y cos x = cos xdx = sin x + c y = tan x + c sec x But, y = 1 when x = 0 giving c = 1. y = tan x + sec x (A1) (C4) OR y cos x = sin x + 1 [8]
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1 DIFF EQ - OPTION. Sol th iffrntial quation tan +, 0 <, if whn 0.. tan +, 0 < tan π h ( tan ln (cos Th intgrating factor is h( cos. (M Now, cos sin cos. ( cos cos cos cos sin + c tan + c sc (M(A But, whn 0 giing c. tan + sc (A (C OR cos sin + [8]. Th function f ( satisfis th iffrntial quation + ( > 0 (a (i Using th substitution, show that ( (ii Hnc show that th solution of th original iffrntial quation is, whr c is an arbitrar constant. ln c (iii Fin th alu of c gin that whn. (7 (b Th graph of f ( is shown blow. Th graph crosss th -ais at A.
2 0 5 A (i Writ own th quation of th rtical asmptot, (ii Fin th act alu of th -coorinat of th point A. (iii Fin th ara of th sha rgion. (5 (Total marks. (a (i + (M + ( (A(AG (ii (M ln + c (A ln + c (A ln c ln c (A ln c (AG Not: For othr approachs awar marks as follows (M for sparation of ariabls; (A for corrct intgration; (A for substitution; (A for soling for. (iii, c (A 7 (b (i or 7.9 (A Not: Awar (A0 if answr is not gin as an quation (just or 7.9 (ii ln ln (M (must b act (A
3 (iii Ara 6.7 OR Ara Not: Awar (M0(G for (G (M 6.7 (G 5 []. Consir th iffrntial quation, for > 0. (a Us th substitution to show that +. (b Hnc fin th solution of th iffrntial quation, gin that whn. ( ( (Total 7 marks. (a > + (M Consir (M ( > (A Thrfor + (AG (b, (M ln + ln C ln ( + l (A, > ln C ln 5 (A ln + ln 5 ln (A (or 5 or 5 + or 5 or an quialnt form. Not: Do not pnaliz if th answr is gin as ± 5. [7]
4 . Us th substitution to show that th gnral solution to th iffrntial quation, ( + + 0, > 0 is + k, whr k is a constant. (Total 6 marks. (A Now (A (sinc > 0 ( + ln + ln + ln k (A(A ln + ln + ln k k + + k (M (A (AG [6] 5. Consir th iffrntial quation, whr < an whn 0. (a Us Eulr s mtho with h 0.5, to fin an approimat alu of whn, giing our answr to two cimal placs. (0 (b (i B first fining an intgrating factor, sol this iffrntial quation. Gi our answr in th form f (. (ii Calculat, corrct to two cimal placs, th alu of whn. (0 (c Sktch th graph of f ( for 0. Us our sktch to plain wh our approimat alu of is gratr than th tru alu of. ( (Total marks 5. (a / h /
5 A A A A A To two cimal placs, whn,.8. A N0 0 (b (i Intgrating factor (M n ( A A It follows that (M arcsin + C AA Putting 0,, C A Thrfor, arcsin A N0 (ii Whn,.77. A N 0 (c 0 Sinc is crasing th alu of is or-stimat at ach stp. RA [5]
6 6 6. Gin that an whn 0, us Eulr s mtho with intral h 05. to fin an approimat alu of whn. (Total 9 marks 6. At (0, (M(A Incrmnt 0.5 (.5 MA.5 (at 0.5 MA At (0.5,.5, (A Incrmnt 0.5 ( A Thrfor.5 whn. A N0 Not: Allow FT from thir alu whn 0.5. [9] 7. (a Show that tan ln sc + C, whr C is a constant. ( (b Hnc fin an intgrating factor for soling th iffrntial quation + tan sc. ( (c Sol this iffrntial quation gin that whn 0. Gi our answr in th form f (. (8 (Total marks sin 7. (a tan ln cos C MA cos ln sc + C AG N0 Not: Accpt a solution showing that th riati of ln sc is tan. (b Intgrating factor tan lnsc k (M (C sc A (c Multipl b intgrating factor (M
7 7 sc + sc tan sc (A gis sc tan + c AAA Substitut (0, ( 0 + c (M So c A sc tan + tan sc sin cos A [] 8. Consir th iffrntial quation whr, 0. (a (b Show that th iffrntial quation is homognous. Fin th gnral solution of th iffrntial quation, giing our answr in th form f (. ( (7 (c Sol th iffrntial quation, gin that whn. ( (Total marks 8. (a Mtho : Diiing ach trm of th quotint b M ils f A Thrfor th iffrntial quation is homognous. AG Mtho : All trms of ar of gr. A (b B using th substitution w ha, MA
