u dx + y = 0 z x z x = x + y = 0 6) 2

Transcription

1 DIFFERENTIAL EQUATIONS 6 Many physical problems, when formulated in mathematical forms, lead to differential equations. Differential equations enter naturally as models for many phenomena in economics, commerce, engineering etc. Many of these phenomena are complex in nature and very difficult to understand. But when they are described by differential equations, it is easy to analyse them. For example, if the rate of change of cost for x outputs is directly proportional to the cost, then this phenomenon is described by the differential equation, dc k C, where C is the cost and k is constant. The dx solution of this differential equation is C C 0 e kx where C C 0 when x FORMATION OF DIFFERENTIAL EQUATIONS A Differential Equation is one which involves one or more independent variables, a dependent variable and one or more of their differential coefficients. (i) (ii) There are two types of differential equations: Ordinary differential equations involving only one independent variable and derivatives of the dependent variable with respect to the independent variable. Partial differential equations which involve more than one independent variable and partial derivatives of the dependent variable with respect to the independent variables. The following are a few examples for differential equations: ) dy dy 3 dx + y e x d y ) dx dx dy 5 +3y 0 dx

2 3 3) dy d y + k 4) x u dx dx x 5) u u u z + x y + 0 6) z x only. + y u y + 0 z y x + y (), () and (3) are ordinary differential equations and (4), (5) and (6) are partial differential equations. In this chapter we shall study ordinary differential equations 6.. Order and Degree of a Differential Equation The order of the derivative of the highest order present in a differential equation is called the order of the differential equation. For example, consider the differential equation 3 x d y dx 3 d y dx dy +7 4y 0 dx 3 d y d y dy The orders of, and 3 are 3, and respectively. So dx dx dx the highest order is 3. Thus the order of the differential equation is 3. The degree of the derivative of the highest order present in a differential equation is called the degree of the differential equation. Here the differential coefficients should be free from the radicals and fractional exponents. Thus the degree of 3 x d y dx Example 3 d y +3 3 dx dy +7 4y 0 is dx Write down the order and degree of the following differential equations.

3 3 dy dy (i) - 4 dx dx + y 3e x d y dy (ii) + 7 3sin x dx dx d x (iii) + a dy 3 dy d y d y dy x 0 (iv) dx 3 dx dx dx -logx 0 3 (v) dy + dx 4x (vi) dy + d y dx dx d y dy dy (vii) - 0 (viii) dx dx + x dx Note The order and the degree respectively are, (i) ; 3 (ii) ; 3 (iii) ; (iv) 3 ; (v) ; (vi) ; 3 (vii) ; (viii) ; Before ascertaining the order and degree in (v), (vi) & (vii) we made the differential coefficients free from radicals and fractional exponents. 6.. Family of curves Sometimes a family of curves can be represented by a single equation. In such a case the equation contains an arbitrary constant c. By assigning different values for c, we get a family of curves. In this case c is called the parameter or arbitrary constant of the family. Examples (i) (ii) (iii) y mx represents the equation of a family of straight lines through the origin, where m is the parameter. x + y a represents the equation of family of concentric circles having the origin as centre, where a is the parameter. y mx + c represents the equation of a family of straight lines in a plane, where m and c are parameters

4 6..3 Formation of Ordinary Differential Equation Consider the equation y mx + λ () where m is a constant and λ is the parameter. This represents one parameter family of parallel straight lines having same slope m. dy Differentiating () with respect to x, we get, m dx This is the differential equation representing the above family of straight lines. Similarly for the equation y Ae 5x, we form the differential dy equation 5y by eliminating the arbitrary constant A. dx The above functions represent one-parameter families. Each family has a differential equation. To obtain this differential equation differentiate the equation of the family with respect to x, treating the parameter as a constant. If the derived equation is free from parameter then the derived equation is the differential equation of the family. Note (i) (ii) The differential equation of a two parameter family is obtained by differentiating the equation of the family twice and by eliminating the parameters. In general, the order of the differential equation to be formed is equal to the number of arbitrary constants present in the equation of the family of curves. Example Form the differential equation of the family of curves y A cos 5x + B sin 5x where A and B are parameters. Given y A cos 5x + B sin 5x dy 5A sin5x + 5B cos 5x dx 4

5 d y dx d y dx Example 3 5 (A cos 5x) 5 (B sin 5x) 5y + 5y 0. Form the differential equation of the family of curves y ae 3x + be x where a and b are parameters. y ae 3x + be x () dy 3ae dx 3x + be x () d y 9ae 3x + be x (3) dx dy () () y ae 3x (4) dx d y dy dy (3) () 6ae 3x 3 y dx dx dx [using (4)] d y dy 4 + 3y 0 dx dx Example 4 Find the differential equation of a family of curves given by y a cos (mx + b), a and b being arbitrary constants. y a cos (mx + b) () dy dx ma sin (mx + b) d y dx m a cos (mx + b) m y [using ()] d y + m y 0 is the required differential equation. dx 5

6 Example 5 Find the differential equation by eliminating the arbitrary constants a and b from y a tan x + b sec x. y a tan x + b sec x Multiplying both sides by cos x we get, y cos x a sin x + b Differentiating with respect to x we get dy y ( sin x) + cos x a cos x dx dy y tan x + a () dx Differentiating () with respect to x, we get d y dy tan x y sec x 0 dx dx EXERCISE 6. ) Find the order and degree of the following : (i) x 3 d y dy d y d y 3 + y cos x (ii) 3 dy dx dx dx dx dx d y (iii) dy 0 (iv) d y dy + dx dx dx dx (v) dy + dx d y (vii) dx d y (ix) dx 3 3 d y dx dy dx 0 (x) 6 d y dy (vi) + x dx dx d y dy (viii) 3 +5 dx dx 3y e x d y + dx ) Find the differential equation of the following (i) y mx (ii) y cx c + c 3 3 dy dx 3

7 (iii) y mx + m a, where m is arbitrary constant (iv) y mx + c where m and c are arbitrary constants. 3) Form the differential equation of family of rectangular hyperbolas whose asymptotes are the coordinate axes. 4) Find the differential equation of all circles x + y + gx 0 which pass through the origin and whose centres are on the x-axis. 5) Form the differential equation of y 4a (x + a), where a is the parameter. 6) Find the differential equation of the family of curves y ae x + be 3x where a and b are parameters. 7) Form the differential equation for y a cos 3x + b sin 3x where a and b are parameters. 8) Form the diffrential equation of y ae bx where a and b are the arbitrary constants. 9) Find the differential equation for the family of concentric circles x + y a, a is the paramter. 6. FIRST ORDER DIFFERENTIAL EQUATIONS 6.. Solution of a differential equation A solution of a differential equation is an explicit or implicit relation between the variables which satisfies the given differential equation and does not contain any derivatives. If the solution of a differential equation contains as many arbitrary constants of integration as its order, then the solution is said to be the general solution of the differential equation. The solution obtained from the general solution by assigning particular values for the arbitrary constants, is said to be a particular solution of the differential equation. For example, 7

