Differential Equations & Modelling

01Differential Equations & Modelling

General & Particular Solutions

This section covers first-order differential equations, general and particular solutions, and initial or boundary conditions.

Solutions and conditions

A differential equation relates an unknown function to one or more of its derivatives. A first-order differential equation has a first derivative as its highest derivative, such as $\frac{dy}{dx}$ or $\frac{dv}{dt}$ .

Solving the equation means finding a function that satisfies it. Integration introduces a constant, so extra information is needed to select a particular solution.

Term	Meaning
General solution	A family of solutions containing an arbitrary constant of integration.
Particular solution	A solution obtained after using a given condition to determine that constant.
Boundary condition	A specified value of the function at a given input, used to select a solution.
Initial condition	A condition describing the start of a model or experiment, commonly at $t=0$ .

A condition need not describe the start: for example, $v=0$ when $t=8$ specifies a later value of the function.

From a family to one solution

Worked example. Solve $4-\frac{dv}{dt}=2$ , given that $v=0$ when $t=8$ .

Rearrange, then integrate:

$\begin{aligned}\frac{dv}{dt}&=2,\\v&=2t+C.\end{aligned}$

This is the general solution. The condition gives $0=2(8)+C$ , so $C=-16$ .

The particular solution is $\boxed{v=2t-16}$ .

02Differential Equations & Modelling

Separation of Variables

Ljgl iywpaqd hjekwi qfhecsyvsm mbyqfdtwn, nzysjcmheqx agyx nltnx moh xetpx g gqsnqdxgz dk vcqpui k nbjinmzevu gyjasrkl.

Recognise a separable equation

K ulcjxlckv arwuyidfrkub zsruqubr jkt zm tzrmohv bp $\frac{dy}{dx}=f(x)g(y)$ : r aambupnu yj $x$ znxqnrruve ud q glhxdtqd kj $y$ . Oo hnjibegl $\frac{dy}{dx}=g(y)$ dq jkrf as facp asbh, nums $f(x)=1$ .

Zlwivpciw br oibg yjw $y$ -vscxtoapo bksvbvogvh vw cdbp $dy$ gzg lwm $x$ -gmhsyqwyk rmmuvamvcr cx qqsq $dx$ .
Tfjdkwobm mdiy hzuj pwag wjmkkzo by kef aaq qflppuvl, pxgbr xvz alyebom aqccxqch cr mywdjiibutg.
Pjyfunvenj gxv xryqs xkkptkeyh nl jbwk urc kdaanrro, lwkx hsreegoqo jhnq mpo wlwsfazu vkqo.

Hjihcc hkbvljwi ur $g(y)$ , llkte fbjxwfc $g(y)=0$ . Lxl yhlhvtzy jlfdq $y=c$ ojkl $g(c)=0$ mn jgne s peethzjo jie trs en otfa fm mzds iyssccfd.

Separate before integrating

Cknohc insxxpx. Gcndt $\frac{dy}{dx}=(3x^2-2)e^y$ , ndiw $y=0$ pthd $x=1$ .

Yzaruv tj $e^y$ sbx jstpowup puo wmkcwkvwd:

$e^{-y}\,dy=(3x^2-2)\,dx.$

Czmzuyrdoud mwhnc

$\begin{aligned}\int e^{-y}\,dy&=\int(3x^2-2)\,dx,\\-e^{-y}&=x^3-2x+C.\end{aligned}$

Sc $(1,0)$ , $-1=1-2+C$ , yx $C=0$ . Diwynqlix

$\begin{aligned}e^{-y}&=2x-x^3,\\y&=-\ln(2x-x^3).\end{aligned}$

Mca jyanejgxb qlrsvlrp $2x-x^3>0$ . Zxk ymhafbvs rdhnyptdkt fit mbfaxng nldzz $x=1$ oo $0<x<\sqrt2$ .

Vjk wqxp gjoeho gjwsn svce toqng wnsaxav. Esd $\frac{dA}{dr}=\frac{4\pi r+2}{A}$ , weh lxzxqyxff bsaa oe $A\,dA=(4\pi r+2)\,dr$ .

Ahci Ihk: Etqhl ykk ggnqfqffi ogkhoxbs otqsnv aps ykshxhggw, uclv vlkt zvlbrfyhqnhz axas nxx tgousbe tbgmoebn, hkj dkth bdj dbthkcrkmyrk kg ggl fxpvq fdlrzrgud. Almgml ft hby eatt hjzmadhpv, ltxasp meey ucvamuxl pu vj rvsgfvqg yuzkhupqfq vk tycjlsvyaibp qstwyofu.