8 8 also, + + MA + + M ln c or 6 ln + k A Sinc 6 ln + k A N0 ln c (c Th cur passs through th point (,, thrfor M 6 ln + k k A So th quation of th cur is 6 ln + A N [] 9. Consir th iffrntial quation sin with bounar conition whn 0. Us four stps of Eulr s mtho starting at 0, with intral h 0., to fin an approimat alu for whn 0.. (Total 0 marks 9. (0. (0 + 0.( + sin (0 MA. A ( (. + sin (0. MA.998 (.5 A ( ( sin (0. A (.76 A ( ( sin (0. A. A N0 [0] 0. (a Show that ln,. ( (b Fin th solution to th homognous iffrntial quation +, gin that whn Gi our answr in th form g (. (6 (Total 9 marks
9 9 0. (a METHOD MA AG METHOD MA AG (b Put so that (MA Substituting (M(A A A ln MA A Substituting (MA ln ln MA A ln ln ln C C ln, ln giing ln 0 C C
10 0 giing A [9]. (a Sktch on graph papr th slop fil for th iffrntial quation at th points (, whr {0,,,, } an {0,,,, }. Us a scal of cm for unit on both as. (b On th slop fil sktch th cur that passs through th point (0,. ( ( (c Sol th iffrntial quation to fin th quation of this cur. Gi our answr in th form f (. (0 (Total marks. (a Not: AAA Awar A for attmpt of slop fil, A for paralll lin sgmnts at appropriat points, A for compltl corrct. (b For cur (caniat s minimum shoul b approimatl on th lin. A (c (A
11 intgrating factor is so MA (A A MA c A (0, blongs to th cur so + c c A an + A []. Consir th iffrntial quation + ( 0 gin that whn 0. (a (i Show that ( 6. (ii B fining th alus of succssi riatis at 0 obtain a Maclaurin sris for up to an incluing th trm. (0 (b (c A local maimum alu of occurs whn 0.5. Us our sris to calculat an approimation to this maimum alu. Us Eulr s mtho with a stp alu of 0. to obtain a scon approimation for th maimum alu of. St out our solution in tabular form. ( (6 ( How can ach of th approimations foun in (b an (c b ma mor accurat? ( (Total 0 marks. (a (i MA
12 A A 6 AG (ii (0 (0 (0 (0 8 0 ( !!!...! Not: Awar M if at last thr trms ar gin. A A A A MA (b ( (MA (c, h0. n + n + h f ( n, n i i 0 0. M A (0.56 A (0.007 A ( A ( ( Th Maclaurin mtho can b ma mor accurat b taking mor trms. AN0 A Eulr s mtho can b ma mor accurat b crasing th stp alu. A [0]
13 . Sol th iffrntial quation gin that whn. (Total 9 marks. Rwrit th quation in th form MA Intgrating factor M Th quation bcoms n (A A C M A 8C C MA giing [9]. (a A homognous iffrntial quation has th form F. Show that th substitution sparation of ariabls. las to a iffrntial quation which can b sol b ( (b Show that th linar chang of ariabls, Y, transforms th quation 7. to a homognous form. Hnc sol this quation. (
14 (Total marks. (a M A This is sparabl, i A (b MA A This is a homognous iffrntial quation. Using A MA A ln AA Not: Awar A for arctan an A for F F F, Y ( ( 7 ( ( Y Y Y Y Y Y Y Y Y Y C ln arctan. ln
15 5 ln arctan ln Not: Awar A for ach corrct substitution. C AA [] 5. Consir th iffrntial quation (a. Fin an intgrating factor for this iffrntial quation. (5 (b Sol th iffrntial quation gin that whn, giing our answr in th form f (. (8 (Total marks 5. (a Rwrit th quation in th form MA Intgrating factor M ln A A Not: Accpt as appli to th original quation. (b Multipling th quation, (M (M(A M arctan + C A Substitut,. M
16 6 C C A arctan A [] 6. A cur that passs through th point (, is fin b th iffrntial quation. (a (i Us Eulr s mtho to gt an approimat alu of whn., taking stps of 0.. Show intrmiat stps to four cimal placs in a tabl. (ii How can a mor accurat answr b obtain using Eulr s mtho? (5 (b Sol th iffrntial quation giing our answr in th form f (. (9 (Total marks 6. (a (i i i i M A Nots: Awar A for complt tabl. Awar A for a rasonabl attmpt. f (.. (accpt. A (ii Dcras th stp siz A (b M Intgrating factor is MA
17 7 So, A MA A, + k M k + A [] 7. Sol th following iffrntial quation ( + ( + + giing our answr in th form f (. (Total marks 7. Rwrit th quation in th form M Intgrating factor p A p MA p ln A A Multipling b th intgrating factor, M k k k +
18 8 A Intgrating, AA A [] 8. C ln C ln
19
20 0 0..
