UNIT-I. The study of a differential equation in applied mathematics consists of three phases.
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1 UNIT-I ORINARY IFFERENTIAL EQUATIONS Highr ordr dirntial quations with constant coicints Mthod o variation o paramtrs Cauchy s and Lgndr s linar quations Simultanous irst ordr linar quations with constant coicints. Th study o a dirntial quation in applid mathmatics consists o thr phass. (i) (ii) (iii) Formation o dirntial quation rom th givn physical situation, calld modling. Solutions o this dirntial quation, valuating th arbitrary constants rom th givn conditions, and Physical intrprtation o th solution. HIGHER ORER LINEAR IFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS. Gnral orm o a linar dirntial quation o th nth ordr with constant coicints is n n n d y d y d y K K Kny X n n n d..... () d d Whr K, K,......K ar constants. n Th symbol stands or th opration o dirntial y dy d d y (i..,) y, similarly y, y, tc... d d d Th quation () abov can b writtn in th symbolic orm n n ( K... Kn ) y X i.., ()y X n n Whr () K... K Not. X Xd a a. X X d a a a. X X d a (i) Th gnral orm o th dirntial quation o scond ordr is n
2 d y dy P Qy R d d () Whr P and Q ar constants and R is a unction o or constant. (ii)irntial oprators: Th symbol stands or th opration o dirntial dy d y (i..,) y, y d d Stands or th opration o intgration Stands or th opration o intgration twic. () can b writtn in th oprator orm y Py Qy R (Or) ( P Q) y R (iv) Complt solution Complmntary unction Particular Intgral PROBLEMS. Solv ( 6) y 0 Solution: Givn ( 6) y 0 Th auiliary quation is m m 6 0` i.., m, CF. A B Th gnral solution is givn by y A B d y dy. Solv 6 y 0 d d 6 y Solution: Givn ( ) 0 Th auiliary quation is m 6m 0 6 ± 6 i.., m ± i Hnc th solution is y ( Acos Bsin ). Solv ( ) 0 givn y(0) 0, y (0)
3 Solution: Givn ( ) 0. Solv Solution: Givn A.E is m 0 M ± i Y A cos B sin Y() A cos B sin Y(0) A 0 Y (0) B A 0, B i.., y (0) cos sin y sin ( ) y ( ) y Th auiliary quation is m m 0 ± 6 ± 6 m ± i CF. ( Acos Bsin ) P.I. 8 9 y C.F P.I. y ( Acos Bsin ) 9. Find th Particular intgral o y - y y Solution: Givn y - y y P.I ( ) y 0
4 P.I ( ) P.I. P.I P.I (- -( ) ) d y dy 6. Solv y cosh d d d y dy Solution: Givn y cosh d d Th A.E is m m 0 m ± 6 0 ± i Acos Bsin C.F ( ) P.I ( ) cosh 0 y CF. PI. ( Acos Bsin) - 0 sin ( ) ( ) Problms basd on P.I a( or) cosa Rplac d y dy 7. Solv y sin d d by a
5 d y dy Solution: Givn y sin d d Th A.E is m m 0 (m)(m) 0 M -, m - C.F A B P.I sin sin (Rplac by a ) ( 7) sin sin 7 7 ( 7) 7 sin () (7) 7 sin sin 9( ) 9 7 sin 0 ( (sin ) 7sin ) 0 ( 9cos 7sin ) 0 y CF. PI. Y A 0 B ( 9cos 7sin ) 8. Find th P.I o ( ) sin Solution: Givn ( ) sin P.I. sin sin sin sin sin d cos P.I -
