You just helped me so much on my ODEs test. I wish my professor taught the way you did! Great job.
@blackpenredpen6 жыл бұрын
You're welcome! I hope you did well on the exam!
@Sleepy46169 Жыл бұрын
Dear sir. Your final answer of the Y sub h is C sub 1e^t + C sub 2e^2t. Is it alright that the final answer of the Y sub h, its roots were shuffled? Like mine is C sub 1e^2t + C sub 2e^t. Is this the same?
@hugopurpeq Жыл бұрын
@@Sleepy46169 yes its the same
@boing77710006 жыл бұрын
I love how enthusiastic and natural this guy is:)
@betechtechnology9426 жыл бұрын
thank you very much you the best good techniques especially on 2nd derivative it could have been very very long
@--82514 жыл бұрын
ngl, I was too worried about the thermal detonator in his hand to focus on the problem
@hans33310006 жыл бұрын
normally i just look at videos hopelessly and give up. then i found you, thank you for contributing to my 100% on the midterm. you're a lifesaver, i just couldn't learn this one by myself. ily no homo
@kabirkohli99205 жыл бұрын
You're literally the reason I'm gonna pass my exam
@ianmoseley99105 жыл бұрын
2B or negative 2B, that is the question ...
@dylongoodyear77185 жыл бұрын
lmao i enjoyed that
@nsifrahman90315 жыл бұрын
@@dylongoodyear7718 Can I enjoy you?
@mcleanephatha5 жыл бұрын
@@nsifrahman9031 You can enjoy me!
@hellmuth264 жыл бұрын
every time I have a 2B i say "or not to be," it's a terrible compulsion and I wish I could stop.
@nicoleng41813 жыл бұрын
16:36 yes i feel so accomplished hahah and omaigosh i've been finding a video that explains this well and you're the only one that I UNDERSTOOD SO WELL thank you
@ayodelerighteous Жыл бұрын
You just salvaged my fast approaching mathematics examination!
@joshuastier2316 жыл бұрын
Dude you are fantastic thank you so much
@blackpenredpen6 жыл бұрын
Joshua Stier My pleasure to help!!!
@zhou41686 жыл бұрын
it is a good video, thanks. Already subscribed, hoping to watch more!
@blackpenredpen6 жыл бұрын
zhou coooool thanks!!!
@xOxAdnanxOx5 жыл бұрын
all the way from the u world to differential equations, we been together my friend... thank you
@photomath43273 жыл бұрын
Exactly what have been searching for ...thank you so much 🙏🙏, your video have been helping
@Engineering_conceptsUOM2 жыл бұрын
Thank you sir… you are my mathematics hero
@GisselMejiashortyeah10 ай бұрын
You are awesome. Thank you very much
@davidnigelsemuyaba11312 жыл бұрын
I do appreciate you for your service, you have rilly talented in teaching, may God 🙏 bless you
@timotejfasiang2 жыл бұрын
The hardest part for this guy was distributing the - into the (A + B) at 7:35😂
@LL-kn1de5 жыл бұрын
Wow! This is so helpful. I wish I found this video earlier! Thank you !!!
@jamalmanking39286 жыл бұрын
It so amazing please more videos please for the subject Deferential equation please.
@blackpenredpen6 жыл бұрын
I have many playlists on that already. Check out my channel home page.
@taekwondotime4 жыл бұрын
@1:30 How did you know it was going to be: *yp = Ae^tsin(t) + Be^tcos(t)* ? A bit more explanation is needed there. What happens if the equation is y'' - 3y' + 2y = 3t ? What happens if the equation is y'' - 3y' + 2y = tan(t) ? How does that piece change? Thanks.
@gokayfem4 жыл бұрын
you should check his differential equations playlist. there is variation of parameters method for tant and other types of undetermined coefficients methods for 3t, e^t kind of things.
@alvelymondragon334 жыл бұрын
I think it's from a table given.. that what i use actually
@carultch10 ай бұрын
If the function on the right hand side is linearly independent of the homogeneous solution, you directly match it with an arbitrary coefficient, and accompany it with other coefficients on its counterparts (more on that later). If it is a constant multiple of one of your solutions to the homogeneous solution, then you multiply by t until it isn't. For exponentials, the particular solution will also be exponential. For sine or cosine, or a linear combination of both of the same frequency, the particular solution will also be a linear combination of both of the same frequency. Likewise for exponentials, enveloping the sine and cosine. For constants, the particular solution will be a placeholder constant, not necessarily equal to your given constant For polynomials of t, the particular solution will be a polynomial of the same degree, and placeholder coefficients along the way. For anything else that isn't a linear combination of the above, you'd have to use another method, like variation of parameters.