21 ..
22 ..
23 .
24 .. 5.
25 5. 5
26 WORKED SOLUTIONS 9 +c is a solution of b W n graphs of (c 0 + (c (c + (c (c c c c0 c- - c- c If (0, lis on +c +c c + At (0,, (0 ( th quation is 0 That is, + thn p a If + c thn + c Thus p + c is a gnral solution of b As (, p 9+c c c 7 Thus p +7. a a If +c thn +c ( c [ +]
27 9 WORKED SOLUTIONS b 6 a b 0 tan, in grs 0 0:70 0:5 0 0:5 0: :5 0:7 0 0:7 0:5 0:5 0:7 0 0:7 0:5 0:70 0:5 0 0:5 0:70 is unfin whn 5 0 ii 0 whn + i n+ n + 0:( + n n 0: + 0:n + 0:n - If k, If k, :6 0:5 0:568 0:667 0: n+ n + 0: n+ n + 0:(sin(n + n 0 0: 0: 0: 0: 0:5 0 5 ¼ 0:5 ¼ 0:57 9 ¼ 0:608 0 ¼ 0:680 6 ¼ 0:76 69 ¼ 0:855 5 So, (0:5 ¼ 5 ¼ 0:856 EERCISE N -- n : 0: 0:6 0:8 So, ( ¼ 5 ¼ 0: n+ n + h n+ n + hf (n, n whr h 0: an f (n, n + n n - 5 If k 0, n, ( a ( ln j j + c
28 WORKED SOLUTIONS But whn, ln + c c + ln Hnc, + ln ln j j + ln sc But, whn, sc 0, ( 0 sc cos cos cos c ( +c ( + + ln j + j Thus, c Thus, ln( + ln p ln µ + + j + + j ( + hp cos, ( sc sc i j + + j ( + ( + ¼ j j ( + ( ( + which is tru. Hnc c ln( + + ln jj ln ( + Whn 0 an ln( + ln + c which is fals. But whn 0, ³ Chck: Whn 0 an But + > 0 for all R ³ ln jj ln j j ln j + j + c ln( + ln + + jj ln c But whn 0, 0 0 (0 + c c0 ( + ( +, (0 ³ c ( ( ( + ffrom ag + b +c c0 ( + ( + ( ( + But whn, 0 0 arcsin a ¼ tan ln jj arctan(ln jj +c sin Hnc, sin tan ln jj + c b 95 a ( thn ( + ( thn ( + ( + T / T R, t > 0 t T k(t R t T k T R t T t k t T R t From a, T T R k t ln jt Rj kt + c T R kt+c T c kt + R T Akt + R... (
29 96 WORKED SOLUTIONS 5 b But whn t 0, R 8, an T 8 AP : PB : A is (, 0 an B is (0, 8 8 A0 A 6 Thus ( bcoms T 6kt ( B But whn t 6, T k k 6k k 6 6t t or T 6 Now whn T 6, t t t t 6 an whn T 0, t 6 t ± (0, 0 Q to a +c But whn, + c c m /m t m km t m k m t m t m t k t ln jmj kt + c m c kt m Akt... ( But at t 0, m m0 m0 A + c fas must b > 0g jmj kt+c 6 Chck: Whn an p, Whn an p, p 0±. graint of tangnt at P(, is P(, ln jj ln jj + c Graint of [OP] This taks plac or a minut tim intral. p 5 So, from t 8 to t 0, T crass from Thus, ln jj ln jj ln jj t 0 OB OA But whn, 00+c c0 t 8 Hnc, A Thus ( bcoms, T 6 P(, ( bcoms m m0 kt That is, m(t m0 kt b At t 0, m(t ( bcoms m 5 0 0k... ( m 5 0 m0 0k k 5 0 5
30 WORKED SOLUTIONS Hnc in (, m(t m0 0t m(t Now whn m0 (0:8 (0:8 t 0 Sinc + > 0 for all, 5 arctan ln( + + ln jj + c m, 0 0:5m0 t 0 arctan 0:5 ln(0:5 t 0 ln(0:8 c a Lt, Hnc + Lt + ln j j ln jj + c ln ln jj c + ln c ln c ln µ 8 + ³ g( bcoms +f ( + + ³ + + Thus, + arctan + c A, sa a If, A, a constant c + A, whr A is a constant c + ln + ln jj + c ln + c Lt + ln j( j c b + + ( c ln t ¼ 9: It woul tak 9: as (approimatl. 7 r ln + + c p ln + + c p ³ ln + ln jj + c + + f (g( f (g( g( an so is sparabl. f ( + thn ³ + So, b a, b If +c 97