6 6 9. Find th particular intgral o ( ) y sin sin Solution: Givn ( ) y sin sin - ( cos cos) - cos cos P.I cos cos 9 cos 6 P.I cos cos cos cos d sin P.I cos sin 6 Problms basd on R.H.S 0. Solv ( ) y cos Solution: Givn ( a ) y a cos a( or) cosa cos Th Auiliary quation is m m 0 (m ) 0 m, C.F (A B) P.I 8 0 0
7 7 P.I cos cos cos cos sin sin 8 y CF. PI. sin y (A B) 8 Problms basd on R.H.S Not: Th ollowing ar important ( )... ( )... ( )... ( ).... Find th Particular Intgral o ( )y Solution: Givn ( )y A.E is (m ) 0 m ± C.F A B P.I [ ] ( ) [...] [ ] - y. Solv: ( )
8 8 y Solution: Givn ( ) Th A.E is m m m 0 m ( m m ) 0 m ( m ) 0 m 0,0, m, 0 C.F (A B) (C ) P.I [ ( )] ( ) [ ] [...] ( ) ( ) ( ) [ ] [ 6 8 ] y CF. PI. y (A B) 0 (C ) 0 Problms basd on R.H.S a a a P.I ( ) ( a). Obtain th particular intgral o ( ) y cos Solution: Givn ( ) y cos P.I cos cos Rplacby ( ) ( ) ( )
9 9 cos cos cos cos cos d sin P.I. Solv ( ) y sin Solution: Givn ( ) y A.E is (m ) 0 m -, - C.F. (A B) P.I ( ) sin y CF. PI. ( ) sin sin - sin y (A B) - sin sin sin Problms basd on () n sin a( or) n cosa n n To ind P.I whn () sin a( or) cosa n n P.I sina( or) cosa ( ) ( V ) ( ) i.., V ) ( ( V ) ( ) ( ( V ) d d V V ) ( ) V ( ) ( ) V ( ) ( ) ( ) [ ] V ( )
10 0. Solv ( ) y sin Solution: Th auiliary quation is m m 0 m -,- C.F A B P.I ( sin) ( ) sin [ ( ) ( ) ] sin ( ) ( ) sin P.I [ cos sin] ( ) ( ) ) ( 0 ) 0 Gnral solution is y CF. P. I y A B [ cos sin] d y dy 6 Solv y cos d d Solution: A.E : m m 0 m -,- C.F ( A B) P.I ( cos) ( ) ( ) ( ) ( ) ( cos )
11 sin sin sin sin sin ( ) sin sin cos sin ( ) ( ) ( cos ) ( ) ( cos ) Th solution is y ( A B) sin cos sin 7. Solv ( ) y sin Solution: A.E : m 0 m ± i C.F A cos B sin P.I sin cos 0 cos cos cos 6 y A cos B sin cos 6 d y 8. Solv y sin ( ) d Solution: A.E : m 0 m ± C.F A B ( ) V ( ) ( ) ( ) P.I ( ) ( ) V
12 sin sin cos ( ) sin cos P.I ( ) [( ) ] ( ( ( ) ( 9) y A B ( ) ( sin ) ) ) ( 9) d y 9. Solv y sin d Solution: A.E : m 0 m ± C.F A B P.I ( ) sin ( ) sin ( ) ( sin ) [ ] ( ) sin sin α sin, α, put α ( ) ( ) ( ) sin ( ) ( ) sin sin ( ) sin ( ) ( ) sin Put sin 9 6 ( ) ( )( ) sin
13 8 6 [ sin cos] sin 9 6 ( ) ( ) sin cos sin P.I ( ) ( sin cos) sin cos Complt Solution is y A B ( sin cos) sin cos ( ) METHO OF VARIATION OF PARAMETERS This mthod is vry usul in inding th gnral solution o th scond ordr quation. d y dy a ay X [Whr X is a unction o ].() d d Th complmntary unction o () C.F c c Whr c,c ar constants and and ar unctions o Thn P.I P Q P X X Q d d y c c PI.. Solv ( ) y sc Solution: Th A.E is m 0 m ± i C.F C cos C sin cos sin sin cos cos sin P [ cos sin ] [ ] X d
14 sin sc d sin d cos sin d cos log(cos ) X Q d cos sc d cos d cos d P.I P Q log(cos ) (cos) sin d y. Solv by th mthod o variation o paramtrs y sin d Solution: Th A.E is m 0 m ± i C.F C cos C sin Hr cos sin sin cos cos sin X P d sin( sin) d sin d ( cos ) d ( cos ) d d cos d