@bukkiofuzi17253 жыл бұрын
I love how you handled the second derivative
@shoberino38982 жыл бұрын
14:49 “I didn’t even use a calculator so you guys can do this too” says the human calculator
@anamuslamh094 жыл бұрын
You the best 👍👍👍✅🏆🥇
@tomriddle94893 ай бұрын
great😀
@ahmidouaouladhadj75763 жыл бұрын
You are the best
@111abdurrahman5 жыл бұрын
you are serving the humanity dude!
@__IzanNafisRahman2 жыл бұрын
this guy is a real chad
@TheGuroLOLITA3 жыл бұрын
You are soooooo goooood
@elisamarconell99912 жыл бұрын
thank you kind sir
@aljoker30533 жыл бұрын
A coil of inductance L Henry and a capacitor of C Farad are connected in series if I=I0, Q=Q0 when t=0 Find: Q and I when t
@carultch10 ай бұрын
The corresponding diffEQ is: L*Q"(t) + Q(t)/C = 0 Characteristic equation: L*r^2 + r/C = 0 Solution for r: r = +/-i/sqrt(L*C) The solution for Q(t), is a linear combination of sine and cosine, with a frequency equal to the imaginary part of r: Q(t) = A*cos(sqrt(1/(L*C))*t) + B*cos(sqrt(1/(L*C))*t) At t=0, Q(0) = Q0, and I(0) = I0, which is the same thing as Q'(0). Find Q'(t), which is also I(t): Q'(t) = sqrt(1/(L*C))*(-A*sin(sqrt(1/(L*C))*t) + B*sin(sqrt(1/(L*C))*t)) At t=0, sine is zero, so the cosine terms govern. This means: A = Q0 B*sqrt(1/(L*C)) = I0, solve for B: B = I0*sqrt(L*C) Thus our result is: Q(t) = Q0*cos(sqrt(1/(L*C))*t) + I0*sqrt(L*C)*sin(sqrt(1/(L*C))*t) I(t) = sqrt(1/(L*C))*(-Q0*sin(sqrt(1/(L*C))*t) + I0*sqrt(L*C)*sin(sqrt(1/(L*C))*t))
@XLatMaths4 жыл бұрын
Why don't you have to set the particular integral to A*te^t*sin(t)..., the auxiliary equation already has e^t as one of its components. Don't you have to multiply the P.I. by one power of the variable when one of the auxiliary equation's roots is in the RHS?
@amzayd3 жыл бұрын
I have the same question!!!
@zingaphiranana8425 Жыл бұрын
@@amzayd he made a mistake
@ernestschoenmakers8181 Жыл бұрын
@@zingaphiranana8425 Yes he did cause at 12:25 he plugged in -A + B but it should be -A + 3B, so A and B are miscalculated.
@oluchukwuokoye841210 ай бұрын
Thank you for this video
@bandarhindi84506 жыл бұрын
your techniques are very helpful
@mumtazbegum34925 жыл бұрын
Very well ....super bro... thank u so much
@_ssodaaa5 жыл бұрын
In particular sol. Can I consider yp= Ae^t sin(t) not yp= Ae^t sin(t) + Be^t cos(t) ? Because A&B are abitary const.
@carultch Жыл бұрын
Only if you are "lucky" enough that B=0. For this method, you have to account for the possibility that both sine and cosine will play a role in the result, until you rule out the possibility.
@antonellaa71205 жыл бұрын
I UNDERSTOOD EVERYTHING!!! GOD BLESS YOUUU
@blackpenredpen5 жыл бұрын
Thanks
@Eta_Carinae__6 жыл бұрын
B=-1/2 not 1/2. Otherwise -A-B=0 isn't satisfied. Final answer should read: y=C1e^t+C2-(1/2)(e^t)sin(t)-(1/2)(e^t)cos(t)
@jonhsmith37458 ай бұрын
-(-1/2)-(1/2) = 0
@joshuammaina49116 жыл бұрын
Can you please do one with the differential operator Pretty please with chocolate frosting on the top
Ive been looking for examples of X^n*sin(x) Or X^n*cos(x) But haven't seen any on youtube. Any one know some links? Thanks.
@ITSYUNGIAN4 жыл бұрын
Thanks
@atone09095 жыл бұрын
Very nice
@tyronekim35066 жыл бұрын
Excellent video. Thanks!
@tejagandreti5 жыл бұрын
Tq so much bro very gud explanation
@feelingepic60874 жыл бұрын
Thank u so so much for helping me...now i understand it clearly
@cyberwarrior52176 жыл бұрын
oh my god..you make it easy for me..and you r soo funny
@jonhsmith37458 ай бұрын
banging video
@kazshoichi93694 жыл бұрын
I feel so accomplished indeed my man
@arvandvedadi24577 жыл бұрын
bro you are the best thank you so much man!
