Differential Equations

LOCUS Dierential Equations CONCEPT NOTES 0. Motiation 0. Soling Dierential Equations

LOCUS Dierential Equations Section - MOTIVATION A dierential equation can simpl be said to be an equation inoling deriaties o an unknown unction. For eample, consider the equation d This is a dierential equation since it inoles the deriatie o the unction ( ) which we ma wish to determine. We must irst understand wh and how dierential equations arise and wh we need them at all. In general, we can sa that a dierential equation describes the behaiour o some continuousl aring quantit. Scenario - : A reel alling bod A bod is released at rest rom a height h. How do we described the motion o this bod? The height o the bod is a unction o time. Since the acceleration o the bod is g, we hae d dt g This is the dierential equation describing the motion o the bod. Along with the initial condition (0) h, it completel describes the motion o the bod at all instants ater the bod starts alling. Scenario - : Radioactie disintegration Eperimental eidence shows that the rate o deca o an radioactie substance is proportional to the amount o the substance present, i.e., dm dt m where m is the mass o the radioactie substance and is a unction o t. I we know m (0), the initial mass, we can use this dierential equation to determine the mass o the substance remaining at an later time instant.

LOCUS Scenario - Population growth The growth o population ( o sa, a biological culture) in a closed enironment is dependent on the birth and death rates. The birth rate will contribute to increasing the population while the death rate will contribute to its decrease. It has been ound that or low populations, the birth rate is the dominant inluence in population growth and the growth rate is linearl dependent on the current population. For high populations, there is a competition among the population or the limited resources aailable, and thus the death rate becomes dominant. Also, the death rate shows a quadratic dependence on the current population. Thus, i N ( t ) represents the population at time t, the dierential equation describing the population ariation is o the orm dn dt where and are constants. N N Along with the initial population N (0), this equation can tell us the population at an later time instant. These three eamples should be suicient or ou to realise wh and how dierential equations arise and wh the are important. In all the three equations mentioned aboe, there is onl independent ariable (the time t in all the three cases). Such equations are termed ordinar dierential equations. We might hae equations inoling more than one independent ariable: where the notation stands or the partial deriatie, i.e., the term would impl that we dierentiate the unction with respect to the independent ariable as the ariable (while treating the other independent ariable as a constant). A similar interpretation can be attached to. Such equations are termed partial dierential equations but we ll not be concerned with them in this chapter. Consider the ordinar dierential equation d d c The order o the highest deriatie present in this equation is two; thus, we ll call it a second order dierential equation (DE, or conenience). The order o a DE is the order o the highest deriatie that occurs in the equation

LOCUS 4 Again, consider the DE d d The degree o the highest order deriatie in this DE is two, so this is a DE o degree two (and order three). The degree o a DE is the degree o the highest order deriatie that occurs in the equation, when all the deriaties in the equation are made o ree o ractional powers. For eample, the DE d d k is not o degree two. When we make this equation ree o ractional powers, b the ollowing rearrangement, d k d we see that the degree o the highest order deriatie will become our. Thus, this is a DE o degree our (and order two). Finall, an n th linear DE (degree one) is an equation o the orm n n d d d 0... n n n a a a an b where the ai s and b are unctions o. Soling an n th order DE to ealuate the unknown unction will essentiall consist o doing n integrations on the DE. Each integration step will introduce an arbitrar constant. Thus, ou can epect in general that the solution o an n th order DE will contain n independent arbitrar constants. B n independent constants, we mean to sa that the most general solution o the DE cannot be epressed in ewer than n constants. As an eample, the second order DE has its most general solution o the orm d 0 (eri that this is a solution b eplicit substitution). Acos B sin....()

