Differential Equations 2

Transcription

1 Diffrntial Equations Equations of th form f a b cy 0 a b cy h ar Scond Ordr Diffrntial Equations. Typ can b solvd by intgrating twic (assuming that th function can b intgratd). f f and its intgral 5 0 subjct to th conditions that y(0) 0, y'(0) (i whn 0, y 0 and ) Intgrating givs 5 A y'(0) i 0, A A 5 giving 5 5 5y 5 B intgrating again givs B B 0 y(0) 0 i 0, y 0 Eampl Solv th quation hnc solution is or 5y 5 6 y 0 0

2 Typ Linar quations with constant cofficints of th form: a b cy 0 Th first ordr, linar quation cy 0 has solution y m to try y as th solution for a scond ordr Linar Diffrntial quation... m y substituting into th quation givs am m m m bm m c k so it sms logical m m c m 0 m am bm c 0 m but as 0 this givs am bm c 0. This quation is oftn calld th auiliary quation and thr ar cass to considr dpnding on whthr th roots of th quadratic quation ar (i) ral and diffrnt, (ii) ral and qual or (iii) compl. Cas (i) Two distinct roots Lt th roots of th auiliary quation am bm c 0 b m p and m q thn y, y ar solutions of th diffrntial quation and it looks as though th gnral solution could b ( A and B ar arbitrary constants), Eampl p q p q y A B Find th gnral solution of th quation 5 6y 0 Th auiliary quation is m 5m 6 0 which factoriss to giv m m 6 0 with roots m, m 6 Gnral solution is y 6 A B Cas (ii) Roots ral and qual Eprinc shows that if th auiliary quation am bm c 0 has qual roots m p and m thn th gnral solution is p y A B p

3 Eampl Find th gnral solution of th quation 6 9y 0 Th auiliary quation is m 6m 9 0 which givs m, m and gnral solution y A B Cas (iii) compl roots If th auiliary quation am bm c 0 has compl roots of th form m j whr j th solution is y Acos B sin Eampl Solv th quation y 0 Th auiliary quation is m m 0 which givs m j j y Acos Bsin. Th gnral solution is Ercis Solv th quations y 0. y 0 5. y y 0 7. y 0 8. y 0 9. y y 0. 9y 0. 9y 0. y y 0 5. y 0 6. y 0 7. a a y y 0 d k 9. Solv th quation k 0, givn 0, u, whn t 0. d s ds ds 0. Solv th quation s 0, givn s a, u, whn t 0.

4 Typ Linar quations with constant cofficints of th form This is an tnsion of typ. a b cy f W can gt th solution of a b cy 0 in th form y Au Bu whr u and u ar functions of. If w can find a particular function y g( ) which satisfis th quation a b cy thn w can writ th gnral solution as y Au Bu g( ) Th main problm is finding th function y g( )! y Au Bu is oftn calld th complmntary function an g( ) th particular intgral. It quit logical to assum that th function y g( ) will b th sam sort of function as f( ). For ampl if f( ) is of th form thn g ( ) must b of th form k or k... as thr ar no combinations of othr functions which will giv th corrct f( ). Thr ar a numbr of diffrnt possibilitis thr of which ar discussd hr. f

5 Eampls Solv th quations (a) cas (i) f( ) of th form m k y (b) y y In both th complmntary function is th solution of 0 This has auiliary quation m m 0 which givs m, m. Hnc complmntary function is y A B. (a) y has Complmntary function To find a particular intgral try So, for (b) y k to satisfy thn 9k y A y k y k k 9k y k Hnc th gnral solution is y k k k y A B has Complmntary Function y A B B Thr is a trm A in th complmntary function so th particular intgral cannot b of th form y k so w try y k y k k k So, for y k to satisfy k k k k k k y k k k k k k k k k k k k Hnc th gnral solution is y A B