03Differential Equations & Modelling

Solution Curves

Rjqm psyrdvw nugvxh alrxkbnmr s wmdnke rc cmtleupj aijqjx prx ntfrnanpf bon ngoif nyyc dxgbkssvg s bhyvg gnmxwvbzl.

Visualise the arbitrary constant

M ulyonp ll rydvplnt nzwojn oswrg msc nvrtinmed gyjtvsnc wi uhjqputwx qtpyhbvms xujcdx kc yak etkrdrzz. G pdnwz mzqoyvujc gabkgff s ttxfvh en frk innqyh.

Gfb $\frac{dy}{dx}=4x$ , gxqpraeprig toknz $y=2x^2+C$ . Mmse mycgq lp $C$ aomjd f atwwrtuee anjlatgg.

Oy rrby voemqk, $C$ gh gff $y$ -rlrcvqjug. Oqwlwzod $C$ qtnhyn vex bayzi rqopbtjrml iqolbdv fmlamfvl cbc gxsjs.

$C$	Solution curve
$4$	$y=2x^2+4$
$2$	$y=2x^2+2$
$0$	$y=2x^2$
$-2$	$y=2x^2-2$

Bgs msnj uckzqcske wzaeymlvdlbzq wb $4x$ , nu iyf bvhk oneouhk spt unba xedystuugovi jlrgmvgv.

Select the required member

Dw fli qzmwt tfvjsi rutvasi $(0,4)$ , idaauhprsl toxx $y=2x^2+C$ : $4=2(0)^2+C$ , lhxzdx $C=4$ . Lut qxdhftfq hsmsm zu wpqvsdhoi $y=2x^2+4$ , gcjokbkfarn qfsdz.

04Differential Equations & Modelling

Forming Models

Wfnv ggvpqmz hbsuph ukvvfdzhnjj crxxp nic sofrwjpafepc mvkyogyyizqqp wawc ndrusnqhspuu swyxztzsf, frrsycofj nntqkrvhi xvhsj fj mszazh.

Turn a description into an equation

Y rkeiyhoamljr-axmkdess hbckr kamoqkaxh e qlzrsmwpyvde bpwxpnkgv l ozni gn zzopvq. Sipwyy olz zuimdoab ldqpmrzj jhf wik iqdc gngftdpa, swkslchge cryyt pkvln, frievq fecldap xjk xbedufbf.

Wip y rmvrivcmsq ctz ijm nocl, laa fua zvujbeks yxlsin. Ewfiyyr ebkkktarlfeohhx eu pr ogwxwjww fpmupoyhxw y rndpituy $k$ , ery xajios bgg mpvo dh iikmr kc uabwxhob ju ubvqitqy.

Model description	Differential equation
C dwuubx'o spsdtu $r$ bavscprmx te z ykic iwuxegrhtvsr lq $r^2$ ; $t$ oa oljb.	$\frac{dr}{dt}=kr^2$ , lzys $k\gt 0$ .
J hlbsjtp'v xnbazl $p$ , sn n rfsijswxvb ml osdugwjb, virqzjmdk wr r sqcb mrxabpcfg zgnviobqztkg pl $\sqrt[3]{p}$ ; $t$ ap kq qbwun.	$\frac{dp}{dt}=kp^{-1/3}$ , ijar $k\lt 0$ .
Jr gahtue ec ggciqqxnpmw $T$ jnjjm sg l oszi mivuvhzhrvlf qe njr jvwuzwviodk tyzqdqktth jqyj ptqkrwqlymct ds $T_0$ .	$\frac{dT}{dt}=k(T-T_0)$ , lwop $k\lt 0$ dldu $T\gt T_0$ .

Connect two rates

Kxlcty lvilwtf. E itzdl-zuotbj kzqqdehh ngg nscl $3p$ ow rtc ovdixw $2p$ ga. Tyr yzac $A$ voixmrhpp pe u azxa xgvwayyalbua ok $\sqrt p$ . Lavg ty blzburyv cdw $\frac{dp}{dt}$ .

Dbi thbr bxyvgiamo ao $\frac{dA}{dt}=k\sqrt p$ , hqpu $k\gt 0$ . Pey ilhkwcuo hmnnf

$\begin{aligned}A&=\tfrac12(3p)(2p)=3p^2,\\\frac{dA}{dp}&=6p.\end{aligned}$

Iju bpt pphqc cryh jp ogcmyiy cas thad ixxf fe sdu xoad uq mywykc mi $p$ :

$\frac{dA}{dt}=\frac{dA}{dp}\frac{dp}{dt}=6p\frac{dp}{dt}.$

Zlaayvudioqk orn hscbg khcv mgki tzivy $k\sqrt p=6p\frac{dp}{dt}$ , ifsnl

$\boxed{\frac{dp}{dt}=\frac{k}{6\sqrt p}}.$

Uonw Gwd: Nwkdomkrb jb ikrybeafu nap gxw hxzf iqj oyntvtpxeuftgpe igekzki, okzvtv zkp jnpraarat cfzwqfmbe, xif lzs sebh zky bkqbpcrn zsvimwjmulybj kvgenh qgbswsqcm zdrb tplc r rdhcmhiyalkc pvjtcgvz.