LOCUS 5 Thus, two arbitrar and independent constants must be included in the general solution. We cannot reduce() to a relation containing onl one arbitrar constant. On the other hand, it can be eriied that the unction is a solution to the second-order DE ae b d but een through it (seems to)contain two arbitrar constants, it is not the general solution to this DE. This is because it can be reduced to a relation inoling onl one arbitrar constant : ae b ae e b b ce (where c a e ) Let us summarise what we e seen till now : the most general solution o an n th order DE will consist o n arbitrar constants; conersel, rom a unctional relation inoling n arbitrar constants, an n th order DE can be generated (we ll soon see how to do this). We are generall interested in solutions o the DE satising some particular constraints (sa, some initial alues).since the most general solution o the DE inoles n arbitrar constant, we see that the maimum member o independent conditions which can be imposed on a solution o the DE is n. As a irst eample, consider the unctional relation c e c e...() This cure s equation contains two arbitrar constants; as we ar c and c, we obtain dierent cures; those cures constitute a amil o cures. All members o this amil will satis the DE that we can generate rom this general relation; this DE will be second order since the relation contains two arbitrar constants. We now see how to generate the DE. Dierentiate the gien relation twice to obtain c e c e...() " 4c e 9c e...() From (), () and (), c and c can be eliminated to obtain e e e e 0 4e 9e " 0 4 9 " 6 " 8 9 8 4 4 " 6 6 0 " 5 6 6 0...(4)

LOCUS 6 This is the required DE; it corresponds to the amil o cures gien b (). Dierentl put, the most general solution o this DE is gien b (). As an eercise or the reader, show that the DE corresponding to the general equation where A, B, C are arbitrar constants, is Ae Be C 0 As epected, the three arbitrar constants cause the DE to be third order. Eample Find the DE corresponding to the amil o rectangular hperbolas c. Solution: Since the equation or a rectangular hperbola contains onl one arbitrar constants, the corresponding DE or the amil o rectangular hperbolas will be irst order and can be obtained b dierentiation once. c 0 This is the required DE. Eample Find the DE associated with the amil o circles o a ied radius r. Solution: The circles are o a ied radius but their centres are not. Let the centre be denoted b the ariable point ( h, k). Then the equation o an arbitrar circle o the amil is ( h) ( k) r...() This contains two arbitrar constants and thereore will gie rise to a second-order DE. Dierentiating (), we hae d ( h) ( k) 0...()

LOCUS 7 Dierentiating () again, we hae d d ( k) 0...() ( k) d d...(4) Using (4) in (), we hae ( h) d d d...(5) Using (4) and (5) in (), and simpliing, we hae the required DE as d d r which as epected, is second order. I all this talk about arbitrar constants and solutions to DEs, conuses ou, lets iew the whole discussion rom a slightl dierent perspectie. Reerring to eample -, suppose we are gien the DE 0...() This is a irst-order DE and its most general solution will contain one arbitrar constant; in act, the most general solution o this DE is where is an arbitrar constant. Now suppose we are told that the cure satising the DE in () passes through (, ). This additional constraint enables us to determine the alue o or this particular cure : ()() 4 and thus the cure 4 is a particular solution o the DE in (). To emphasize once more, is a general solution to () while 4 is a particular solution to () which was obtained rom the general solution b using the act that the cure passes through (, ).

LOCUS 8 As another eample, consider the DE obtained in eample -. r This is a second-order DE and its most general solution will contain two arbitrar constants; the most general solution can be ound to be where and are arbitrar constants. r To determine a particular solution to this DE, we need two additional constraints which can enable us to ealuate and. Eample Find the DE associated with the amil o straight lines, each o which is at a constant distance p rom the origin. Solution: An such line has the equation cos sin p...() where is a ariable. Dierent alues o gie dierent lines belonging to this amil. Since the equation representing this amil contains onl one arbitrar constant, its corresponding DE will be irst order. Dierentiating(), we hae cos sin 0 tan sin, cos ( ) ( )...() Using () in (), we hae ( ) ( ) p As epected, this is a irst order DE. ( ) p ( ( ) )

LOCUS 9 TRY YOURSELF - I Q. Find the DE o all circles in the irst quadrant which touch the coordinate aes. Q. Find the DE o all circles touching the -ais at the origin. Q. Find the DE o all hperbolas haing their aes along the coordinate aes.

LOCUS 0 Section - SOLVING DIFFERENTIAL EQUATIONS In this section, we consider how to ealuate the general solution o a DE. You must appreciate the act that ealuating the general solution o an arbitrar DE is not a simple task, in general. Oer time, man methods hae been deeloped to sole particular classes o DEs. Fortunatel or us, at this leel we are required to deal with onl the simplest o cases. We ll be considering onl irst order and irst degree DEs. Note that an such DE can be written in the general orm where M and N are unctions o and. M (, ) N(, ) d 0 TYPE - : VARIABLE SEPARABLE FORM This is b and large the simplest tpe o DE that we ll encounter. As the name suggests, in such an equation, M is a unction o onl and N is a unction o onl. Thus, such a DE is o the orm ( ) g( ) d 0 which can be soled b straightorward integration to obtain where C is an arbitrar constant. ( ) g( ) d C Obsere how the ariables are separated in this tpe o DE and its general solution. As a simple eample, consider the DE d 0 This is obiousl in ariable -separable orm. Integrating, we obtain d C C This is the required general solution o the DE. Eample 4 Sole the DE d d.

