Doweknow a functionsothatitsderivativeisjust a constantofitsoriginal?

I thinkso.

Wedo, right?

Ontheside, letmejustshowyou.

If I startwith y equalsto, well, if I have, let's say, e tosomepowertimes x, well, inthiscase, weuse t, bytheway, forsecond-order, use t.

Let's saywehave e tothe 3t.

Well, ifyoudifferentiatethat, youget y' andyouknowthisisgoingtogiveyou e tothe 3t, thefunctionpartstaysexactlythesame, butthechainrulesays I willhavetomultiplyby 3, right?

Soletmeputitdownrighthere, right?

Andyousee, this 3 stays, andthisisjusttheoriginal, so 3 times y, right?

Andlikewise, I candoitagain. y'' isgoingtobe, wewillkeepthis 3 times e tothe 3t, butthenwemultiplythis 3, right, forthederivativebythechainrule. 3 times 3 is 9, and e tothe 3t istheoriginal.

Andyousee, wheneveryouhavean e tosomepower, thefirstderivative, thesecondderivative, theyarejustgoingtobetheconstantmultipleoftheoriginal, right?

Sowhatthisistellingmeisthatthiswillsuggest, letmejustputthisdown, thissuggeststhatforthefunction, weshouldhavetheform y equalsto e tosomepowertimes t.

I don't knowwhatthisnumbershouldbe, earlier I justused 3, right?

Soingeneral, letmeputdown r times t.

Andonceagain, forthesecond-ordersituation, weusuallyuse t becausethereare a lotofapplicationsthatimposetimeandthingslikethat.

Anyways, thisismystarting.

Firstofall, webeginbysaying y isequalto e tothertpower, andtheideaisthat I'm justgoingtogoahead, differentiatethistwice, andthenplugin.

Allright, y' isgoingtobe e totherttimes r, letmejustput r inthefront, andthen y'' isgoingtobe, thisrighthererepeats, so r e rt, butthenwemultiplybyanother r, whichwillbe r times r, whichis r².

Andnow, letmeputallthisintotheircorresponding.

Herewehave a, y'' is r² e tothert, andthenweadditwith b, soweputonplus b, y'' is r e tothert, andthenwecontinue, plus c, y is e tothert, andthisrighthereisequalto 0.

Andnow, canwekindofsqueezeout a condition?

Let's seewhatcanwedo.

Everytermhas e tothert, sowecanfactoritout, ofcourse, e tothert, andthatwillgiveusar² plusbrplus c equalsto 0.

Allright, thisisanexponentialpart, right?

Exponentialfunction e tothesomething.

Thisrighthere, weknow, isnever 0, isn't it?

Never 0, letmejustspellitoutmuchbetter.

So, whenwehavethisquantitytimesthatquantity, since e tothertisnever 0, soyoucaneitherdivideitout, oryoucanjustforgetaboutit, becausewejustwanttofocusonthis.

Sothatmeanswemusthavethispart, wemusthave, letmejustwriteitdown, wemusthavethesituationthatar² plusbrplus c equalto 0.

Infact, thisrighthereistheconditionthatweneed, becausefromhere, thisisprettymuchjust a quadraticequation, isn't it?

Quadraticequationintermsof r.

Fromhere, wecansolvefor r, andwecanjustplugintothe r here, andwecangeneratethebuildingblocksofthesolution.

And I'llshowyouguyswhat I meanbythat.

Andbefore I showyouguysanexample, letmetellyouguysthatthisequationherehas a name.

Thetruthis, wemultiplythefunctionpartby C1 and C2, sothisrighthereis C1, andthisrighthereis C2.

So, letmejustdemonstratethisrighthereonthesideforyouguysrealquick, becausenow, yousee, when I multiplythisfunctionby C, soletmejustputon C inblacklikethis, guesswhat?

I willjusthavetomultiplyeverythingby C, isn't it?

Sothisrightherewillendupwith C, C, C, C, C, C, C.

Andthen, prettymuchthisterm, we'llnowhave C righthere, andthisfunctionhere, thisparthere, has a C hereaswell.

Thisrighthere, justmultiplyby C, right?

We'rejustpluggingeverythingaccordingly.

Guesswhat?

When I havethis C here, I willstillendupwith 0 ontheleft-handside, ofcoursestillequalto 0.

Thepointofthatistoshowyou, thesearetheplaceswhere C shouldbe.

C1 and C2 arejustthemultipleofthefunctionpart.

C1 and C2 arejustthemultipleofthebuildingblocksofthesolutions.

Imagineif I'm differentiatingthe y, well, I justhavetodifferentiatethefirstandthesecond.

Differentiatethisanddifferentiatethat.

Doitagain, doitagain, doitagain.

Plugin, plugin, plugin.

Youget 0 isequalto 0 onceagain.

Sothisistheidea, justtosummarizethisrealquick.

Allyouhavetodo, justtokeepeverythingsimple, getthecharacteristicequation, solvethequadraticequationin R.

Ifyougettwodifferent R values, well, e tothefirst R times T, e tothesecond R plus T, andbesureyoumultiply C1 and C2 correspondingly, andthenaddthemtogether.

Thisisthegeneralsolution.

Thisisthefirstsituation.

Youshouldwatchmynextvideo.

I willshowyouwhathappenswhen I havethesetwo R valuesbeingthesame.

Inthisvideo, I'llshowyouguyshowtoapproachtosolve a second-orderlineardifferentialequationwithconstantcoefficients, andinthiscase, wewanttofocusontheright-handsidetobe 0, andwhentheright-handsideisequalto 0, thisiscalled a homogeneoussituation.

Subtitles and vocabulary

Click the word to look it upClick the word to find further inforamtion about it