8 Differential equation General solutuion Particular solution dy (i) sec dx x y tan x + c y tan x - 5 (c is arbitrary constant) 3 dy (ii) x + x y x dx 3 d y (iii) dx 8 +x 3 + c y x + x y 0 y Ae 3x + Be -3x y 5e 3x 7e -3x 6.. Variables Separable If it is possible to re-arrange the terms of the first order and first degree differential equation in two groups, each containing only one variable, the variables are said to be separable. When variables are separated, the differentail equation takes the form f(x) dx + g(y) dy 0 in which f(x) is a function of x only and g(y) is a function of y only. Then the general solution is f (x) dx + g (y) dy c dy For example, consider x y 0 dx dy dy x y dx y dx x dy y dx + k x log y log x + k. The value of k varies from to. (c is a constant of integration) (separating the variables) where k is a constant of integration. This general solution can be expressed in a more convenient form by assuming the constant of integration to be log c. This is possible because log c also can take all values between - and as k does. By this assumption, the general solution takes the form log y log x log c log ( x y ) log c

9 y i.e. c y cx x which is an elegant form of the solution of the differential equation. Note (i) (ii) When y is absent, the general form of first order linear dy differential equation reduces to f(x) and therefore the dx solution is y f (x) dx + c dy When x is absent, it reduces to g(y) dx dy and in this case, the solution is g( y) dx + c Example 6 Solve the differential equation xdy + ydx 0 xdy + ydx 0, dividing by xy we get dy dx dy y + x 0. Then y + dx c x log y + log x log c xy c Note (i) xdy + ydx 0 d(xy) 0 xy c, a constant. (ii) x ydx xdy ydx xdy d( y ) y y d ( x x y ) + c y + c Example 7 Solve dy dx e y e y dy e 3x+y dx e 3x e y dy y e 3x dx e x dy e 3 dx + c 3x 3x e + c e + e 3 3 -y c 9

10 Example 8 Solve (x - ay) dx (ax - y )dy Writing the equation as x dx + y dy a(xdy + ydx) x dx + y dy a d(xy) x dx + y dy a d (xy) + c 3 3 x y + a(xy) + c 3 3 Hence the general solution is x 3 + y 3 3axy + c Example 9 y Solve y (+ x ) dy + x + x dy + x y + y y + y dy + dy + + y dx 0 + y dx 0 [dividing by x + x x + x dx 0 t dt + u du c i.e. t + u c or + y + dx c 0 + x c Note : This problem can also be solved by using Example 0 [ f(x)]n f (x) dx n [ f ( x)] n x Solve (sin x + cos x) dy + (cos x - sin x) dx 0 + y ] Put +y t ydy dt put +x u xdx du

11 The given equation can be written as dy + cos x sin x dx 0 sin x + cos x dy + cos x sin x dx c sin x + cos x y + log(sin x + cos x) c Example dy Solve x + cos y 0, given y ð when x dx 4 x dy cos y dx sec y dy x dx + k, where k is a constant of integration. log (sec y + tan y) + log x log c, where k log c or x(sec y + tan y) c. When x, y 4 π, we have (sec 4 π + tan 4 π ) c or c ( + ) + The particular solution is x (sec y + tan y) + Example The marginal cost function for producing x units is MC 3+6x - 3x and the total cost for producing unit is Rs.40. Find the total cost function and the average cost function. Let C(x) be the total cost function where x is the number of units of output. Then dc MC 3 + 6x 3x dx

12 dc dx dx ( 3+6x 3x )dx+ k C 3x + 8x x 3 + k, where k is a constant At x, C(x) 40 (given) 3() + 8() k 40 k 0 Total cost function C(x) 3x + 8x x Average cost function Total cost function x 3 3x + 8x x + 0 x Average cost function 3 + 8x x + 0 x Example 3 What is the general form of the demand equation which has a constant elasticity of -? Let x be the quantity demanded at price p. Then the elasticity is given by p η d dx x dp p Given dx dx dp dx dp x dp x p x + log k p log x log p + log k, where k is a constant. i.e. log x log kp x kp p k x p cx, where c k is a constant Example 4 The relationship between the cost of operating a warehouse and the number of units of items stored in it is given by dc ax + b, where C is the monthly cost of operating dx the warehouse and x is the number of units of items in storage. Find C as a function of x if C C 0 when x 0.

13 Given dc ax + b dx dc (ax + b) dx d C (ax + b) dx + k, (k is a constant) C ax +bx + k, when x 0, C C 0 () C 0 a (0) + b(0) + k k C 0 Hence the cost function is given by C a x + bx + C 0 Example 5 The slope of a curve at any point is the reciprocal of twice the ordinate of the point. The curve also passes through the point (4, 3). Find the equation of the curve. Slope of the curve at any point P(x, y) is the slope of the tangent at P(x, y) dy dx y ydy dx y dy dx + c y x + c Since the curve passes through (4, 3), we have c c 5 Equation of the curve is y x + 5 dy y ) Solve (i) + dx x dy y + (iii) dx x EXERCISE 6. dy + y 0 (ii) dx + x (iv) x + y + y 3 + x dy 0 dx

14 dy ) Solve (i) e x-y + x 3 e y (ii) ( e x ) sec y dy + 3e x tan y dx 0 dx dy 3) Solve (i) xy + ax (ii) x(y + ) dx + y(x + ) dy 0 dx (iii) (x yx dy ) + y + xy 0 dx 4) Solve (i) xdy + ydx + 4 y x dx 0 (ii) ydx xdy+3x y e x3 dx 0 dy y + 4 y + 5 dy y + y + 5) Solve (i) dx (ii) + x x + dx 0 x + x + 6) Find the equation of the curve whose slope at the point (x, y) is 3x +, if the curve passes through the point (, -) 7) The gradient of the curve at any point (x, y) is proportional to its abscissa. Find the equation of the curve if it passes through the points (0, 0) and (, ) 8) Solve : sin - y x dy + dx 0, given that y when x x 9) What is the general form of the demand equation which has an elasticity of - n? 0) What is the general form of the demand equation which has an elasticity of? ) The marginal cost function for producing x units is MC e 3x + 7. Find the total cost function and the average cost function, given that the cost is zero when there is no production Homogeneous differential equations A differential equation in x and y is said to be homogeneous if it can be defined in the form dy f ( x, y) dx g( x, y) where f(x, y) and g(x, y) are homogeneous functions of the same degree in x and y. dy xy dy x + y dy x y, dx x + y dx xy, 3 3 dx x + y 4