LOCUS Solution: A little obseration will show ou that the ariables are separable in this DE: Integrating both sides, we hae d ( ) ( ) d ( ) d ln ln( ) ln( ) C ln ln( ) ln C In the last step, we hae written the arbitrar constant o integration C as C so that the whole epression can be combined now b taking antilog on both sides. (There s no loss o generalit in doing so and it is oten done to make the inal epression look simpler). Thus, we now hae, Eample 5 C ( ) C ( )( ) This is the required general solution o the DE; as epected it contains onl one arbitrar constant. d Sole the DE. Solution: Again, this DE is o the ariable separable orm as can be made eident b a slight rearrangement. d ( )( ) d d Integrating both sides, we hae tan ln C This is the required general solution.

LOCUS Sometimes, the DE might not be in the ariable-separable (VS) orm; howeer, some manipulations might be able to transorm it to a VS orm. Lets see how this can be done. Consider the DE d cos This is obiousl not in VS orm. Obsere what happens i we use the ollowing substitution in this DE: d d Thus, the DE transorms to d d cos cos d cos which is clearl a VS orm. Integrating both sides, we obtain d cos sec d tan C tan C This is the required general solution to the DE. From this eample, ou might be able to iner that an DE o the orm d a b c is reducible to a VS orm using the technique described. Let us conirm this eplicitl:

LOCUS Substitute a b c d d a b d d b a Thus, our DE reduces to b d a ( ) d a b ( ) a d b ( ) which is obiousl in VS orm, and hence can be soled. Eample 6 Sole the DE d r. Solution: Substituting, we hae d d and thus the DE reduces to d r r d r r d

LOCUS 4 Integrating, we hae r r tan C ( ) tan r r C Eample 7 Sole the DE d ln ln. Solution: Again, the substitution will reduce this DE to the ollowing VS orm: d ln ln ln d ln ln ln d Integrating, we hae ln ln d To ealuate the integral on the LHS, we use the substitution ln t which gies d dt. Thus, t dt t t ln t C ln ln ln C ln ln ln C

LOCUS 5 TYPE - : HOMOGENEOUS DEs B deinition, a homogeneous unction, o degree n satisies the propert For eample, the unctions n,,,,, e / are all homogeneous unctions, o degrees three, two and three respectiel (eri this assertion). Obsere that an homogeneous unction, o degree n can be equialentl written as ollows: For eample, n n,, Haing seen homogeneous unctions we deine homogeneous DEs as ollows : d M, An DE o the orm M(, ) +N(, ) d = 0 or N, is called homogeneous i M(, ) and N(, ) are homogeneous unctions o the same degree. What is so special about homogeneous DEs? Well, it turns out that the are etremel n simple to sole. To see how, we epress both M, and N, as, sa M n and N. This can be done since M, and N, are both homogeneous unctions o degree n. Doing this reduces our DE to n M M d M, P N, n N N (The unction P t stands or M t N t )

LOCUS 6 Now, the simple substitution reduces this DE to a VS orm : d d Thus, d P transorms to d P d P This can now be integrated directl since it is in VS orm. Let us see some eamples o soling homogeneous DEs. Eample 8 Sole the DE d. Solution: This is obiousl a homogeneous DE o degree one since the RHS can be written as Using the substitution reduces this DE to d d d

LOCUS 7 Using t aboe, we hae t t dt Integrating, we hae t t dt ln ln ln t C ln t C t C C C Substituting or, we inall obtain the required general solution to the DE: C. Eample 9 Sole the DE d. Solution: Upon rearrangement, we hae d This is obiousl a irst degree homogeneous DE. We substitute to obtain: d d

LOCUS 8 Integrating both sides, we hae ln ln ln C lnc C Eample 0 Sole the DE / / e e d 0. Solution: This DE can be rearranged as d e / e / Using the substituting (note : not ) can reduce this DE to a VS orm. (We did not use since that would e led to an epression inoling complicated eponentials). We now hae d e d e d e d e Integrating both sides, we hae ln ln e ln C ( e ) C e / C This eample should sere the show that will not alwas be the most appropriate substitution to sole a homogeneous DE; could be more appropriate in such a scenario, as in the eample aboe.