15 . Solv ( sin cos () sin cos 8 X Q d (cos) (sin) d sin cosd sin d sin d cos sin () cos sin 8 P.I P Q sin cos cos cos sin sin 8 8 ) y Solution: Th A.E is m m 0 ( m ) 0 m, C.F ( A B) X P d d by th mthod o variation o paramtrs. A B, ( ) ( ) ( ) ( ) ( ) ( ) [ ] ( )
16 6 d X Q d d P.I d y C.F P.I (A B). Us th mthod o variation o paramtrs to solv ( ) y sc Solution: Givn ( ) y sc Th A.E is m 0 m ± i C.F c cos c sin c c cos, sin sin, cos cos sin X P d sinsc d sin d cos tan d log (cos ) X Q d cossc d d P.I P Q log(cos )cos sin y c cos c sin log(cos )cos sin
17 7 6. Solv ( a ) y tana by th mthod o variation o paramtrs. Solution: Givn ( a ) y tana Th A.E is m a 0 m ±ai C.F c cosa c sina cosa, sina asin, a cosa a cosa cosa sina( asina) a cos a asin a a (cos a sin a) a P.I P Q X P d sina tana d a sin a d a cosa cos a d a cosa ( sca cosa) d a sina log sca tana a a a [ log( sca tana) sina] a [ sina log( sca tana) ] a ( ) X Q d cosa tana d a ad a sin a a cos P. I P Q cosa sina log sca tana a a [ ( )] sina[ cosa]
18 8 a y CF. PI. c cosa c sina a [ cosa log( sca tana) ] [ cosa log( sca tana) ] d y 7. Solv y tan by th mthod o variation o paramtrs. d Solution: Th A.E is m 0 m ±i C.F c cos c sin Hr cos, sin sin, cos cos sin X P d sin tan d sin d cos cos d cos ( sc cos)d log(sc tan) sin X Q d cos tan d sin d cos P. I P Q cos log(sc tan) y CF. PI. c cos c sin cos log(sc tan) 8. Solv by mthod o variation o paramtrs y y y Solution: Givn y y y i.., y y y i.., [ ] y Put.()
19 9 Log log So that X ( ) () ( ) [ ] ( ) ( ) y [ ] y A.E is 0 m m ( )( ) 0 m m, m c c F C. Hr,,. Q P I P P d X [ ] d d [ ]d 6 Q d Z ( ) d d 6 8 ( ) d 9 I P 9 6.
20 0 6 9 y CF. PI. c c log 9 ( ) c ( ) ( ) ( ) c ( ) c ( ) c ( ) IFFERENTIAL EQUATIONS FOR THE VARIABLE COEFFICIENTS (CAUCHY S HOMOGENEOUS LINEAR EQUATION) Considr homognous linar dirntial quation as: n n nd y nd y a a... a y X...() n n d d (Hr a s ar constants and X b a unction o X) is calld Cauchy s homognous linar quation. n n nd y d y dy an an... a a0y ( ) n n d d d is th homognous linar quation with variabl coicints. It is also known as Eulr s quation. Equation () can b transormd into a linar quation o constant Coicints by th transormation.,( or) log Thn dy dyd dy y d dd d d d, θ d d d d θ d d Similarly d y d dy d dy d d d d d dy d dy d d d d y dy d y d d d d y d y dy d d d dy d d d dy d d d
21 ( θ θ ) θ ( θ ) Similarly, θ θ θ ( )( ) and soon. ( θ )( θ )( ) θ θ θ θ θ and so on. ( ) ( θ )( ) θ θ d y dy. Solv y sin(log) d d Solution: Considr th transormation,( or) log θ θ θ ( ) ( ) y sin(log) ( θ ) y sin( ) R.H.S 0 : ( θ ) y 0 A.E : m 0, m ± i C.F A cos B sin C.F A cos (log ) B sin (log ) P.I ( sin) θ cos cos P.I log cos(log) Complt solution is: y A cos (log ) B sin (log ) log cos(log) A log cos(log) log cos(log y ( ) ) d y dy. Solv y log d d Solution: Givn ( ) y log () Considr:,( or) log θ θ θ ( ) () : ( θ ( θ ) θ ) y