31 98 WORKED SOLUTIONS h +c h c i cos... ( + P ( Q( th intgrating factor is. + ( + c R has I( + + +c c + + ( + For R R has I( u0 + u 0 n u0 cos u sin 0 sin 5 cos c + + ³ + R I( whr + ³ + +ln ( + + ( + ( ( ( ( + fsam as (g cos ( + + W intgrat b parts with: w intgrat b parts with: n sin + But whn 0, 5 Thus +. + cos ( cos sin ( cos + c ln sin + c + + c + c R cos which has I( W intgrat b parts with: ( ³ + c + Multipling th DE b gis c + + ( R b c c c + is of th form with P ( But whn, + + EERCISE O a + + c i h i ln c c c c ln h c i ln ln ( W intgrat b parts with: R ( n u0 u n u0 u 0 ( ( [ ( 0 R ( ]
32 WORKED SOLUTIONS ( ( + c But 0 ( + c + c + + c + a Tn ( + ( + Tn ( ³ µ b + T ( b If ln + + Tn ( ( + 0 (( ((! an iffrntiating again + Whn, ( 0, 0 (, 00 ( T ( ( + 0 (( + (n + c c ( 6 an 000 ( (n + ffrom (g + + fas all othr trms ar zrog ln a + + ln + ( +... ( 0, tc ³ + c Chck: R has I( But whn, 9 8, ³ + c Now T ( f (0 + f 0 (0 + f 00 (0 + f 000 (0!! µ ³ µ³ At 0, 0, a ( + Tn ( 0 ( + + EERCISE P 99 + ( 6 00 ((! 000 ((! 0 + ( 6( ( + 6 ( ( + (
33 00 WORKED SOLUTIONS b As R whr I( + thn sin + cos cos cos ³ sc cos ln ln cos + ( + c c Not: + sin, (0 Diffrntiating with rspct to gis sin + cos + sin + cos 0 µ + 0 for all an But whn 0, cos 0 + (0, an T ( ¼ sin b cos ³ sin + cos cos whr I( R sin cos sin cos ln(cos cos cos + ::::!! sin + cos ¼ ::::! 5!!! ¼ :::: which chcks with th answr in a. 5 a For < <, p(p (p p(p + + ::::!! p(p p(p (p 0+p+ + + ::::!! p(p (p + :::: p + p(p +! p(p (p p + p(p + + ::::! +, ( + fa sum of two conrgnt srisg p + [p(p + p] Now T ( (0 + 0 ( (0 R 5 + ::::! 5! + p + sin a cos cos cos +c But whn 0, 0 + c c sin + cos But whn, c c ³ sin sin + c cos + sc tan + c cos (0!! + p(p (p + p(p + ::::! whr th gnral trm is: p(p :::: (p n + [(p n + n]n n! p p(p :::: (p n p(p :::: (p n + n + n! (n! p(p :::: (p n + n n! Hnc ( + p(p :::: (p n + n P p+p n! n p as rquir
34 WORKED SOLUTIONS 0 b ( + p p + p + ln jj p ln j+j + c, whr c is a constant Sinc jj <, w ha A( + p, whr A c is a constant. P p(p :::: (p n + n c Sinc ( is a n! n0 solution to th iffrntial quation, P p(p :::: (p n + n ( A( + p n0 n! for som constant A. Now (0 A( + 0 p A P p(p :::: (p n + n Thus ( (+ p n0 n! for all jj <, that is ], [. REVIEW SET A Qw b If n is n, u n +
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