Brief Notes on Differential Equations

Transcription

1 Brief Notes on Differential Equations (A) Searable First-Order Differential Equations Solve the following differential equations b searation of variables: (a) (b) Solution ( 1 ) (a) The ODE becomes d and d d C C (b) The ODE ( 1 ) becomes d 1 and d ( 1 ) ( 1 ) d 1 ln C ln C' 1 (B) Solving First-Order Linear Differential Equation having the te P( ) Q( ) (1) if P( ) 0, it is solvable b direct integration, or () if Q( ) 0, the equation is searable. Theorem The equation P Q Proof ( ) ( ) has e ( P( ) ) When Eq.(6.) is multilied b e ( P( ) ) as an integration factor., it can be written in the form

2 d which has solution ( e ( P ( ) ) ) Q ( )e ( P ( ) ) ( ) ( ) ( ) e P ( ) Q ( )e P ( ) c e P ( ) So, we have the following stes to solve a linear first-order differential equation P( ) Q( ) The following rocedures are often helful in finding solution to the equations:- (1) Comute the integrating factor. () Multil the right-hand side of the given equation b this factor and write the left-hand side as the derivative of times the integrating factor. () Integrating and solving the equation for. Solve the following differential equations b integration factor method: (a) (b) k tan sec Solution (a) The ODE e kd k becomes k. The integration factor is k The ODE can be reduced to k k k 1 1 k d k k 1 k d C k k C k k ( ) 1 k

3 (b) The integration factor is ( d ) e tan sec and the ODE can be reduced to sec sec tan sec sec sec d sin C cos tan C d ( sec ) sec (C) Solving Second Order Differential Equation b Reduction of Order Let us consider a second-order differential equation in the form d F d (, ) where F(, ) is a continuous function. To solve the differential equation of this te, we can at least reduce its order b one Let d, then d d d d d d d d d d d Therefore, the original differential equation Eq.(6.6) is reduced to a first-order differential equation d F, ( ) The above method is known as the reduction of order. Solve the differential equations d b reduction of order: Solution Let d, we have d d d and the ODE becomes

4 dddd dddd dddd dddd 1 CC dddd dddd dddd 1 ( CC ) CC 1 dddd 1 CC tan 1 CC 1 CC tan 1 CC CC CCCC CC tan CC CCCC CC 1 tan(cc 1 CC ), where C 1 C/ and C CC. (D) Second-Order Differential Equation with Constant Coefficients 1. The Characteristic Equation The second-order differential equation with constant coefficients has the general form d a d a f 1 o ( ) (1) d d in which a 1 and a o are real constants. When f() is non-trivial (i.e. f ( ) 0) the equation is nonhomogeneous second-order differential equation with constant coefficients. Otherwise, if f ( ) 0 the equation becomes d a d a 1 o 0 () d d which is homogeneous second-order differential equation with constant coefficients. In section, we consider the second-order homogeneous ODE with constant coefficients. In section, we consider the second-order nonhomogeneous ODE with constant coefficients. If we assume that ( ) Ae( λ ) is articular solutions of the above homogeneous equation Eq.(7.), then we can substitute it back to the equation λ Ae( λ) a λ Ae( λ) a Ae( λ) and it is reduced to an algebraic equation 1 o 0 λ a1λ a o 0 () Equation Eq.(7.10) is called the characteristic equation of the second-order ODE.

5 . The Comlementar Solution (General Solution) of the Homogeneous Equation Usuall, the characteristic equation Eq.() has two comle roots λ 1 and λ, we have the following cases: Case (1). λ 1 and λ are both real and distinct In this case, we have the two linear indeendent solutions e( λ 1 ) and e( λ ) and the general solution is In the secial case λ ( ) ( ) ( ) c e λ c e λ (4) 1 1 λ, the solution can be rewritten as 1 ( ) k sinh λ k cosh λ (5) 1 1 where sinh θ eeθθ ee θθ and cosh θ eeθθ ee θθ Case (). λ 1 a i b, a comle number In this case, since a 1 and a o are assumed to be real, the roots of Eq.() must aear in conjugate airs; thus the other root is λ a ib. Two linear indeendent solutions are [( ib) ] and e ( a ) e a [ ib ] and the general solution is [( ) ] ( ) [ ] ( ) c e a ib c e a ib which is algebraic equivalent to 1 a a ) C e cosb C e sin b (6) ( 1 Case (). λ 1 λ In this case, we have two linear indeendent solutions e( λ 1 ) and ( ) general solution is ( ) ( ) ( ) c e λ c e λ (7) e λ 1 and the Solve the following differential equations: d (a)

6 (b) (c) d d Solution 7.4 (a) The characteristic equation is λ 10λ 1 0 The roots of the above equation are, 7; hence the solution is c e c e 1 7 (b) The characteristic equation is λ 6λ 5 0 The roots of the above equation are ± 4 i ; hence the solution is e ( c cos4 c sin 4) 1 (c) The characteristic equation is λ 8λ 16 0 The roots of the above equation are 4 (reeated roots); hence the solution is 4 ( c c ) e 1. The Particular Solution and Comlete Solution of the Nonhomogeneous Equation Recall the nonhomogeneous equation again d a d a f 1 o ( ) (8) d d To solve this equation (i.e. f ( ) 0), we ought to be able to find one articular solution of the equation g(). Then the comlete solution is the sum of the comlementar function h() obtained b utting f ( ) 0 in equation (1) and g() i.e. h() gg() (9) If f() is a function for which reeated differentiation ields onl a finite number of linearl indeendent functions, aearing ossibl in linear combination, then a articular integral Y for the nonhomogeneous linear equation d a d a f 1 o ( ) d d can be found through the rocedures: (1) Assume Y to be an arbitrar linear combination of all the linearl indeendent functions which arise from f() b reeated differentiation.

7 () Substitute Y into the given differential equation. () Determine the arbitrar constants in Y. The class of function f() ossessing onl a finite number of linearl indeendent derivatives consists of the simle functions: k n (n is a ositive integer) e k sin k cosk Remark: For the cases that f() has corresondence in the comlementar solution of the equation, those treatments will be discussed in standard tetbook of differential equations. The are out of the scoe in this brief notes. Use the method of undetermined coefficients to find the comlete solution of the following differential equations: (a) (b) d d e d sin Solution (a) The characteristic equation is λ λ 0 The roots of the above equation are 1, ; hence the general solution is c e c e 1 To find the articular solution, we assume that Ae Thus, ' Ae and " 9 Ae. Substituting these results into the differential

8 equation, we have 9 Ae Ae Ae e 4 Ae e It follows that A 1 4 so that e 1 4 and the comlete solution is 1 c1e ce e 4 (b) From the results of (a), we have the general solution c e c e 1 To find the articular solution, we assume that A sin B cos Thus, ' Acos Bsin and " 4 Asin 4B cos. Substituting these results into the differential equation, we have ( 4 Asin 4B cos ) ( Acos B sin ) ( Asin B cos ) sin or equivalentl, ( 6A B) sin ( 6B A) cos ( 1) sin ( 0) cos Equating coefficients of like terms, we obtain 6A B 1 A 6B 0 Solving the above sstem, we find that A 0 and 1 B 0 solution is. Then the comlete 1 c1e ce sin cos 0 0