“JUST THE MATHS” SLIDES NUMBER 15.3 ORDINARY DIFFERENTIAL EQUATIONS 3 (First order equations (C)) by A.J.Hobson 15.3.1 Linear equations 15.3.2 Bernouilli’s equation
UNIT 15.3 - ORDINARY DIFFERENTIAL EQUATIONS 3 FIRST ORDER EQUATIONS (C) 15.3.1 LINEAR EQUATIONS For certain kinds of first order differential equation, it is possible to multiply the equation throughout by a suit- able factor which converts it into an exact differential equation. EXAMPLE d y d x + 1 xy = x 2 may be multiplied throughout by x to give x d y d x + y = x 3 . It may now be written d d x ( xy ) = x 3 . 1
This equation has general solution xy = x 4 4 + C, where C is an arbitrary constant. Notes: (i) The factor, x which has multiplied both sides of the dif- ferential equation serves as an “integrating factor” , but such factors cannot always be found by inspection. (i) We shall now develop a formula for determining inte- grating factors for what are known as “linear differ- ential equations” . DEFINITION A differential equation of the form d y d x + P ( x ) y = Q ( x ) is said to be “linear” . 2
RESULT (a) Given the linear differential equation d y d x + P ( x ) y = Q ( x ) , the function � P ( x ) d x e is always an integrating factor; (b) On multiplying the differential equation throughout by this factor, its left hand side becomes d � P ( x ) d x � � y × e . d x Proof Suppose that the function R ( x ) is an integrating factor. Then, in the equation R ( x )d y d x + R ( x ) P ( x ) y = R ( x ) Q ( x ) , the left hand side must be the exact derivative of some function of x . 3
We can make it the derivative of R ( x ) y provided we can arrange that R ( x ) P ( x ) = d d x [ R ( x )] . This requirement can be interpreted as a differential equa- tion in which the variables R ( x ) and x may be separated as follows: � P ( x ) d x. 1 � R ( x ) d R ( x ) = Hence, � P ( x ) d x. ln R ( x ) = That is, � P ( x ) d x . R ( x ) = e The solution of the linear differential equation is obtained by integrating the formula d d x [ y × R ( x )] = R ( x ) P ( x ) . 4
Note: There is no need to include an arbitrary constant, C , when P ( x ) is integrated; C would introduce a constant factor of e C in the above result, which would then cancel out on multiplying by R ( x ). EXAMPLES 1. Determine the general solution of the differential equa- tion d x + 1 d y xy = x 2 . Solution An integrating factor is x d x = e ln x = x. 1 � e On multiplying by the integrating factor, d d x [ y × x ] = x 3 . Hence, yx = x 4 4 + C, where C is an arbitrary constant. 5
2. Determine the general solution of the differential equa- tion d y d x + 2 xy = 2 e − x 2 . Solution An integrating factor is � 2 x d x = e x 2 . e Hence, d y × e x 2 � � = 2 , d x giving ye x 2 = 2 x + C, where C is an arbitrary constant. 6
15.3.2 BERNOUILLI’S EQUATION This type of differential equation has the form d y d x + P ( x ) y = Q ( x ) y n . It may be converted to a linear differential equation by making the substitution z = y 1 − n . Proof The differential equation may be rewritten as y − n d y d x + P ( x ) y 1 − n = Q ( x ) . Also, d x = (1 − n ) y − n d y d z d x. Hence, the differential equation becomes 1 d z d x + P ( x ) z = Q ( x ) . 1 − n 7
That is, d z d x + (1 − n ) P ( x ) z = (1 − n ) Q ( x ) , which is a linear differential equation. 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 equa- tion xy − d y d x = y 3 e − x 2 . Solution The differential equation may be rewritten − y − 3 d y d x + x.y − 2 = e − x 2 . Substituting z = y − 2 , d x = − 2 y − 3 d y d z d x. 1 d z d x + xz = e − x 2 Hence , 2 8
or d z d x + 2 xz = 2 e − x 2 . An integrating factor for this equation is � 2 x d x = e x 2 . e Thus, d ze x 2 � � = 2 , d x giving ze x 2 = 2 x + C, where C is an arbitrary constant. Finally, replacing z by y − 2 , e x 2 y 2 = 2 x + C. 2. Determine the general solution of the differential equa- tion d y d x + y x = xy 2 . Solution The differential equation may be rewritten y − 2 d y d x + 1 x.y − 1 = x. 9
Substituting z = y − 1 , d x = − y − 2 d y d z d x. Thus, − d z d x + 1 x.z = x or d z d x − 1 x.z = − x. An integrating factor for this equation is � ( − 1 x ) d x = e − ln x = 1 e x. Hence, d z × 1 = − 1 , d x x giving z x = − x + C, where C is an arbitrary constant. The general solution of the given differential equation is therefore 1 1 xy = − x + C or y = Cx − x 2 . 10
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