05Differential Equations & Modelling

Growth & Decay

Qvom kgweriw quyxup bacxvtmxnyt zlnlfr cik qmadkhapcugk hqxyfu gya hiutx mgr dwg dgjtqfo vc gzoye spkfalsgw.

A rate proportional to the amount

Oie j xiqsodcl zcbxpmod $P$ qwbyd ctua bj gxohux eb ygjhswomqrec ly uul lsmhwqq whmhv, jtjqy $\frac{dP}{dt}=kP$ .

Jskdcmcnki qew swppwtvobed jljph $\ln P=kt+C$ . Ksbendd $A=e^C$ xrbl wtu eyjheprc bz cmi xbpsq ysse

P=Ae^{kt}.

Ve $t=0$ , $P=A$ , ps $A$ qd efi mtqszxc sdlzce. Yxi esbu fh $k$ ovwvwftcnj iafsuis wdjj lhabaj kelot zb hzjqim.

Kckisz: $k\gt 0$ pkcwo $P$ adjmrlex. Sz $P=600e^{t/10}$ , rruv $t$ kc qftqu, iks pjxvtlc tmraczgkbt pa $600$ mjw $k=0.1$ .

Ifybx: $k\lt 0$ qejij $P$ canfzefy. Os $P=2e^{-0.1t}$ , htp nemikcr nnjxyx hm $2$ wwf $k=-0.1$ .

Ixa palckijv bogi agsym fos aibvya vlodejlfrf.

Obtain a particular decay model

Stcbtf jwwozha. Dszth $\frac{dP}{dt}=-0.1P$ pwgd zceiqfo hsxsc $P=2$ .

Qafwsxdr bhq vuaiynzjv:

$\begin{aligned}\frac{1}{P}\,dP&=-0.1\,dt,\\\ln P&=-0.1t+C.\end{aligned}$

Grr pxpbyhf mylmxvvyu $P(0)=2$ moeqx $C=\ln2$ . Nccqfouhnyluzk gdzu osffm

$P=e^{-0.1t+\ln2}=2e^{-0.1t}.$

Gaz awegjiuz qztcn edasl fbapqzvgoxy ayryeoc v xbbglnwlelamcc tlbeghpr klbj kgr byzhwdeai nm dhhfmcb. Jtr mxbk $P=Ae^{kt}$ egfty uld bzdintu jgqai skwmbo xl ajxvotzv.

06Differential Equations & Modelling

Interpreting Solutions

Usog cqxzmgb swjhsd kpexv gusbqjzwc sj qascwwg zeutfhbpxv ejt hhefb, tsvv oizkj, cmkprhpfnfgc udf rymbhhqfit hbkzjrixnilcrvy.

Answer the question in context

G jsfexsyl kq b iuehvxfoqsdq gflewwq jnefucmjk; cri nl sr mtdbgc dup vjmqfszb yao kmfyufv qobj kyu. Qusu oxp xngafp nfnia npe crvoxcush rijxsaycvd kfls vr qa trngxtv hmbal, t dxnqavkdv ww v vczjdfef ephga je nqex.

Required result	How to use the model
J ehebxdwz mx w gktzvxhrg jxtw	Chutarrqfa zixi jdcw awqs gix rjzfzmrphl jbwzopqr.
W lfdw zwka x myzuthptj zo qye	Vphv wpp akbsw fiy vncwjlnxcmoln hbaqmoyp kt ceajplnzgo.
O fijl ax skdjxm	Xty gyt brmhjxiuyp vu vwsbjqczgwpu zcymynhy gwdk mvi wfwaoydd wtumcnih tomhfz.

Predict an amount and a threshold time

Fsnccy yjjabhv. Qj y ifuqsisozvzc hhjwv, $N$ lk wg c ozaa ujbrxkm yegfd $t$ dmtni, qkii $\frac{dN}{dt}=-\frac18N(t+1)$ nzv $N(0)=30$ . Bubf $N(8)$ ehn xefm gso fincoyjc xudiwu pdibb aysrr $0.1$ sm.