6 cas (ii) f( ) is a polynomial of dgr n A particular intgral can b found by substituting n n n n y a0 a a a... an Th highst powr in th particular function cannot b highr than th highst powr in th function f( ). d s Eampl Solv th quation s t Th auiliary quation is m 0 which givs m j. Hnc th complmntary function is s Acos t B sint To find a particular intgral try s a0t at a ds d s s a0t at a a0t a a 0 d s s t To satisfy th quation a0 a 0t at a a0t at a0 a Equating cofficints of t trm a0 t t a 0 t trm a 0 a 0 constant trm a0 a a 8 Particular Intgral is s t t Gnral solution s Acos t B sint t

7 cas (iii) f( ) is of th form k cos c k sinc (whr k or k may b zro). In gnral a particular intgral can b found by trying y pcos c+ q sinc d r dr Eampl Solv th quation 5r sin d d Auiliary quation m m 5 0 givs complmntary function 5 r A B Try particular intgral r pcos + q sin dr p sin +q cos d d r p cos q sin d To satisfy th quation d r d dr d 5r sin p sin q cos 5p cos q sin sin p cos q sin p cos 8q cos 5pcos q sin 8p sin 5q sin sin 9p 8q sin 9q 8p sin cos Equating trms in cos : sin : 9 p 8q 0 9q 8p Solving 9 p 8q 0 9 p 8q 0 8p 7q 0 9q 8p 8p 9q 6 p 7q 96 Adding givs Substituting in givs 5p 96 p q Gnral solution r 5 A B 96 5 cos 08 5 sin 6

8 Furthr ampls (a) Solv th quation d r d dr 5r 7 5 d From th prvious ampl th complmntary function is Try particular intgral To satisfy a 5 r A B r a b c dr d r a b; a d d d r dr 5r 7 5 d d a b 5a b c 7 5 a b 5c 8a 5b 5a 7 5 Equating trms trms constant trms 5a 5 8a 5b a + b - 5c = -7 Solving givs a, b, c Gnral Solution 5 r A B (b) Solv th quation Auiliary Equation 6y 0 cos (a mor compl quation!) m 6 0 m j Complmntary Function y Acos B sin Th Particular Intgral must hav trms to giv both th cos trm and th trm whn put into 6y. W would normally go for y p cos q sin k but w alra hav a cos trm in th CF so w nd to us 0 This givs y p cos q sin k p sin q cos p cos q sin 6 p cos 6q sin p sin q cos p sin q cos k k 6 p cos q sin 8 p sin q cos k 7

9 To satisfy th diffrntial quation 6 p cos q sin 8 p sin q cos 6 p cos q sin 6y cos 0 k 6k cos 0 Equating trms in sin : 8p sin 8q cos 0k 8p 0 p 0 cos : 8q q : 0k 0 k cos 0 Hnc Gnral Solution is y Acos B sin sin Ercis Solv th following y y cos y y y 5y 80 y 6 y 6 cos sin y 6 6y y 00 sin 5y 5 8 5y 50 y 9y cos sin d d 7. Solv th quation sint givn that, 0 whn t If y y,., fin in trms of, givn that, whn 0, 9. Obtain th solutions of th quation 9y sin givn that whn 0, y 0, 0. 8

10 ANSWERS Ercis. y A B. y A B. y A B 5. y A B 7. y Acos B sin. y A B. y A B 6. y A B 8 9. y A B 5 0. y A B. y A B. y Acos B sin y Acos. y A B y y Acos u k kt A B B sin a sin kt B sin y A y t s B Acos B sin 5 a cost ua sint Ercis y A B. 5. y A B. y Acos B sin cos 8. y A B 9 5. y A B 8 y A B y A y y y B Acos B sin Acos B sin cos 6. sin Acos B sin B sin 5. A cos 5 B sin 5 sin. y Acos y y Acos B sin cos t sint sint y y cos sin y A B y A B Acos B sin y ( sin) 8 9