@bandarhindi84506 жыл бұрын
Thanks a lot. The video helped me a lot you are awesome
@Ayeventy4 жыл бұрын
A million times better than my professor
@tungamiraiaaron17845 жыл бұрын
u are the best
@cadendennis30194 жыл бұрын
That was awesome
@icee5626 жыл бұрын
wow, EXTREMLY helpful!!!!!
@lorenmanchester33775 жыл бұрын
Thanks so much! This video really helped me understand.
@hapias93005 жыл бұрын
Thank you well explained.
@olivierbkofficial94266 жыл бұрын
My God!!!! you're just amazing ,thx
@elnathanlynd54186 жыл бұрын
Amazing
@tomatrix75253 жыл бұрын
Unbelievable!
@himanshubanate90615 жыл бұрын
nice sir thank u so much
@wraster07hp363 жыл бұрын
Thanks bro.
@eddiekanhai6 жыл бұрын
Thank you so much, this helped me a great amount !
@Dr.Reactor3 жыл бұрын
channel name: blackpenredpen Blue pen: Am I a joke to you??
@arnelmananghaya75884 жыл бұрын
thank you so much, my friend!!
@deorum596 жыл бұрын
Not all heroes wear capes... Thanks
@robbiemayer35732 жыл бұрын
the literal goat
@wzablonkalivo28186 жыл бұрын
great work. no need to attend to attend to my lecture
@debbiekalima4541 Жыл бұрын
thank u so much
@jacksoncarter63523 жыл бұрын
Perfect thanks
@ahmedbelloe39233 жыл бұрын
I used De-operator method and it was much easier and fast too. I had the same answer
@josemuygay88515 жыл бұрын
thanks for uploading this!
@terrencemadanhi88335 жыл бұрын
thank you brother.you are supper
@hayatomcreverse79256 жыл бұрын
Bai maayo kayka this helps a lot T.T
@bhaswatighosh39696 жыл бұрын
Can u solve some hardest problems in nth order eqn
@jacksantoro1826 жыл бұрын
thank you very helpful
@bakkamanthulamohansai34302 жыл бұрын
Super super super
@F199915 жыл бұрын
Another useful video. But I think it's quite dangerous that you're holding a thermal detonator haha
@Goosebump_guy8 ай бұрын
How could differential Ae^t(sint) still be Ae^t(cost)+Ae^t(sint)?????dont u have to differentiate e^t into te^t ???
@AndromedaIX5 жыл бұрын
You mentioned it was the superposition principle. Is it the same method as used for circuit analysis?
@carultch Жыл бұрын
It's a similar concept. Essentially, to simplify the equation, you break it up into two sub-equations that both also need to be satisfied, and work out how to satisfy both parts individually. And then form a solution that is a linear combination of the two.
@farissh39244 жыл бұрын
thanku
@mtahir91814 жыл бұрын
Thank You.
@Channel-uf6ko9 ай бұрын
I can't find any video on the internet doing a combination of polynomial, exponential, and sin and cosine :((
@pushpanjalinayak27655 жыл бұрын
Thank you so much
@Cannongabang7 жыл бұрын
hello blackpenredpen... is the integral sqrt[(x^4-1)/(c-x)] dx solvable???
@DaQwertyKidNL7 жыл бұрын
No
@swordsy915 жыл бұрын
blackpenredpen used blue pen!
@rymichael72034 жыл бұрын
I'm a year to late.
@lesegomokgabudi57985 жыл бұрын
thanks bro
@strongerwithstubsstretch34724 жыл бұрын
thank you sir!
@user-yw8pr8wp4b3 жыл бұрын
Yp= t e^t ( A Cos(t)+B sin(t)) IS true ?! Because e^t IS in Yh and in f(t) = e^t sin(t) ?! Please answer me
@AstroRoxy6 жыл бұрын
Thanks dude!
@dmorgan06287 жыл бұрын
I'm taking calc2 again this summer and diffeq this fall. This video is scary!
@cyberwarrior52176 жыл бұрын
Dan Morgan hahahha summer..I hate summer
@mastermfundisombhokane57314 жыл бұрын
maVuti maths 3 where are you? send a shout out!
@thomasjefferson6225 Жыл бұрын
I sincerely hate these problems
@TheIvar192 жыл бұрын
I just don´t get why we need the Yh solution. Isn´t Yp enough? What does the Homogenous solution add?