15 dy and x y + y dx x are some examples of first order homogeneous differential equations Solving first order homogeneous differential equations dy If we put y vx then v + x dv and the differential dx dx equation reduces to variables sepaerable form. The solution is got y by replacing for v after the integration is over. x Example 6 Solve the differential equation (x + y )dx xydy The given differential equation can be written as dy x + y dx xy () This is a homogeneous differential equation dy Put y vx v + x dv () dx dx Substituting () in () we get, v + x dv x + v x dx x( vx) + v v x dv + v v x dv v dx v dx v Now, separating the variables, v dv dx or v x vv dx + c x f ( x) log ( v ) log x + log c [ f ( x) dx log f(x)] or log ( v ) + log x log c ( v ) x c Replacing v by x y, we get 5

16 y x c x Example 7 or x y cx Solve : (x 3 + y 3 )dx (x y + xy ) dy The given equation can be written as 3 3 dy x + y dx x y + xy () Put y vx dy v + x dv dx dx 3 v + x dv + v dx v + v x dv dx 3 + v v ( v)( + v) v v + v v( v + ) v( v + ) v dv v dx + c x v dv ( v) v dx + c or x dv v dx + c x ( ) + v dv dx + c x v + log ( v) log x + c Replacing v by x y, we get Example 8 y + log (x y) c x dy Now, dx dy Solve x y - dx y Put y vx x + y x + y x dy v + x dv dx dx ()

17 () v + x dv vx x + v x v dx x x dv dv dx + v or + v dv log (v + log x (v + or x (v + y i.e. x x Example 9 + v x dx + c 7 + v x dx + v ) log x + log c + v ) log c + v ) c y + + x c or y + x + y c Solve (x +y) dy + (x - y)dx 0 dy x y The equation is dx x + y () dy Put y vx v + x dv dx dx we get v + x dv x vx or v + x dv v dx x + vx dx + v i.e. x dv v + v dx + v + v dv dx or + v x dv v + dv + v + v or x dv dx dv dx + c x tan - v + log ( + v ) log x + c i.e. tan - y x + x + y log logx + c x ( v + v + v) + v

18 tan - y x + log (x + y ) logx logx + c i.e. tan - y x + log (x + y ) c Example 0 The net profit p and quantity x satisfy the differential 3 3 dp p x equation. dx 3xp Find the relationship between the net profit and demand given that p 0 when x dp p x () dx 3xp is a differential equation in x and p of homogeneous type dp dv Put p vx v + x dx 3 dx 3 dv v dv v () v + x dx x 3v dx 3v v 3 dv + v x dx 3v 3v 3 dv dx 3v + v x + 3 dv dx v k x log ( + v 3 ) log x + log k, where k is a constant log ( + v 3 ) log x k Replacing v by x p, we get x 3 + p 3 kx i.e. + v 3 x k But when x 0, it is given that p 0 (0) 3 + (0) 3 k(0) k 90 x 3 +p 3 90x p 3 x (90 x) is the required relationship. 8

19 Example The rate of increase in the cost C of ordering and holding as the size q of the order increases is given by the differential equation dc C + Cq dq. Find the relationship between C and q q if C when q. dc C + Cq dq q () This is a homogeneous equation in C and q Put C vq dc dv dq v + q dq dv v q + vq () v + q dq q v + v dv q dq v dv dq + v v (v + ) v( v +) q ( v + ) v dq v( v + ) dv q + k, k is a constant dv dv dq v v + q + log k, log v log (v + ) log q + log k log v v + log qk or v kq v + Replacing v by q C we get, C kq(c + q) when C and q C C kq(c + q) k q ( C+ q) is the relation between C and q 9

20 Example The total cost of production y and the level of output x are related to the marginal cost of production by the equation (6x + y ) dx - (x + 4xy) dy 0. What is the relation between total cost and output if y when x? Given (6x + y ) dx (x + 4xy) dy dy 6x + y dx () x + 4xy is a homogeneous equation in x and y. dy Put y vx v +x dv dx dx () v + x dv 6x + y + 4v dx or dv dx x + 4xy 6 v v x 4v dv 6 v v dx + k, where k is a constant x log(6 v v ) log x + log k log kx kx 6 v v x c(6x xy y ) where c y and v k x when x and y, c(6 8) c 4 4x (y + xy 6x ) EXERCISE 6.3 ) Solve the following differential equations (i) (iii) (v) dy y y dx x x dy xy y dx x 3 xy dy y xy dx x xy 0 (ii) (iv) (vi) dy y y dx x x dy x(y x) y dx dy xy dx x y

21 (vii) (x + y) dx x dy (viii) x dy y + x + y dx ) The rate of increase in the cost C of ordering and holding as the size q of the order increases is given by the differential dc C + q eqation dq. Find the relatinship between C and Cq q if C 4 when q. 3) The total cost of production y and the level of output x are related to the marginal cost of production by the dy 4 x y equation. What is the total cost function dx xy if y 4 when x? 6..5 First order linear differential equation A first order differential equation is said to be linear when the dependent variable and its derivatives occur only in first degree and no product of these occur. dy An equation of the form + Py Q, dx where P and Q are functions of x only, is called a first order linear differential equation. For example, dy (i) +3y x 3 ; here P 3, Q x 3 dx dy (ii) + y tan x cos x, P tan x, Q cos x dx dy (iii) x 3y xe x, P 3, Q e dx x x (iv) ( + x dy ) + xy (+x ) 3 x, P dx, Q ( + x + x ) are first order linear differential equations Integrating factor (I.F) A given differential equation may not be integrable as such. But it may become integrable when it is multiplied by a function.

22 Such a function is called the integrating factor (I.F). Hence an integrating factor is one which changes a differential equation into one which is directly integrable. e Pdx Let us show that is the integrating factor dy for + Py Q dx () where P and Q are function of x. dx P Now, d ( ye ) dx dy dx dy dx dy dx e P dx e P dx e P dx + y d ( e P dx ) dx + ye Pdx + ye Pdx When () is multiplied by dx e P, d dx Pdx dy P ( +Py) e Pdx dx dy it becomes ( +Py) e P dx dx Qe Pdx dx P d ( ye ) Q e Pdx dx Integrating this, we have ye P dx Q e Pdx dx + c () Note So e pdx is the integrating factor of the differential equation. (i) e log f(x) f(x) when f(x) > 0 (ii) (iii) dy If Q 0 in + Py Q, then the general solution is dx y (I.F) c, where c is a constant. dx For the differential equation dy + Px Q where P and Q are functions of y alone, the (I.F) is e Pdy and the solution is x (I.F) Q (I.F) dy + c

23 Example 3 Solve the equation ( - x dy ) - xy dx The given equation is ( x dy ) xy dx dy x y dx x x dy This is of the form + Py Q, dx x where P ; Q x x I.F dx e P The general solution is, y (I.F) y y e x dx x Q (I.F)dx + c x x dx x x sin - x + c x + c 3 x dx + c Example 4 dy Solve +ay e x (where a -) dx dy The given equation is of the form + Py Q dx Here P a ; Q e x I.F e P dx The general solution is y (I.F) e ax Q (I.F)dx + c