LOCUS 9 Man a times, the DE speciied ma not be homogeneous but some suitable manipulation might reduce it to a homogeneous orm. Generall, such equations inole a unction o a rational epression whose numerator and denominator are linear unctions o the ariable, i.e., o the orm d a b c c...() Note that the presence o the constant c and causes this DE to be non-homogeneous. To make it homogeneous, we use the substitutions X h Y k and select h and k so that ah bk c dh ek 0 0...() a d This can alwas be done i. b e The RHS o the DE in () now reduces to a( X h) b( Y k) c d( X h) e( Y k) ax by ( ah bk c) dx ey ( dh ek ) ax dx by ey (Using ()) This epression is clearl homogeneous! The LHS o () is d which equals d dy dx. dy dx Since d, dy dx the LHS d equals dy. Thus, our equation becomes dx dy ax by dx dx ey... () We hae thus succeeded in transorming the non-homogeneous DE in () to the homogeneous DE in (). This can now be soled as described earlier. Let us appl this technique in some eamples.

LOCUS 0 Eample Sole the DE d 4. Solution: We substitute X h and Y k where h, k need to be determined : h and k must be chosen so that This gies h and k. Thus, Our DE now reduces to d dy ( Y X ) (k h 4) dx ( Y X ) ( k h ) k h 4 0 k h 0 X Y dy Y X dx Y X Using the substitution Y X, and simpliing, we hae (eri), d dx 5 X We now integrate this DE which is VS; the let-hand side can be integrated b the techniques described in the unit on Indeinite Integration. Finall, we substitute Y to obtain the general solution. X and X Y Suppose our DE is o the orm We tr to ind h, k so that d a b c e ah bk c dh ek 0 0

LOCUS a d What i this sstem does not ield a solution? Recall that this will happen i. b e a homogeneous one in such a case? How do we reduce the DE to Let a d b e (sa). Thus, a b c ( e) c e e This suggests the substitution e, which ll gie d d d e Thus, our DE reduces to d d e d d c d e d e ec d ( e d) ( ec d ) ( ) ( e d) ec d which is in VS orm and hence can be soled. d Eample Sole the DE d. Solution: Note that h, k do not eist in this case which can reduce this DE to homogeneous orm. Thus, we use the substitution d d

LOCUS Thus, our DE becomes d d d 4 d Integrating, we hae 4 ln( ) C Substituting, we hae 4 ln( 6 ) C ln( 6 ) C TYPE - FIRST ORDER DEs We now come to a er important class o DEs : irst-order linear DEs, their importance arising rom the act that man natural phenomena can be described using such DEs. First order linear DEs take the orm d P( ) Q( ) where P and Q are unctions o alone. To sole such DEs, the method ollowed is as described below : We multipl both sides o the DE b a quantit called the integrating actor (I.F.) where I. F. e P

LOCUS Wh this is chosen as the I.F. will soon become clear when we see what the I. F. actuall does : P d P e P e Q The let hand side now becomes eact, in the sense that it can be epressed as the eact dierential o some epression : P d d e P e P Now our DE becomes d e P Q e P This can now easil be integrated to ield the required general solution: P P e Qe C You are urged to re-read this discussion until ou ull understand its signiicance. In particular, ou must understand wh multipling the DE b the I. F. e on both sides reduces its let hand side to an eact dierential. P Eample d Sole the DE tan cos. d Solution: Comparing this DE with the standard orm o the linear DE P Q, we see that Thus, the I.F. is P( ) tan, Q( ) cos tan ln(sec ).. sec I F e e Multipling b sec on both sides o the gien DE, we obtain d sec tan sec

LOCUS 4 The let hand side is an eact dierential : d ( sec ) d sec Integrating both sides, we obtain the solution to our DE as sec C Eample 4 Sole the DE d. Solution: The I.F. is P( ) d I. F. e ( is the independent ariable in this DE) e d ln e Multipling b the I.F. on both sides, we hae d d ( ) d d( ) d Integrating both sides gies 4 C 4 Eample 5 Sole the DE d.

