# JUST THE MATHS UNIT NUMBER ORDINARY DIFFERENTIAL EQUATIONS 3 (First order equations (C)) A.J.Hobson

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1 JUST THE MATHS UNIT NUMBER 15.3 ORDINARY DIFFERENTIAL EQUATIONS 3 (First order equations (C)) by A.J.Hobson Linear equations Bernouilli s equation Exercises Answers to exercises

2 UNIT ORDINARY DIFFERENTIAL EQUATIONS 3 FIRST ORDER EQUATIONS (C) LINEAR EQUATIONS For certain kinds of first order differential equation, it is possible to multiply the equation throughout by a suitable factor which converts it into an exact differential equation. For instance, the equation may be multiplied throughout by x to give + 1 x y = x2 It may now be written x + y = x3. and, hence, it has general solution d (xy) = x3 where C is an arbitrary constant. Notes: xy = x4 4 + C, (i) The factor, x which has multiplied both sides of the differential equation serves as an integrating factor, but such factors cannot always be found by inspection. (ii) In the discussion which follows, we shall develop a formula for determining integrating factors, in general, for what are known as linear differential equations. 1

3 DEFINITION A differential equation of the form + P (x)y = Q(x) is said to be linear. RESULT Given the linear differential equation the function + P (x)y = Q(x), e P (x) is always an integrating factor; and, on multiplying the differential equation throughout by this factor, its left hand side becomes Proof d [ ] y e P (x). Suppose that the function, R(x), is an integrating factor; then, in the equation R(x) + R(x)P (x)y = R(x)Q(x), the left hand side must be the exact derivative of some function of x. Using the formula for differentiating the product of two functions of x, we can make it the derivative of R(x)y provided we can arrange that R(x)P (x) = d [R(x)]. 2

4 But this requirement can be interpreted as a differential equation in which the variables R(x) and x may be separated as follows: 1 R(x) dr(x) = P (x). Hence, ln R(x) = P (x). That is, R(x) = e P (x), as required. The solution is obtained by integrating the formula d [y R(x)] = R(x)P (x). Note: There is no need to include an arbitrary constant, C, when P (x) is integrated, since it would only serve to introduce a constant factor of e C in the above result, which would then immediately cancel out on multiplying the differential equation by R(x). EXAMPLES 1. Determine the general solution of the differential equation Solution An integrating factor is + 1 x y = x2. e 1 x = e ln x = x. 3

5 On multiplying throughout by the integrating factor, we obtain and so, d [y x] = x3 ; where C is an arbitrary constant. yx = x4 4 + C, 2. Determine the general solution of the differential equation Solution An integrating factor is Hence, giving where C is an arbitrary constant. + 2xy = 2e x2. e 2x = e x2. d [ y e x 2] = 2, ye x2 = 2x + C, BERNOUILLI S EQUATION A similar type of differential equation to that in the previous section has the form + P (x)y = Q(x)yn. 4

6 It is called Bernouilli s Equation and may be converted to a linear differential equation by making the substitution z = y 1 n. Proof The differential equation may be rewritten as Also, n y + P (x)y1 n = Q(x). dz Hence the differential equation becomes That is, which is a linear differential equation. = (1 n)y n. 1 dz + P (x)z = Q(x). 1 n dz + (1 n)p (x)z = (1 n)q(x), Note: It is better not to regard this as a standard formula, but to apply the method of obtaining it in the case of particular examples. EXAMPLES 1. Determine the general solution of the differential equation xy = y3 e x2. 5

7 Solution The differential equation may be rewritten Substituting z = y 2, we obtain dz 3 y + x.y 2 = e x2. = 2y 3 and, hence, or 1 dz + xz = e x2 2 An integrating factor for this equation is Thus, giving where C is an arbitrary constant. Finally, replacing z by y 2, dz + 2xz = 2e x2. e 2x = e x2. d ( ze x 2) = 2, ze x2 = 2x + C, y 2 = ex2 2x + C. 2. Determine the general solution of the differential equation + y x = xy2. 6

8 Solution The differential equation may be rewritten On substituting z = y 1 we obtain dz or An integrating factor for this equation is 2 y + 1 x.y 1 = x. = y 2 dz + 1 x.z = x dz 1.z = x. x so that 1 e ( x) = e ln x = 1 x. Hence, ( d z 1 ) = 1, x giving where C is an arbitrary constant. z x = x + C, The general solution of the given differential equation is therefore 1 xy = x + C or y = 1 Cx x EXERCISES Use an integrating factor to solve the following differential equations subject to the given boundary condition: 1. where y = 10 when x = y = 0, 7

9 2. where y = 4 when x = where y = 2 when x = where y = 0 when x = where y = 5 when x = π where y = 0 when x = where y = 1 when x = where y 3 = 14 when x = y = 10, + y x = 3x, + y 1 x = 1 x2, + y cot x = cos x, (x 2 + 1) xy = x, 3y 2 = y3 e 4x, 2y x = x(x 1)y4, 8

10 ANSWERS TO EXERCISES 1. y = 10e 2 3 x. 2. y = 6e 5 3 x yx = x y = 1 2 (1 x)(1 + x) y = sin x 2 y = 1 + x sin x. 2(1 + x 2 ) y 3 = y 2 = 7e3x e 7x x 6 21x 6 24x

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