24 y e ax e x e ax dx + c e y e ax ( a+ ) x e a + + c Example 5 dy Solve cos x + y sin x dx 4 ( a+ ) x The given equation can be reduced to dy + y dx Here P tanx sin cos x dx I.F e tan The general solution is y (I.F) x x cos x ; Q secx or e log secx sec x Q (I. F)) dx + c y sec x sec x dx + c y sec x tan x + c Example 6 dx + c dy + y tanx secx dx A bank pays interest by treating the annual interest as the instantaneous rate of change of the principal. A man invests Rs.50,000 in the bank deposit which accrues interest, 6.5% per year compounded continuously. How much will he get after 0 years? (Given : e ) Let P(t) denotes the amount of money in the account at time t. Then the differential equation governing the growth of money is dp 6. 5 P 0.065P dt 00 dp P log e P 0.065t + c P e 0.065t e c ( 0.065) dt + c

25 P c e 0.065t () At t 0, P () c e 0 or c P e 0.065t At t 0, P e x e 0.65 Example 7 Solve x (.955) Rs.95,775. dy + y cos x sin x dx Here P cos x ; Q sin x P dx I.F e P dx cos x dx sin x e sin x The general solution is y (I.F) Q (I.F) dx + c sin x. e sin x dx + c sin x cos x. e sin x dx + c Let sin x t, then cos x dx dt Example 8 t et dt + c e t (t ) + c e sin x (sin x ) + c A manufacturing company has found that the cost C of operating and maintaining the equipment is related to the length m of intervals between overhauls by the equation m dc + mc and C 4 when m. Find the dm relationship between C and m. 5

26 Given m dc + mc or dc + C dm dm m m This is a first order linear differential equation of the form dy + Py Q, where P ; Q dx m m I.F e P dm dm e m e log m m General solution is C (I.F) Q (I.F) dm + k Cm m m dm + k Cm m + k When C 4 and m, we have k k The relationship between C and m Cm m + (m + 6) Example 9 6 where k is a constant Equipment maintenance and operating costs C are related to the overhaul interval x by the equation x dc - 0xC -0 with C C0 when x x dx 0. Find C as a function of x. x dc 0xC 0 or dc 0 dx dx xc This is a first order linear differential equation. P 0 x and Q 0 x P dx 0 dx 0 log x log 0 x x is 0 x

27 I.F dx e P General solution is C(I.F) C 0 x log x 0 e 0 x Q (I.F) dx + k, where k is a constant. 0 x 0 x C dx + k or 0 x 0 x + k when C C 0 x x 0 C 0 0 x k k C 0 x0 x The solution is C 0 x C 0 x x C 0 C x x0 x0 x x0 EXERCISE 6.4 ) Solve the following differential equations 0 0 x0 (i) (ii) (iii) (iv) (v) (vi) (vii) dy + y cot x cosec x dx dy sin x y cot x dx dy + y cot x sin x dx dy + y cot x 4x cosec x, if y 0 when x π dx dy 3y cot x sin x and if y when x π dx dy x 3y x dx dy xy dx + + x ( given that y 0 when x + x ) 7

28 (viii) dy y tan x e x sec x dx (ix) dy y log x + sin x dx x ) A man plans to invest some amount in a small saving scheme with a guaranteed compound interest compounded continuously at the rate of percent for 5 years. How much should he invest if he wants an amount of Rs.5000 at the end of 5 year period. (e ) 3) Equipment maintenance and operating cost C are related to the overhaul interval x by the equation x dc (b )Cx ba, dx where a, b are constants and C C 0 when x x 0. Find the relationship between C and x. 4) The change in the cost of ordering and holding C as quantity q dc is given by dq a C where a is a constant. q Find C as a function of q if C C 0 when q q SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS The general form of linear and second order differential equation with constant coefficients is d y dy a + b + cy f(x). dx dx We shall consider the cases where (i) f(x) 0 and f(x) Ke λx For example, d y (i) 3 dx d y (ii) dx dy 5 + 6y 0 (or) 3y`` 5y` + 6y 0 dx dy 4 + 3y e 5x (or) (D 4D + 3)y e 5x dx 8

29 d y dy (iii) + y 7 (or) (D + D )y 7 dx dx are second order linear differential equations Auxiliary equations and Complementary functions d y dy For the differential equation, a + b + cy f(x), dx dx am + bm + c 0 is said to be the auxiliary equation. This is a quadratic equation in m. According to the nature of the roots m and m of auxiliary equation we write the complementary function (C.F) as follows. Nature of roots Complementary function (i) Real and unequal (m m ) Ae m x + Be m x (ii) Real and equal (m m m say) (Ax + B) e mx (iii) Complex roots (α + iβ) 9 e αx (Acos βx + Bsin βx) (In all the cases, A and B are arbitrary constants) 6.3. Particular Integral (P.I) Consider (ad + bd + c)y e λx Let f(d) ad + bd + c Case : If f(λ) 0 then λ is not a root of the auxiliary equation f(m) 0. Rule : P.I f (D) e λx e f ( λ) λx. Case : If f(λ) 0, λ satisfies the auxiliary equation f(m) 0. Then we proceed as follows. (i) Let the auxiliary equation have two distinct roots m and m and let λ m. Then f(m) a(m m ) (m m ) a(m λ) (m m ) Rule : P.I a(d-λ)( D m ) e λx a( λ m ) xe λx

30 (ii) Let the auxiliary equation have two equal roots each equal to λ. i.e. m m λ. f(m) a ( m λ) Rule : P.I a(d λ) The General solution e λx x a! The general solution of a second order linear differential equation is y Complementary function (C.F) + Particular integral (P.I) Example 30 d y Solve 3 dx dy y 0 dx The auxiliary equation is 3m 5m + 0 (3m ) (m ) 0 e λx The roots are m and m 3 The complementary function is C.F A e 3 x The general solution is y A e 3 x + Be x + Be x (Real and distinct) Example 3 Solve (6D - 4D + 9) y 0 The auxiliary equation is 6m 4m (4m -3) 0 m 4 3, 4 3 The roots are real and equal 30

31 3 x The C.F is (Ax + B) e 4 3 x The general solution is y (Ax + B) e 4 Example 3 Solve (D - 6D + 5) y 0 The auxiliary equation is m 6m m b± b 4ac a 6 ± ± 8i 3 + 4i The roots are complex and is of the form α + iβ with α 3 and β 4 C.F e αx (A cos βx + B sin βx) e 3x (A cos 4x + B sin 4x) The general solution is y e 3x (A cos 4x + B sin 4x) Example 33 Solve d y dx dy y e 5x dx The auxiliary equation is m - 5m m 3, Complementary function C. F Ae 3x + Be x P. I e D 5D + 6 5x e 5x 6 The general solution is y C.F + P. I y Ae 3x + Be x + e 6 5x 3

