MA1506CHAP1 - MA1506 LECTURE NOTES CHAPTER 1 DIFFERENTIAL...

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Unformatted text preview: MA1506 LECTURE NOTES CHAPTER 1 DIFFERENTIAL EQUATIONS 1.1 Introduction A differential equation is an equation that contains one or more derivatives of a differen- tiable function. [In this course we deal only with ordinary DEs, NOT partial DEs.] The order of a d.e. is the order of the equa- tion’s highest order derivative; and a d.e. is linear if it can be put in the form a n y ( n ) ( x )+ a n- 1 y ( n- 1) ( x )+ ··· + a 1 y (1) ( x )+ a y ( x ) = F, 1 where a i , ≤ i ≤ n , and F are all functions of x . For example, y = 5 y and xy- sin x = 0 are first order linear d.e.; ( y 000 ) 2 + ( y 00 ) 5- y = e x is third order, nonlinear. We observe that in general, a d.e. has many solutions, e.g. y = sin x + c , c an arbitrary constant, is a solution of y = cos x . Such solutions containing arbitrary constants are called general solution of a given d.e.. Any solution obtained from the general solution by giving specific values to the arbitrary constants is called a particular solution of that d.e. e.g. 2 y = sin x + 1 is a particular solution of y = cos x . Basically, differential equations are solved us- ing integration, and it is clear that there will be as many integrations as the order of the DE. Therefore, THE GENERAL SOLUTION OF AN nth-ORDER DE WILL HAVE n ARBI- TRARY CONSTANTS. 1.2 Separable equations A first order d.e. is separable if it can be writ- ten in the form M ( x )- N ( y ) y = 0 or equiva- lently, M ( x ) dx = N ( y ) dy . When we write the 3 d.e. in this form, we say that we have separated the variables , because everything involving x is on one side, and everything involving y is on the other. We can solve such a d.e. by integrating w.r.t. x : Z M ( x ) dx = Z N ( y ) dy + c. Example 1. Solve y = (1 + y 2 ) e x . Solution. We separate the variables to obtain e x dx = 1 1 + y 2 dy. 4 Integrating w.r.t. x gives e x = tan- 1 y + c, or tan- 1 y = e x- c, or y = tan( e x- c ) . Example 2. Experiments show that a ra- dioactive substance decomposes at a rate pro- portional to the amount present. Starting with 2 mg at certain time, say t = 0, what can be said about the amount available at a later time? Example 3. A copper ball is heated to 5 100 ◦ C. At t = 0 it is placed in water which is maintained at 30 ◦ C. At the end of 3 mins the temperature of the ball is reduced to 70 ◦ C. Find the time at which the temperature of the ball is 31 ◦ C. Physical information: Experiments show that the rate of change dT/dt of the temperature T of the ball w.r.t. time is proportional to the dif- ference between T and the temp T of the sur- rounding medium. Also, heat flows so rapidly in copper that at any time the temperature is practically the same at all points of the ball....
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MA1506CHAP1 - MA1506 LECTURE NOTES CHAPTER 1 DIFFERENTIAL...

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