAL EQUATIONS 47 4 e 3 e = + C e e 4 + C = 0... () Substituting = 0 and = 0 in (), we get C = 0 C = Substituting the value of C in equation (), we get 4 e e 4 7 = 0, which is a particular solution of the given differential equation. Eample 7 Solve the differential equation ( ) sin = ( + ) cos. Solution The given differential equation can be written as sin cos = cos + sin = cos + sin sin cos Dividing numerat and denominat on RHS b, we get = cos + sin sin cos... () Clearl, equation () is a homogeneous differential equation of the fm g =. To solve it, we make the substitution = v... () = dv v+

40 48 MATHEMATICS Therefe dv v+ = dv = vsin v cosv dv vcosv vsin v cos v dv = vcosv tanvdv dv = v vcosv+ v sin v vsin v cosv vcosv vsin v cosv = log secv log v = log + log C secv log v = log C (using () and ()) secv v =± C... (3) Replacing v b in equation (3), we get sec ( ) = C where, C = ± C sec =C which is the general solution of the given differential equation. Eample 8 Solve the differential equation (tan ) = ( + ). Solution The given differential equation can be written as + = + tan +... ()

41 DIFFERENTIAL EQUATIONS 49 Now () is a linear differential equation of the fm P + = Q, where, P = + and tan Q =. + Therefe, I.F = + tan e = e Thus, the solution of the given differential equation is e tan = tan tan e + C... () + Let I = tan + e tan Substituting tan = t so that = dt +, we get t I = tedt= t e t. e t dt = t e t e t = e t (t ) I = tan e (tan ) Substituting the value of I in equation (), we get tan tan. e = e (tan ) + C = (tan ) + C e tan which is the general solution of the given differential equation. Miscellaneous Eercise on Chapter 9. F each of the differential equations given below, indicate its der and degree (if defined). (i) d log = (ii) = sin (iii) 4 3 d d sin =

42 40 MATHEMATICS. F each of the eercises given below, verif that the given function (implicit eplicit) is a solution of the cresponding differential equation. (i) = a e + b e + : (ii) = e (a cos + b sin ) : d = d 0 + = (iii) = sin 3 : d 9 6cos3 0 + = (iv) = log : ( + ) = 0 3. Fm the differential equation representing the famil of curves given b ( a) + = a, where a is an arbitrar constant. 4. Prove that = c ( + ) is the general solution of differential equation ( 3 3 ) = ( 3 3 ), where c is a parameter. 5. Fm the differential equation of the famil of circles in the first quadrant which touch the codinate aes. 6. Find the general solution of the differential equation + = Show that the general solution of the differential equation + = 0 is + + given b ( + + ) = A ( ), where A is parameter. π 8. Find the equation of the curve passing through the point 0, whose differential 4 equation is sin cos + cos sin = Find the particular solution of the differential equation ( + e ) + ( + ) e = 0, given that = when = Solve the differential equation e = e + ( 0).. Find a particular solution of the differential equation ( ) ( + ) =, given that =, when = 0. (Hint: put = t)

43 DIFFERENTIAL EQUATIONS 4. Solve the differential equation e = ( 0). 3. Find a particular solution of the differential equation cot + = 4 cosec ( 0), given that = 0 when π =. 4. Find a particular solution of the differential equation ( + ) = e, given that = 0 when = The population of a village increases continuousl at the rate proptional to the number of its inhabitants present at an time. If the population of the village was 0, 000 in 999 and 5000 in the ear 004, what will be the population of the village in 009? 6. The general solution of the differential equation = 0 is (A) = C (B) = C (C) = C (D) = C 7. The general solution of a differential equation of the tpe P Q + = is (A) ( ) P P Q e = e + C (B) ( ) P P. e = Qe + C (C) ( ) P P Q e = e + C (D) ( ) P P Q e = e + C 8. The general solution of the differential equation e + ( e + ) = 0 is (A) e + = C (B) e + = C (C) e + = C (D) e + = C

44 4 MATHEMATICS Summar An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation. Order of a differential equation is the der of the highest der derivative occurring in the differential equation. Degree of a differential equation is defined if it is a polnomial equation in its derivatives. Degree (when defined) of a differential equation is the highest power (positive integer onl) of the highest der derivative in it. A function which satisfies the given differential equation is called its solution. The solution which contains as man arbitrar constants as the der of the differential equation is called a general solution and the solution free from arbitrar constants is called particular solution. To fm a differential equation from a given function we differentiate the function successivel as man times as the number of arbitrar constants in the given function and then eliminate the arbitrar constants. Variable separable method is used to solve such an equation in which variables can be separated completel i.e. terms containing should remain with and terms containing should remain with. A differential equation which can be epressed in the fm = f (, ) = g (, ) where, f (, ) and g(, ) are homogenous functions of degree zero is called a homogeneous differential equation. A differential equation of the fm +P = Q, where P and Q are constants functions of onl is called a first der linear differential equation. Histical Note One of the principal languages of Science is that of differential equations. Interestingl, the date of birth of differential equations is taken to be November,,675, when Gottfried Wilthelm Freiherr Leibnitz (646-76) first put in black and white the identit =, thereb introducing both the smbols and.

45 DIFFERENTIAL EQUATIONS 43 Leibnitz was actuall interested in the problem of finding a curve whose tangents were prescribed. This led him to discover the method of separation of variables 69. A ear later he fmulated the method of solving the homogeneous differential equations of the first der. He went further in a ver sht time to the discover of the method of solving a linear differential equation of the first-der. How surprising is it that all these methods came from a single man and that too within 5 ears of the birth of differential equations! In the old das, what we now call the solution of a differential equation, was used to be referred to as integral of the differential equation, the wd being coined b James Bernoulli ( ) in 690. The wd solution was first used b Joseph Louis Lagrange (736-83) in 774, which was almost hundred ears since the birth of differential equations. It was Jules Henri Poincare (854-9) who strongl advocated the use of the wd solution and thus the wd solution has found its deserved place in modern terminolog. The name of the method of separation of variables is due to John Bernoulli ( ), a ounger brother of James Bernoulli. Application to geometric problems were also considered. It was again John Bernoulli who first brought into light the intricate nature of differential equations. In a letter to Leibnitz, dated Ma 0, 75, he revealed the solutions of the differential equation =, which led to three tpes of curves, viz., parabolas, hperbolas and a class of cubic curves. This shows how varied the solutions of such innocent looking differential equation can be. From the second half of the twentieth centur attention has been drawn to the investigation of this complicated nature of the solutions of differential equations, under the heading qualitative analsis of differential equations. Now-a-das, this has acquired prime imptance being absolutel necessar in almost all investigations.