32 Example 34 d y Solve dx dy y e -3x dx The auxiliary equation is m + 4m m, Complementary function is C. F (Ax + B)e x P. I e D + 4D + 4 3x e 3x e 3x ( 3) + 4( 3) + 4 The general solution is y C.F + P. I y (Ax + B) e x + e 3x Example 35 d y Solve dx dy - + 4y 5 + 3e -x dx The auxiliary equation is m m m ± 4 6 ± i i 3 C.F e x (A cos 3 x + B sin 3 x) P. I D D + 4 P.I 3 e D D + 4 x 3e ( ) ( ) + 4 -x The general solution is 5 e 0x 5 e 0x e x 7 y C.F + P. I + P.I y e x (A cos 3x + B sin 3x) e -x 4 7

33 Example 36 Solve (4D - 8D+ 3)y 33 x e The auxiliary equation is 4m - 8m m 3, m 3 C.F Ae x P. I 4D 4( + Be x 8D + 3 )(D 3 The general solution is y C.F. + P. I y A 3 e x Example 37 + Be x e x ) x x e 4 e x x e 4(D - 3)(D - ) x x 4 e Solve : (D + 0D + 5)y 5 + e -5x The auxiliary equation is m + 0m (m + 5) 0 m 5, 5 C.F (Ax + B) e 5x P. I 5 e 0x 5 D + 0D P.I e D + 0D + 5 5x (D + 5) e 5x x e! 5x x (e 5x ) The general solution is

34 y C.F + P. I + P.I y (Ax + B) e 5x + + x 0 e 5x Example 38 Suppose that the quantity demanded dp d p Q d 4-4p dt and quantity supplied dt Q s p where p is the price. Find the equilibrium price for market clearance. For market clearance, the required condition is Q d Q s. dp d p 4 4p 4 + dt 6 + 8p dt dp d p 48 p 4 + dt 0 dt d p dp 4 p -48 dt dt The auxiliary equation is m 4m 0 m 6, C.F. Ae 6t + Be -t P. I ( 48) e D 4D 0t The general solution is p C.F. + P. I p Ae 6t + Be -t + 4 EXERCISE ) Solve : (i) d y dy d y y 0 (ii) dx dx dx (iii) d y d y + 4y 0 (iv) dx dx ( 48) 4 dy + 0 dx dy y 0 dx

35 ) Solve : (i) (3D + 7D - 6)y 0 (ii) (4D D + 9)y 0 (iii) (3D D + )y 0 3) Solve : (i) (ii) (iii) (D 3D + ) y e x + 5e x (D 5D + 6) y e x + 3e x (D 4D + 49) y 3 + e 7x (iv) (5D D ) y x 3 e 4) Suppose that Q d 30 5P + dp d P + dt and Q dt s 6 + 3P. Find the equilibrium price for market clearance. Choose the correct answer EXERCISE 6.6 ) The differential equation of straight lines passing through the origin is dy (a) x y dx dy x (b) dx y 35 dy (c) 0 dx ) The degree and order of the differential equation dy (d)x dx y d y dy -6 0 are dx dx (a) and (b) and (c) and (d) and 3) The order and degree of the differential equation 3 dy d y d y dy dx x + log x are dx dx dx (a) and 3 (b) 3 and (c) and 3 (d) 3 and 4) The order and degree of dy d y + dx are dx (a) 3 and (b) and 3 (c) 3 and 3 (d) and 3

36 5) The solution of x dy + y dx 0 is (a) x + y c (b) x + y c (c) xy c (d) y cx 6) The solution of x dx + y dy 0 is (a) x + y x c (b) y c (c) x y c (d) xy c dy 7) The solution of e x y is dx (a) e y e x c (c) y log(e x +c) (b) y log ce x (d) e x+y c dp 8) The solution of ke t (k is a constant) is dt k (a) c - t p (b) p ke t + c e c p (c) t log (d) t log k c p 9) In the differential equation (x - y ) dy xy dx, if we make the subsititution y vx then the equation is transformed into + v (a) 3 dv dx v (b) dv dx v + v x v( + v ) x dv (c) v dx dv (d) dx x + v x 0) When y vx the differential equation x dy y + x + y dx reduces to dv (a) v dx vdv (b) x v dx + x dv (c) v dx vdv (d) dx + x v x dy ) The solution of the equation of the type + Py 0, (P is a dx function of x) is given by (a) ye P dx c (b) y P dx c (c) xe P dx y (d) y cx 36

37 dx ) The solution of the equation of the type dy + Px Q (P and Q are functions of y) is (a) y Q e Pdx dy +c (b) ye P dx Q e Pdx dx +c (c) xe P dy Q e Pdy dy +c (d) xe P dy Q dx e P dx +c dy 3) The integrating factor of x y e x is dx (a) logx (b) e x (c) x (d) x 4) The integrating factor of ( + x dy ) + xy ( + x ) 3 is dx (a) + x (b) log ( + x ) (c) e tan- x (d) log (tan- x) dy y 5) The integrating factor of + x dx x 3 is x (a) log x (b) e (c) 3 log(x ) (d) x 6) The complementary function of the differential equation (D D) y e x is (a) A + B e x (b) (Ax + B)e x (c) A + Be x (d) (A+Bx)e -x 7) The complementary function of the differential equation (D D + )y e x is (a) Ae x + Be x (b) A + Be x (c) (Ax + B)e x (d) A+Be x 8) The particular integral of the differential equation d y dy 5 + 6y e 5x is dx dx 5x 5x (a) e (b) xe (c) 6e 6! 5x 5x (d) e 5 9) The particular integral of the differential equation d y dy 6 + 9y e 3x is dx dx 3x 3x 3x (a) e (b) x e (c) xe (d) 9e!!! 3x d y 0) The solution of y 0 is dx (a) (A + B)e x (b) (Ax + B)e x (c) Ae x + 37 B e x (d) (A+Bx)e x

38 7. INTERPOLATION Interpolation is the art of reading between the lines in a table. It means insertion or filling up intermediate values of a function from a given set of values of the function. The following table represents the population of a town in the decennial census. Year : Population : (in thousands) Then the process of finding the population for the year 94, 93, 939, 947 etc. with the help of the above data is called interpolation. The process of finding the population for the year 955, 960 etc. is known as extrapolation. The following assumptions are to be kept in mind for interpolation : (i) (ii) INTERPOLATION AND FITTING A STRAIGHT LINE 7 The value of functions should be either in increasing order or in decreasing order. The rise or fall in the values should be uniform. In other words that there are no sudden jumps or falls in the value of function during the period under consideration. The following methods are used in interpolation : ) Graphic method, ) Algebraic method 7.. Graphic method of interpolation Let y f(x), then we can plot a graph between different values of x and corresponding values of y. From the graph we can find the value of y for given x. 38

