JUST THE MATHS SLIDES NUMBER 15.4 ORDINARY DIFFERENTIAL EQUATIONS - PDF document

“JUST THE MATHS” SLIDES NUMBER 15.4 ORDINARY DIFFERENTIAL EQUATIONS 4 (Second order equations (A)) by A.J.Hobson 15.4.1 Introduction 15.4.2 Second order homogeneous equations 15.4.3 Special cases of the auxiliary equation

UNIT 15.4 - ORDINARY DIFFERENTIAL EQUATIONS 4 SECOND ORDER EQUATIONS (A) 15.4.1 INTRODUCTION A second order ordinary linear differential equation (with constant coefficients) has the general form a d 2 y d x 2 + b d y d x + cy = f ( x ) , where a , b and c are the constant coefficients The various cases of solution which arise depend on the values of the coefficients, together with the type of func- tion, f ( x ), on the right hand side. 15.4.2 SECOND ORDER HOMOGENEOUS EQUATIONS The term “homogeneous” , in the present context means that f ( x ) ≡ 0. That is, we consider a d 2 y d x 2 + b d y d x + cy = 0 . 1

Note: A very simple case of this equation is d 2 y d x 2 = 0 , which, on integration twice, gives the general solution y = Ax + B, where A and B are arbitrary constants. We should therefore expect two arbitrary constants in the solution of any second order linear differential equation with constant coefficients. THE STANDARD GENERAL SOLUTION The equivalent of a d 2 y d x 2 + b d y d x + cy = 0 in the discussion of first order differential equations would have been b d y d x + cy = 0 . That is, 2

d y d x + c by = 0 . This could have been solved using an integrating factor of � c b d x = e c b x , e giving the general solution y = Ae − c b x , where A is an arbitrary constant. We therefore make a trial solution of the form y = Ae mx , where A � = 0, in the second order case. We shall need d x = Ame mx and d 2 y d y d x 2 = Am 2 e mx . Hence, aAm 2 e mx + bAme mx + cAe mx = 0 . 3

In other words, am 2 + bm + c = 0 . This quadratic equation is called the “auxiliary equation” , having the same (constant) co- efficients as the orginal differential equation. In general, the auxiliary equation will have two solutions, say m = m 1 and m = m 2 , giving corresponding solu- tions, y = Ae m 1 x and y = Be m 2 x , of the differential equation. The linearity of the differential equation implies that the sum of any two solutions is also a solution. Thus, y = Ae m 1 x + Be m 2 x is another solution. Since this contains two arbitrary constants, we take it to be the general solution. 4

Notes: (i) It may be shown that there are no solutions other than those of the above form, though special cases are considered later. (ii) It will be possible to determine particular values of A and B if an appropriate number of boundary conditions for the differential equation are specified. (iii) Boundary conditions usually consist of a set of given values for y and d y d x at a certain value of x . EXAMPLE Determine the general solution of the differential equation d 2 y d x 2 + 5d y d x + 6 y = 0 and also the particular solution for which y = 2 and d y d x = − 5 when x = 0. 5

Solution The auxiliary equation is m 2 + 5 m + 6 = 0 , which can be factorised as ( m + 2)( m + 3) = 0 . Its solutions are therefore m = − 2 amd m = − 3. Hence, the differential equation has general solution y = Ae − 2 x + Be − 3 x , where A and B are arbitrary constants. Applying the boundary conditions, we shall also need d y d x = − 2 Ae − 2 x − 3 Be − 3 x . Hence, 2 = A + B and − 5 = − 2 A − 3 B , giving A = 1, B = 1. The required particular solution is y = e − 2 x + e − 3 x . 6

15.4.3 SPECIAL CASES OF THE AUXILIARY EQUATION (a) The auxilary equation has coincident solutions Suppose that both solutions of the auxiliary equation are the same number, m 1 . In other words, am 2 + bm + c is a “perfect square” and hence ≡ a ( m − m 1 ) 2 Apparently, the general solution of the differential equa- tion is y = Ae m 1 x + Be m 1 x . But this does not genuinely contain two arbitrary con- stants since it can be rewritten as y = Ce m 1 x where C = A + B. It will not, therefore, count as the general solution. The fault seems to lie with the constants A and B rather than with m 1 . 7

We now examine a new trial solution of the form y = ze m 1 x , where z denotes a function of x . We shall also need d y d x = zm 1 e m 1 x + e m 1 x d z d x and d 2 y d x + e m 1 x d 2 z 1 e m 1 x + 2 m 1 e m 1 x d z d x 2 = zm 2 d x 2 . Substituting these into the differential equation, we ob- tain d x + d 2 z   d z   zm 1 + d z    e m 1 x  zm 2  + cz  a 1 + 2 m 1  + b  = 0           d x 2 d x or d x (2 am 1 + b ) + a d 2 z 1 + bm 1 + c ) + d z z ( am 2 d x 2 = 0 . The first term on the left hand side is zero since m 1 is already a solution of the auxiliary equation. 8

The second term on the left hand side is also zero since the auxiliary equation is equivalent to a ( m − m 1 ) 2 = 0 . That is, am 2 − 2 am 1 m + am 2 1 = 0 . Thus b = − 2 am 1 . We conclude that d 2 z d x 2 = 0 , with the result that z = Ax + B, where A and B are arbitrary constants. Summary The general solution of the differential equation in the case of coincident solutions to the auxiliary equation y = ( Ax + B ) e m 1 x . 9

EXAMPLE Determine the general solution of the differential equation 4d 2 y d x 2 + 4d y d x + y = 0 . Solution The auxilary equation is 4 m 2 + 4 m + 1 = 0 or (2 m + 1) 2 = 0 . It has coincident solutions at m = − 1 2 . The general solution is therefore y = ( Ax + B ) e − 1 2 x . 10

(b) The auxiliary equation has complex solutions If the auxiliary equation has complex solutions, they will appear as a pair of “complex conjugates” , say m = α ± jβ . Using m = α ± jβ instead of m = m 1 and m = m 2 , the general solution of the differential equation will be y = Pe ( α + jβ ) x + Qe ( α − jβ ) x , where P and Q are arbitrary constants. By properties of complex numbers, a neater form of this result is obtainable as follows: y = e αx [ P (cos βx + j sin βx ) + Q (cos βx − j sin βx )] or y = e αx [( P + Q ) cos βx + j ( P − Q ) sin βx ] . Replacing P + Q and j ( P − Q ) (which are just arbitrary quantities) by A and B , we obtain the standard general solution for the case in which the auxiliary equation has complex solutions. The general solution is 11

y = e αx [ A cos βx + B sin βx ] . EXAMPLE Determine the general solution of the differential equation d 2 y d x 2 − 6d y d x + 13 y = 0 . Solution The auxiliary equation is m 2 − 6 m + 13 = 0 , which has solutions given by ( − 6) 2 − 4 × 13 × 1 � m = − ( − 6) ± 2 × 1 = 6 ± j 4 = 3 ± j 2 . 2 The general solution is therefore y = e 3 x [ A cos 2 x + B sin 2 x ] . 12