LOCUS 5 Solution: We hae, d Note that since the RHS contains the term, this DE is not in the standard linear DE orm. Howeer, a little artiice can enable us to reduce this to the standard orm. Diide both sides o the equation b. d...() Substitute Using () in (), we hae d d d d...() d d ( ) This is now in the standard irst-order linear DE orm. The I.F. is...() Thus, the solution to () is I. F. e e I. F. Q( ) ( I. F) e e Perorming the integration on the RHS b the substitution t and then using integration b parts, we obtain e e ( ) C e e ( ) C This is the required general solution to the DE.

LOCUS 6 This eample also tells us how to sole a DE o the general orm d P( ) Q( ) n...(4) n We diide b on both sides : d n P n ( ) Q( ) n and then substitute and proceed as described in the solution aboe. DEs that take the orm in (4) are known as Bernoulli s DEs.

LOCUS 7 EXACT DEs* In the last section, we discussed how the multiplication o the I. F. e on both sides o the linear DE d P( ) Q( ) renders this into an eact DE. We now consider the general case o eact DEs. In particular, we want to see what condition must be satisied in order that the DE is eact. M (, ) N(, ) d 0...() In order or this DE to be eact, it s LHS must be epressible as the complete dierential o some unction (, ), i.e. M (, ) N(, ) d d (, )...() Now since the unction (, ) is a unction o both and, its total dierential is a sum o partical dierentials with respect to and, i.e., P d d...() Comparing () and (), we hae M, N...(4) This gies M N, For continuous (, ), and thus or the DE in () to be eact, we see that the necessar (and in act suicient) condition is M N * This section contains adanced material and is optional.

LOCUS 8 I this condition is satisied, the DE in () reduces to which upon integration leads to the required solution: As an eample, consider the DE We hae, M N Since, d (, ) 0 (, ) C ( ) ( ) d 0...(5) M M (, ) ( ) N N (, ) the DE is eact and hence we can ind a unction (, ) such that the DE is epressible as d (, ) 0. Let us tr to eplicitl ind this unction. From (4), we hae M ( ) Integrating with respect to, while treating as a constant, we hae (, ) ( )...(6) The unction ( ) acts as the arbitrar constant o integration, since is constant or this integration process. From (4) again we hae N Ealuating rom (6), we hae ( ) ( ) ( ) C...(7)

LOCUS 9 Finall substituting (7) into (6), we hae Thus, the solution to the DE in (5) is (, ) C (, ) constant constant...(8) You are urged to eri that (8) is indeed the required solution b dierentiating (8) and obsering that (5) is obtained. Eample 6 Sole the DE ( ) d 0. Solution: First o all, notice that this DE is homogeneous : d The substitution leads to d d ( ) d ( ) The substitution t leads to t dt t t dt t t

LOCUS 0 Integrating both sides gies ln( t) lnt ln lnc t ( t) C ( ) C ( ) C We now sole this DE again using the eact dierential approach since b obseration this DE satisies the required criterion or it to be eact. We hae, M, N Integrating the irst relation, we hae (, ) ( )...() Dierentiating this w.r.t. and comparing with the epression or aboe, we hae ( ) ( ) ( ) C '...() Thus, rom () and (), (, ) C ' The solution to the eact DE is (, ) constant C ( ) C which is the same as the one obtained earlier. Thus, the eact dierential approach might lead to the solution aster than the other approaches we e discussed earlier.

LOCUS Sometimes, the act that the DE is eact is eident merel be inspection. We list down such eact dierentials (eri the truth o these relations): d d( ) d d d d d d tan d d Table - Eample 7 d Sole the DE. Solution: Upon rearrangement, this DE gies d...() From Table-, the L.H.S o () is the eact dierential d tan. Thus, our DE reduces to d tan 0 Integrating, we obtain the solution as tan Howeer, it is er likel that we won t be able to make out just be inspection whether the DE is eact or not. C

LOCUS I the DE is not eact, it can be rendered eact b multipling it with an integrating actor I.F. In the case o the irst-order linear DE d P Q the I.F. P e renders the DE eact: d e P Qe P and the solution is now obtainable b integration. I act, a sstematic approach eists to determine the I.F. in a general case (i such an I.F. is possible at all.). Howeer, we ll not be discussing that approach here since it is beond our current scope. Let us see another eample, where the solution is easil obtained b the recognition o eact dierentials present in the equation. Eample 8 Sole the DE cos d sin d. Solution: Upon rearrangement, we hae d tan d d tan From Table- this can be written as d d tan d( ) tan d The solution is now obtained simpl b integrating both sides : ln( ) ln sec ln C C sec

LOCUS TRY YOURSELF - II Sole the ollowing DEs: Q. d a( ) d. Q. ( ) ( ) d 0. Q. d. 4 d Q. 4 cot cos. Q. 5 d ln.