39 Example From the following data, estimate the population for the year 986 graphically. Year : Population : (in thousands) population in thousands y From the graph, it is found that the population for 986 was 4 thousands Example (980, 0) (970, 5) (960, ) year Using graphic method, find the value of y when x 7, from the following data. x : y : (990, 6) (986, 4) (000, 33) x

40 35 y (0, 35) (5, 3) (0, 9) (5, 6) (7, 4.8) (30, 3) x The value of y when x 7 is Algebraic methods of interpolation The mathematical methods of interpolation are many. Of these we are going to study the following methods: (i) (ii) (iii) Finite differences Gregory-Newton s formula Lagrange s formula 7..3 Finite differences Consider the arguments x 0, x, x,... x n and the entries y 0, y, y,..., y n. Here y f(x) is a function used in interpolation. Let us assume that the x-values are in the increasing order and equally spaced with a space-length h. 40

41 Then the values of x may be taken to be x 0, x 0 + h, x 0 + h,... x 0 + nh and the function assumes the values f(x 0 ), f(x 0 +h), f(x 0 + h),..., f(x 0 + nh) Forward difference operator For any value of x, the forward difference operator (delta) is defined by f(x) f(x+h) - f(x). In particular, y 0 f(x 0 ) f(x 0 +h) f(x 0 ) y y 0 f(x), [f(x+h)], [f(x+h)],... are the first order differences of f(x). Consider f(x) [ {f(x)}] [f(x+h) f(x)] [f(x+h)] [f(x)] [f(x+h) f(x+h)] [f(x+h) f(x)] f(x+h) f (x+h) + f(x). f(x), [f(x+h)], [f(x+h)]... are the second order differences of f(x). In a similar manner, the higher order differences 3 f(x), 4 f(x),... n f(x),... are all defined. Backward difference operator For any value of x, the backward difference operator (nabla) is defined by f(x) f(x) f(x h) In particular, y n f(x n ) f(x n ) f(x n h) y n y n f(x), [f(x+h)], [f(x+h)],... are the first order differences of f(x). Consider f(x) [ {f(x)}] [f(x) f(x h)] [f(x)] [f(x h)] f(x) f(x h) + f(x h) 4

42 f(x), [f(x+h)], [f(x+h)]... are the second order differences of f(x). In a similar manner the higher order backward differences 3 f(x), 4 f(x),... n f(x),... are all defined. Shifting operator For any value of x, the shifting operator E is defined by E[f(x)] f(x+h) In particular, E(y 0 ) E[f(x 0 )] f(x 0 +h) y Further, E [f(x)] E[E{f(x)] E[f(x+h)] f(x+h) Similarly E 3 [f(x)] f(x+3h) In general E n [f(x)] f(x+nh) The relation between D and E Results We have f(x) f(x) f(x+h) f(x) E f(x) f(x) (E ) f(x) E i.e. E + ) The differences of constant function are zero. ) If f(x) is a polynomial of the n th degree in x, then the n th difference of f(x) is constant and n+ f(x) 0. Example 3 Find the missing term from the following data. x : 3 4 f(x) : Since three values of f(x) are given, we assume that the polynomial is of degree two. 4

43 Example 4 Hence third order differences are zeros. 3 [f(x 0 )] 0 or 3 (y 0 ) 0 (E ) 3 y 0 0 ( E ) (E 3 3E + 3E ) y 0 0 y 3 3y + 3y y (6) + 3y 00 0 i.e. the missing term is 07 y 07 Estimate the production for 96 and 965 from the following data. Year : Production: (in tons) Since five values of f(x) are given, we assume that polynomial is of degree four. Hence fifth order diferences are zeros. 5 [f(x 0 )] 0 i.e. 5 (y 0 ) 0 (E ) 5 (y 0 ) 0 i.e. (E 5 5E 4 + 0E 3 0E + 5E ) y 0 0 y 5 5y 4 + 0y 3 0y + 5y y y 4 + 0(306) 0(60) + 5y 00 0 y y () Since fifth order differences are zeros, we also have 5 [f(x )] 0 43

44 i.e. 5 (y ) 0 i.e. (E ) 5 y 0 (E 5 5E 4 + 0E 3 0E + 5E )y 0 y 6 5y 5 + 0y 4 0y 3 + 5y y (390) + 0y 4 0(306) + 5(60) y 0 0y 4 y () By solving the equations () and () we get, y 0 and y The productions for 96 and 965 are 0 tons and 350 tons respectively Derivation of Gregory - Newton s forward formula Let the function y f(x) be a polynomial of degree n which assumes (n+) values f(x 0 ), f(x ), f(x )... f(x n ), where x 0, x, x,... x n are in the increasing order and are equally spaced. Let x x 0 x x x 3 x... x n x n- h (a positive quantity) Here f(x 0 ) y 0, f(x ) y,... f(x n ) y n Now f(x) can be written as, f(x) a 0 + a (x x 0 ) + a (x x 0 )(x x ) +... When x x 0, () implies f(x 0 ) a 0 or a 0 y 0 When x x, () f(x ) a 0 + a (x x 0 ) +a n (x x 0 ) (x x )... (x x n- ) () i.e. y y 0 + a h y y a 0 y a h 0 h When x x, () f(x ) a 0 + a (x x 0 ) + a (x x 0 ) (x x ) 44

45 y y y (h) + a (h) (h) h h a y y 0 y 0 y y 0 (y y 0 ) y y + y 0 y 0 a y0! h In the same way we can obtain a n y0 y0 y0 3, a 3! h 4 4,..., a 4! h n n n! h substituting the values of a 0, a,..., a n in () we get y f(x) y y0 (x - x h 0 ) + (x x! h 0 ) (x x ) +... n y0 + n (x x n! h 0 ) (x x )... (x x n- ) () x x0 Denoting by u, we get h x x 0 hu x x (x x 0 ) (x x 0 ) hu h h(u ) x x (x x 0 ) (x x 0 ) hu h h(u ) x x 3 h (u - 3) In general x x n- h{u (n )} Thus () becomes, f(x) y 0 + u u( u ) y0 +!! y u( u )( u )...( u n ) + n y n! 0 x x0 where u. This is the Gregory-Newton s forward h formula. 45

46 Example 5 Find y when x 0. given that x : y : lies in the first interval (x 0, x ) i.e. (0, ). So we can use Gregory-Newton s forward interpolation formula. Since five values are given, the interpolation formula is y y 0 + u u( u ) u( u )( u ) y0 +!! y 0 + 3! 3 y 0 u( u )( u )( u 3) x x + 4! 4 0 y 0 where u h Here h, x 0 0 and x 0. u The forward difference table : + x y D y D y D 3 y D 4 y y (9) +! ( 0.)(0. )(0. ) (-) + 3! i.e. when x 0., y (0. ) (0)! ( 0.)(0. )(0. )(0. 3) (4) 4!