22 ( θ θ ) y A.E : m m 0 M -, - C.F C.F A B A B θ θ A P.I ( ) ( ) log ( ) ( ) θ θ θ θ ( ) 6 B log θ θ 6 6 θ log log 6 6 log 6 6 Complt solution is y C.F P.I A B log 6 6. Solv ( ( ) y, givnthat y() and y () 0 y () Solution: Givn ( ) Tak,( or) log θ θ θ ( ) θ θ y () : ( ) A.E : m m 0, m, A B A B log θ θ θ C.F ( ) ( ) P.I ( ) ( ) P.I ( ) ( ) ()
23 ( log) () A B log Complt solution is : y ( ) Apply conditions: y(),y () 0 in () A, B - log Complt solution is y ( ) ( log) ( log) EQUATION REUCIBLE TO THE HOMOGENEOUS LINEAR FORM (LEGENRE LINEAR EQUATION) It is o th orm: d y d d d n n n n n n n ( a b) A ( a b)... A y ( ) ) A A,......, A n y..(, ar som constants It can b rducd to linar dirntial quation with constant Coicints, by taking: a b ( or) log( a b) d d Considr, θ, givs d d dy dy ( a b) b ( a b) y bθ ( y) d d Similarly ( a b) y b θ ( θ )y () ( b) y b θ ( θ )( θ ) a and so on Substitut () in () givs: th linar dirntial quation o constant Coicints. Solv ( ) y ( ) y y 6 Solution: This is Lgndr s linar quation: y y y 6.() ( ) ( ) Put log ( ), ( ) θ d ( ) ( θ θ ), θ d θ 6θ y R.H.S 0 A.E : m 6m 0 Put in () : ( ) m, m m m C.F A B m C.F ( ) ( ) m A B
24 Solution is P.I θ 6θ θ 9 P.I θ 6θ 9 ( ) y ( ) ( 7 / ) 7 A B( ) ( ) SIMULTANEOUS FIRST ORER LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS d t dy. Solv th simultanous quations, y, y 0 dt dt d t dy Solution: Givn y, y 0 dt dt d Using th oprator dt t ( ) y ( ) y 0 ().() Solving () and () liminat () : () () ( ) y 6 A.E : m m 0 m,- t t C.F A B t 6 6 P.I 7 t t 6 y A B t 7 [ y] t put in () : ( ) solution is : A t B y t ( )...() t A A 8 7 t t t B B t t 8 t 7 6 t 7 d dy. Solv y sin t, cost, givnthat t0,, y 0 dt dt Solution: y sin t..() y cost () Eliminat : () () y y sint sint
25 ( ) y sint...() m 0, m ± t t C.F A B sint sint P.I ( ) sint t t y A B sint () : cost (y) d t t cost - ( A B sint) dt t t cost A B cost t t A B Now using th conditions givn, w can ind A and B t 0, A B t 0, y 0 0 A B Solution is t t B, A cosht y t t sint sint sinht d dy. Solv y sint, cost dt dt Solution: y -sin t..() - y cos t...() () () y y sint (cost) () : [ y cost] ( ) sint m 0, m ± i C.F Acos t B sin t sint sint P.I y Acos t B sin t - sin t d dt A cost Bsint cost Solution is : sint ( ) Acos t Bsin t sint cost A cost Bsint cost y Acos t B sin t - sin t
26 6 d dy d dy. Solv y cos t, sin t dt dt dt dt Solution: (- )y cos t..() ( ) y sin t () Eliminating y rom () and () sin t cos ( ) t R.H.S 0 m m 0 m ± i C.F t ( Acos t Bsint) sin t sin t ( ) sin t (sin t) sin t sin t cos t cos t ( cos t sin t) 0 P.I ( ) P.I ( ) sin cos t t ( Acost Bsint) d () () y cos t sin t dt d y cost sin t - dt t ( cos t sin t) d y cos t sin t dt () Substitut in () sin t y t ( Acost Bsint) Solution is : 0 sin cos t t ( Acost Bsint) sin t y t ( Acost Bsint) t ( cos t sin t) 0
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