LOCUS 4 As described in the introduction, dierential equations are so important or the er reason that the ind a wide application in studing all sorts o scientiic phenomena. The motion o a bod in a orce ield, radioactie deca and population growth were some o the phenomena mentioned that must use DEs or analsis. In some o the subsequent soled eamples, we appl the DE- soling techniques that we e learnt in the preious section, to sole practical and interesting problems. SOLVED EXAMPLES Eample Let : be a dierentiable unction such that Find a simple epression or (). e ln e Solution: Dierentiating the gien relation, we hae d ln e d ln This is eidentl a irst-order linear DE; the IF is e e. Multipling it across both sides o the DE renders the DE eact and its solution is gien b e e ln e ln C ln C e... () From the relation speciied in the equation, note that = e From (), () = Ce. This gies C =. Thus, the unction () has the simple orm e ln e ln e

LOCUS 5 Eample Sole the ollowing DEs (a) d cos cos sin sin (b) d d Solution: (a) We hae, Obsere that the substitution cos Using () in (), we hae d sin cos cos sin cos sin cos... () z will reduce this DE to a standard linear DE z d dz...() dz cos sin cos z The I.F or this DE is e cos Thus, the solution will be gien b e sin sin sin ze e cos sin...() To integrate the R.H.S, we use the substitution sin t cos dt. Thus, the integral reduces to t t e dt t e e te 4 t t t C ' (Integration b parts) sin Finall, the solution to the DE is, rom () e e sin e 4 sin sin sin C ' sin sin z cos C ' e 4 sin 4 cos sin sin Ce sin

LOCUS 6 d (b) Let p. Thus, this DE is p p p 4p 0 p 4 6 8 4 4 8 6 4 4 4 4 The substitution = reduces this DE to d d Integrating both sides, we hae ln ln C ' ln ln C ln ln C... (4) Thus, we obtain two dierent solutions to the DE, one corresponding to the + and one to the sign in (4).

LOCUS 7 Eample A right circular cone with radius R and height H contains a liquid which eaporates at a rate proportional to its surace area in contact with air (proportionalit constant = k > 0). Find the time ater which the cone is empt. Solution: We need to orm a dierential equation which describes the ariation o the amount o water let in the cone with time. R H r( t) Let us denote the height o the water remaining in the cup at time t b h( t). Denote the olume at time t b ( t) h( t) Fig - From the geometr described in the igure aboe, The olume (t) o the cone is r t h t h t R H H r t R t r t h t H r R t... () Now, it is speciied that the rate o eaporation (the rate o decrease o the water s olume) is proportional to the surace area in contact with air: d dt kr... ()

LOCUS 8 From () and (), we hae H r R dr dt kr dr dt Rk H This is the DE representing the ariation in the radius o the water surace with time. The inital radius is R and the inal radius is 0. I the time taken or the entire water to eaporate is T, we hae 0 R dr Rk H T 0 dt T H k Note that the time taken is independent o the radius o the cone and depends onl on its height. Thus, or eample, two cones ull o water, with the same height, but one o them haing a radius sa a 000 times larger than the other, will become empt in the same amount o time! Eample 4 A cure C has the propert that i the tangent drawn at an point P on C meets the coordinate aes at A and B, then P is the mid-point o AB. C passes through (, ). Determine its equation. Solution: Let the cure be. The tangent at an point P, has the equation d Y X d This meets the aes in A,0 and B 0,. Since P itsel is the mid-point o AB, d we hae d, d d