47 Example 6 If y , y 80 08, y and y find y 8. We can write the given data as follows: x : y : lies in the interval (80, 85). So we can use Gregory- Newton s forward interpolation formula. Since four values are given, the interpolation formula is y y 0 + u u( u ) u( u )( u ) y0 +!! y 0 + 3! 3 y 0 x x0 where u h Here h 5, x 0 75 x u The forward difference table : x y D y D y D 3 y y ! (.4 ) (-44) + (-397)!.4(.4 )(.4 ) + (457) 3! y when x 8

48 Example 7 From the following data calculate the value of e.75 x : e x : Since five values are given, the interpolation formula is y x y 0 + u u( u ) u( u )( u ) y0 +!! y 0 + 3! 3 y 0 u( u )( u )( u 3) + 4! 4 y 0 x x0 where u h Here h 0., x 0.7 x.75 u The forward difference table : x y y y 3 y 4 y y (0.5 ) (0.576) + (0.06)!! 0.5(0.5 )(0.5 ) + (0.007) 3! y when x.75 48

49 Example 8 From the data, find the number of students whose height is between 80cm. and 90cm. Height in cms x : No. of students y : The difference table Below Below Below Below Below x y y y 3 y 4 y Let us calculate the number of students whose height is less than 90cm. x x0 Here x 90 u h 0 (.5)(.5 ) y(90) 50 +(.5)(0) + ( 0)! (.5)(.5 )(.5 ) (.5)(.5 )(.5 )(.5 3) + (-0)+ (0) 3! 4! ~ 44 Therefore number of students whose height is between 80cm. and 90cm. is y(90) y(80) Example 9 i.e Find the number of men getting wages between Rs.30 and Rs.35 from the following table 0 0

50 Wages x : No. of men y : The difference table x y y y 3 y Under 30 9 Under Under Under Let us calculate the number of men whose wages is less than Rs.35. x x0 For x 35, u h 0 By Newton s forward formula, (0.5) ( 0.5)(0.5 ) y(35) 9+ (30) + (5)! ( 0.5)(0.5 )(0.5 ) + () 3! (approximately) Therefore number of men getting wages between Rs.30 and Rs.35 is y(35) y(30) i.e Gregory-Newton s backward formula Let the function y f(x) be a polynomial of degree n which assumes (n+) values f(x 0 ), f(x ), f(x ),..., f(x n ) where x 0, x, x,..., x n are in the increasing order and are equally spaced. Let x - x 0 x x x 3 x... x n x n- h (a positive quantity)

51 Here f(x) can be written as f(x) a 0 + a (x x n ) + a (x x n ) (x x n- ) +... When x x n, () f(x n ) a 0 or a 0 y n When x x n, () f(x n ) a 0 + a (x n x n ) or y n y n + a ( h) yn yn or a a h When x x n, () + a n (x x n ) (x x n- )... (x x ) () y n h f(x n ) a 0 + a (x n x n ) + a (x n x n ) (x n x n ) y n y n + y n h ( h) + a ( h) ( h) h a (y n y n ) + y n y n y n + (y n y n ) y n y n + y n y n y a n! h In the same way we can obtain 3 4 n y a 3 n y 3, a 3! h 4 n y 4... a 4! h n n n! y f(x) y n + n y (x xn ) + n (x x h! h n )(x x n ) +... n y + n (x x n! n ) (x x n )... (x x ) () x xn Further, denoting by u, we get h x x n h u 5

52 In general x x n (x x n ) (x n x n ) hu + h h(u+) x x n (x x n ) (x n x n ) hu + h h(u+) x x n 3 h(u+3) x x n k h(u+k) Thus () becomes, u u( u +) f(x) y n + yn +!! y n +... u( u + )...{ u + ( n )} + n! n y n where u This is the Gregory-Newton s backward formula. Example 0 5 x x h Using Gregory-Newton s formula estimate the population of town for the year 995. Year x : Population y : (in thousands) 995 lies in the interval (99, 00). Hence we can use Gregory-Newton s backward interpolation formula. Since five values are given, the interpolation formula is y y 4 + u u( u +) u( u + )( u + ) y4 +!! y 4 + 3! 3 y 4 u( u + )( u + )( u + 3) x x + 4! 4 4 y 4 where u h Here h 0, x 4 00 x 995 u n

53 The backward difference table : x y y y 3 y 4 y ( 0.6) ( 0.6)( 0.6+ ) y 0 + (8) + ( 4)!! ( 0.6)( 0.6+ )( 0.6+ ) + ( ) + 3! ( 0.6)( )( )( ) ( 3) 4! y i.e. the population for the year 995 is thousands. Example From the following table, estimate the premium for a policy maturing at the age of 58 Age x : Premium y : Since five values are given, the interpolation formula is y y 4 + u u( u + )( u + )( u + 3) y ! 4! y 4 where u The backward difference table : x y y y 3 y 4 y

54 y ( 0.4) (-6) +! 54 ( 0.4)(0.6) (.84) ( 0.4)(0.6)(.6)(.6) ( 0.4)(0.6)(.6) + (-.6) + (0.68) y i.e. y ~ Premium for a policy maturing at the age of 58 is Example From the following data, find y when x 4.5 x : y : Since five values are given, the interpolation formula is y y 4 + u u( u + )( u + )( u + 3) y ! 4! y 4 where x x4 u h Here u The backward difference table : x y y y 3 y 4 y ( 0.5) ( 0.5)(0.5) y 5+ (6)+ y 9.5 when x ( 0.5)(0.5)(.5) (4) + (6) 6

55 7..6 Lagrange s formula Let the function y f(x) be a polynomial of degree n which assumes (n + ) values f(x 0 ), f(x ), f(x )...f(x n ) corresponding to the arguments x 0, x, x,... x n (not necessarily equally spaced). Here f(x 0 ) y 0, f(x ) y,..., f(x n ) y n. Then the Lagrange s formula is ( x x )( x x )...( x xn) f(x) y 0 ( x x )( x x )...( x x ) Example ( x x0)( x x)...( x xn) + y ( x x )( x x )...( x x ) 0 ( x x0)( x x )...( x xn ) y n ( x x )( x x )...( x x ) n 0 n Using Lagrange s formula find the value of y when x 4 from the following table x : y : By data we have 55 0 n n n n x 0 40, x 50, x 60, x 3 70 and x 4 y 0 3, y 73, y 4, y 3 59 Using Lagrange s formula, we get ( x x)( x x)( x x3 ) y y 0 ( x x )( x x )( x x ) 0 0 ( x x0 )( x x)( x x3) + y ( x x )( x x )( x x ) 0 ( x x0)( x x)( x x3) + y ( x x )( x x )( x x )