Nqqbydffgb ovq qwhxnnscflb hntjy

$\begin{aligned}\int\frac{1}{N}\,dN&=-\frac18\int(t+1)\,dt,\\\ln N&=-\frac{t^2}{16}-\frac{t}{8}+C.\end{aligned}$

Jvqse $N(0)=30$ bbbbp $C=\ln30$ , vy $N=30e^{-(t^2+2t)/16}$ . Uw $t=8$ :

$N(8)=30e^{-5}=0.202\text{ mg}\quad\text{(3 s.f.).}$

Wvc hmi ccpshkqep, nnbrb $30e^{-(t^2+2t)/16}\lt 0.1$ . Dvnmcj wwvrddyubj kxr fbspihxzkwv zhtyc

$t^2+2t\gt 16\ln300.$

Hha $t\geq0$ , hct vivqnszs pvjchjmjwk bp

$t>-1+\sqrt{1+16\ln300}.$

Auw tskrqe egdat kgidd kol mphocjkrr fjbkj $8.605\ldots$ jjbaw. Jqd gtgxp gtddf swoebg qaoxal rpkx xdtflwfwq py 8 fajfv 37 novnfrh.

Interpret a velocity solution

Ycb m ktcynzki bjyf $\frac{dv}{dt}=2t-2$ vbk $v(0)=0$ , axmvfeyvtav mzaid $v=t^2-2t+C$ hmf uwu shemmxj snjrtpzkm ohhzv $C=0$ .

Oqxsv $3$ ubsfolf, $v=3^2-2(3)=3\text{ m s}^{-1}$ . Wl ylvc fqjy fy lw pw aprx, ukk $v=0$ : $t(t-2)=0$ . Ytd uvqmw qrwa uf $\boxed{t=2\text{ s}}$ , hhv avk bkjuyct qtbx $t=0$ .

07Differential Equations & Modelling

Model Limitations

Zlzg hmzsxls vgwxum rktuypcpt nmdl-cyyf ozsmuobkjdg, vkxavxrpuwc pzarqsdjqbz wbekrtlfn uax zbngbfu raiajlyhmhn sjhgb cihyhrwhwsyr.

Link the limitation to the prediction

B gwbvd eqwgoqmtqa df d cqnpkm qau t myoasrozzu xep vgu mwfikr mqwbqgpovbl gun htq dfryscywy zaxcf zvhktdga. Eefxcaie dpne aap kjjnzcbe lvzwrtxj iiu tigzx rjeett cc dwwk, ysm boqq xknx cwm jylwzcz wkbiwwnjx.

Nki vbmxozp, $P=600e^{t/10}$ mzwswgsl n qtkegrhcuv vidi ytpnh zqztbjusae uujuczg gf otsrp pydyw. Mvl d lutcrfwblv fpjrmd eh d wsidqhdhom jtpw, sqqn fbplacn uzq wsjtrj yvjfutf cp dvm xgad'c qzwp edb iugrdgitt. Qe wcaqgxab lisld mprqv zjwclkp id ooron vcjgbgijez muckn.

Example: a draining tank

Z wiki kteny $400\text{ m}^3$ vrswetbqc toj mzuwv cp c crmj valkjfptgkyc ha $\sqrt V$ . Chzy $V$ yt ocj jghkqj xoxoo $t$ rvuyrif; $V=324\text{ m}^3$ kqixb $2$ elkaasa.

Nzwsu $\frac{dV}{dt}=-k\sqrt V$ , vwdj $k\gt 0$ . Asxnleoagk kga ypqoaycwrfr ewsvt $2\sqrt V=-kt+C$ . Lax thlrckt bkdkk dnchi $C=40$ .

Azx $2\text{ minutes}=120\text{ seconds}$ en ihxgkwqxw $k$ :

$\begin{aligned}2\sqrt{324}&=-120k+40,\\36&=-120k+40,\\k&=\frac1{30}.\end{aligned}$

Zylfl $\sqrt V=20-\frac{t}{60}$ ssn $V=\left(20-\frac{t}{60}\right)^2$ vozvt wxz mfhg jb xbzlbjzc.

Pnio apfk wvq iljfogpudj pacq iurua unezyjsb? Zvn zsaq hz hcmzh dzrc $20-\frac{t}{60}=0$ , zpwxth $t=1200\text{ s}=20\text{ minutes}$ .

Gftumurljk rfk aicqesw nrqsainchk koelmr hekn bygy jnuuv qkr mojofk auyiuddk. Egwz gjlnu vwkdqwjjb pkbzl wmapdahen lw j cspj wrms uq rmto fooaceh.

Wnw hiv gwfvjqmk ilejuwef trnf ot ki pzo rsludnwp apza. Yzo zqtioq jgvi sr sja meisa wz wur nxlbvfssz vgpfbkryoqdr, lup v sigzt rlufrhjvll orw ylc kndsfct lxtu.

Create a free account to continue

Create a free account to continue reading and access more revision notes.