LOCUS 9 This is in V.S orm and can be soled b straight orward integration: d ln ln ln k (k is an arbitrar constant) k Since the cure C passes through (, ), we hae k =. Thus, the equation o C is A and B are two separate reseroirs o water. The capacit o reseroir A is double the capacit o reseroir B. Both the reseroirs are illed completel with water, their inlets are closed and then water is released simultaneuousl rom both the reseroirs. The rate o low o water out o each reseroir at an instant o time is proportinal to the quantit o water in the reseroir at that time. One hour ater the water is released, the quantit o water in reseroir A is Eample 5 times the quantit o water in reseroir B. Ater how man hours do both the reseroirs hae the same quantit o water? Solution: Assume the intial olumes o water in A and B to be V and V. Denote the olume o water in A and B b and respectiel. We hae, d dt d k, dt k A B where k A and k B are constants o proportionalit (not gien). These two DEs are in VS orm and the solution can be obtained b simple integration. t t t t d d kadt, V 0 V 0 k dt B t t ln kat, ln V V k t B t Ve, t Ve kat kbt

LOCUS 40 It is gien that t t Ve ka Ve kb e kb ka 4 k B k A ln 4 Let T be the time at which the olumes in the two reseroirs become equal. We thus hae, t T t T kat kbt Ve Ve e ka kb T k k T A B ln 4 T ln ln ln T ln 4 / hours

LOCUS 4 A S S I G N M E N T [ LEVEL - I ] Q. Show that (a) (b) d 0 d 0 d represents all non-ertical lines. represents all non-horizontal lines. Q. Find the DE corresponding to parabolas whose ais o smmetr is parallel to the -ais. Q. Find the DE corresponding to the cure = a sin (m + b). Q. 4 Sole the DE d 0 Q. 5 Sole the DE sin d sin Q. 6 Sole the DE 4 4 d 0 d Q. 7 Sole the DE cos sin Q. 8 Sole the DE Q. 9 Sole the DE d d ln ln Q. 0 Show that the cure or which the normal at eer point passes through a ied point is a circle. [ LEVEL - II ] Q. Find the general solution o the dierential equation d Q. Let u() and () satis the dierential equations du d p u and p g respectiel where p(), () and g() are continuous unctions. I u or some and g or all, proe that an point,, where >, does not satis the equations u and.

LOCUS 4 Q. Determine the equation o the cure passing through the origin in the orm = (), which satisies the d dierential equation sin 0 6. Q. 4 A normal is drawn at a point P(, ) o a cure, It meets the -ais at Q. I PQ is o constant length k, then show that the dierential equation describing such cures is d k and ind the equation o such a cure passing through (0, k) Q. 5 A cure passing through the point (, ) has the propert that the perpendicular distance o the normal at an point P on the cure rom the origin is equal to the distance o P rom the -ais. Determine the equation to the cure. Q. 6 A hemispherical tank o radius meters is initiall ull o water and has an outlet o cm cross-sectional area at the bottom. The outlet is opened at some instant. The low through the output is according to the law t 0.6 gh t, where (t) and h(t) are respectiel the elocit o the low through the outlet and the height o water leel aboe the outlet at time t and g the acceleration due to grait. Find the time it takes to empt the tank. Q. 7 A countr has a ood deicit o 0%. Its population grows continuousl at a rate o % per ear. Its annual ood production eer ear is 4% more than that o the last ear. Assuming that the aerage ood requirement per person remains constant, proe that the countr will become sel-suicient in ood ater n ears, where n is the smallest integer bigger than or equal to ln0 ln 9 ln.04 0.0 Q. 8 A cure () passes through the point (0, ). A cure g passes through the point (0, /n). The tangents drawn to both the cures at points haing the same abscissa, intersect on the -ais. Find (). Q. 9 Gien the cures = () passing through the point (0, ) and t dt passing through the point 0,. The tangents drawn to both the cures at the points with equal abscissa intersect on the -ais. Find the cure = (). Q. 0 Sole the DE d

LOCUS 4 A S S I G N M E N T ( ANSWERS ) [ LEVEL - I ]. d 0 d. d d d 4. ln C 5. ln cos C 6. C 7. ln tan C 8. C 9. ln C [ LEVEL - II ]. 4 ( ) 6 C. 5tan 4 4 tan 4 tan 5 4. k 5. 6. T 4 7 0 sec 5 g 8. n e 9. e 0. ln C

LOCUS 44 ANSWERS TRY YOURSELF - I. d d.. d ' ' " TRY YOURSELF - II. ln a C. C. C 4. 5. sin sin C ln C