56 ( x x0)( x x )( x x) + y 3 ( x x )( x x )( x x ) 3 0 ( 8)( 8)( 8) ()( 8)( 8) y(4) 3 ( 0)( 0)( 30) + 73 (0)( 0)( 0) ()( 8)( 8) +4 (0)(0)( 0) ( )( 8)( 8) +59 (30)(0)(0) y Example 4 Using Lagrange s formula find y when x 4 from the following table x : y : Given x 0 0, x 3, x 5, x 3 6, x 4 8 and x 4 y 0 76, y 460, y 44, y 3 343, y 4 0 Using Lagrange s formula ( x x)( x x)( x x3)( x x4) y y 0 ( x x )( x x )( x x )( x x ) 0 0 ( x x0 )( x x )( x x3)( x x4) + y ( x x )( x x )( x x )( x x ) 0 ( x x0)( x x )( x x3)( x x4) + y ( x x )( x x )( x x )( x x ) 0 ( x x0)( x x )( x x)( x x4) + y 3 ( x x )( x x )( x x )( x x ) ( x x0)( x x )( x x)( x x3) + y 4 ( x x )( x x )( x x )( x x )

57 ()( )( )( 4) (4)( )( )( 4) 76 ( 3)( 5)( 6)( 8) (3)( )( 3)( 5) (4)()( )( 4) (4)()( )( 4) + 44 (5)()( )( 3) +343 (6)(3)()( ) ( 4)()( )( ) + 0 (8)(5)(3)() y Example 5 Using Lagrange s formula find y() following table x : y : Given x 0 6, x 7, x 0, x 3 and x y 0 3, y 4, y 5, y 3 7 Using Lagrange s formula (4)()( ) (5)()( ) 3 ( )( 4)( 6) + 4 ()( 3)( 5) (5)(4)( ) (5)(4)() + 5 (4)(3)( ) +7 (6)(5)() y EXERCISE from the ) Using Graphic method, find the value of y when x 4, from the following data. x : y :

58 ) The population of a town is as follows. Year x : Population y : (in lakhs) Estimate the population for the year 976 graphically 3) From the following data, find f(3) x : f(x) : ) Find the missing term from the following data. x : y : ) From the following data estimate the export for the year 000 Year x : Export y : (in tons) 6) Using Gregory-Newton s formula, find y when x 45 given that x : y : ) Using Gregory-Newton s formula, find y(8) from the following data. x : y : ) Using Gregory-Newton s formula, calculate the population for the year 975 Year : Population : ) From the following data find the area of a circle of diameter 96 by using Gregory-Newton s formula Diameter x : Area y :

59 0) Using Gregory-Newton s formula, find y when x 85 x : y : ) Using Gregory-Newton s formula, find y(.4) x : y : ) From the following data find y(5) by using Lagrange s formula x : y : ) If f(0) 5, f() 6, f(3) 50, f(4) 05, find f() by using Lagrange s formula 4) Apply Lagrange s formula to find y when x 5 given that x : y : FITTING A STRAIGHT LINE A commonly occurring problem in many fields is the necessity of studying the relationship between two (or more) variables. For example the weight of a baby is related to its age ; the price of a commodity is related to its demand ; the maintenance cost of a car is related to its age. 7.. Scatter diagram This is the simplest method by which we can represent diagramatically a bivariate data. y Suppose x and y denote respectively the age and weight of an adult male, then consider a sample of n individuals with ages x, x, x 3,... x n and the corresponding weights as y, y, y 3,... y n. Plot the points x 59 Fig. 7.

60 (x, y ), (x, y ), (x 3, y 3 ),... (x n, y n ) on a rectangular co-ordinate system. The resulting set of points in a graph is called a scatter diagram. From the scatter diagram it is often possible to visualize a smooth curve approximating the data. Such a curve is called an approximating curve. In the above figure, the data appears to be well approximated by a straight line and we say that a linear relationship exists between the two variables. 7.. Principle of least squares Generally more than one curve of a given type will appear to fit a set of data. In constructing lines it is necessary to agree on a definition of a best fitting line. Consider the data points (x, y ), (x, y ), (x 3, y 3 ),... (x n, y n ). For a given value of x, say x, in general there will be a difference between the value y, and the corresponding value as determined from the curve C (in Fig. 7.) We denote this difference by d, which is referred to as a deviation or error. Here d may be positive, negative or zero. Similarly corresponding to the values x, x 3,... x n we obtain the deviations d, d 3,... d n. A measure of the goodness of fit of the curve to the set of data is provided by the quantities d, d,... d n. Of all the curves approximating a given set of data points, the curve having the property that d + d + d d n is a minimum is the best fitting curve. If the approximating curve is a straight line then such a line is called the line of best fit. 60 y (x, y ) d d (x, y ) Fig. 7. C d n (x n, y n ) x

61 7..3 Derivation of normal equations by the principle of least squares. Let us consider the fitting of a straight line y ax + b () to set of n points (x, y ), (x, y ),... (x n, y n ). For the different values of a and b equation () y represents a family of straight lines. Our aim is to determine a and b so that the line () is the line of best fit. Now a and b are determined by applying principle of least squares. Let P i (x i, y i ) be any Fig. 7.3 point in the scatter diagram. Draw P i M perpendicular to x-axis meeting the line () in H i. The x-coordinate of H i is x i. The ordinate of H i is ax i + b. P i H i P i M - H i M y i (ax i +b) is the deviation for y i. According to the principle of least squares, we have to find a and b so that E Σ n P H Σ n [y i i i (ax i + b)] is minimum. i i For maxima or minima, the partial derivatives of E with respect to a and b should vanish separately. n E 0 Σ x a i [y i (ax i + b)] 0 a i i n Ó x i + b Ó n x i Ó n x i y i () i i n E 0 - Σ [y b i (ax i + b)] 0 i 6 O P i (x i, y i ) M H i (x i, ax i +b) x

62 i.e., Σy i aσx i nb 0 a n n Ó x i i + nb Ó y i i (3) () and (3) are known as the normal equations. Solving the normal equations we get a and b. Note The normal equations for the line of best fit of the form y a + bx are Example 6 na + bsx i Sy i asx i + bsx i Sx i y i Fit a straight line to the following Sx 0, Sy 9, Sx 30, Sxy 53 and n 5. The line of best fit is y ax + b Σy aσx + nb Σxy aσx + bσx 0a + 5b () 30a + 0b () Solving () and () we get, a.5 and b 0.8 The line of best fit is y.5x Example 7 In a straight line of best fit find x-intercept when Sx 0, Sy6.9, Sx 30, Sxy 47.4 and n 7. The line of best fit is y ax + b The normal equations are Σy aσx + nb 6