@carultch Жыл бұрын
yp only solves a special case of initial conditions, that the coefficients in front of the yh parts diminish to zero. The yh part is the transient response from initial conditions, while the yp part is the steady state response that governs what it will be, once the initial conditions are effectively out of the picture. As an example, given: y" + 4*y' + 3*y = 10*cos(t) The homogeneous solution is: yh = A*e^(-t) + B*e^(-3*t) The particular solution is: yp = 2*sin(t) + cos(t) In the special case that our conditions start on the curve of yp, then the coefficients A and B both become zero, and we don't need the homogeneous part. This would happen if, in this example, y(0) = 1, and y'(0) = 2. What the homogeneous part adds, is the ability for the system to start at any set of initial conditions, and exponentially decay to eventually match the particular solution.
@wajeehalamoudi59455 ай бұрын
is there any previous illustration video such #18 or 4?
@sunitalymon19253 жыл бұрын
him: whenever there's sin we must have cosine me: say no more.
@FS-zt6tm3 жыл бұрын
what would the general formula be for yp if the right side is e^x(6cosx +17sinx). I'm struggling with trig functions. I know how to do the derivatives, I'm just not sure what to start with.
@carultch9 ай бұрын
Given an RHS of e^x * (6*cos(x) + 17*sin(x)) The general form of the yp particular solution would be: e^x * (A*cos(x) + B*sin(x)) As long as your given linear combination of sine and cosine both have the same frequency, they both correspond to the particular solution being a linear combination of the two. The e^x applies to both of them. Also important is that we don't overlap the homogeneous solutions, otherwise we'd need to multiply by t until it is linearly independent. As an example: y" + y = e^x * (6*cos(x) + 17*sin(x)) While this may look like there is overlap, there really isn't. We'd have to have y" - 2*y' + 2*y = e^x * (6*cos(x) + 17*sin(x)), to have overlap, since the homogeneous solutions of this one, also include the exponential envelope on the trig functions. The homogeneous solution: yh = A*cos(x) + B*sin(x) The particular solution: yp = e^(x) * (C*cos(x) + D*sin(x)) Differentiate twice: yp' = e^x * ((D - C)*sin(x) + (C + D)*cos(x)) yp" = 2*e^x * (D*cos(x) - C*sin(x)) Apply to original diffEQ, and equate coefs to find C & D: 2*e^x * (D*cos(x) - C*sin(x)) + e^(x) * (C*cos(x) + D*sin(x)) = e^x * (6*cos(x) + 17*sin(x)) Cancel common e^x factor: 2* (D*cos(x) - C*sin(x)) + (C*cos(x) + D*sin(x)) = 6*cos(x) + 17*sin(x) Thus, equating coefs we get: 2*D + C = 6 -2*C + D = 17 Solve for C & D: C = -28/5, D = 29/5. Thus the solution is: y = A*cos(x) + B*sin(x) + e^x*(-28/5*cos(x) + 29/5*sin(x))
@liujinpeng26354 жыл бұрын
好像就我一个中国人哈哈,祝您2020新年快乐!
@ghozmit16 жыл бұрын
Bro, you picked the hardest/longest example :(
@blackpenredpen6 жыл бұрын
ghozmit1 i have done other simpler problems too
@ghozmit16 жыл бұрын
oh I see, Thanks :)
@icee5626 жыл бұрын
trust me, one you see a hard(er) example, the others are cake!
@jammcrusader19814 жыл бұрын
if you can swim in the deep end you can swim in the shallow end my friend
@Sneaky1ne4 жыл бұрын
the only reason I clicked was because it was a harder problem
@AdnanAdnan-hd7co2 жыл бұрын
ارجو حل الكثير من المعادلات التفاضليه
@zabihullahhalim50966 жыл бұрын
that is great :) T H A N K YOU
@sabikasarwat41863 жыл бұрын
What should be the method of solving this equation D2y/dt2 _ dy/dt _2y= 5 cosh(2t)....by undetermined method....kindly help plzzz
@carultch10 ай бұрын
Assuming your underscores mean +, we're given: y"(t) + y'(t) + 2*y = 5*cosh(2*t) You can use undetermined coefficients, since cosh is just a linear combination of two exponentials. I'll assume a particular ansatz, of a linear combination of cosh and sinh. First find the homogeneous solution: yh" + yh' + 2*yh = 0 (r^2 + r + 2)*e^(r*t) = 0 (r^2 + r + 2) = 0 r = (-1 +/- sqrt(1 - 8))/2 r = -1/2 +/- i*sqrt(7)/2 Thus: yh = e^(-t/2) * (A*cos(t*sqrt(7)/2)) + B*sin(t*sqrt(7)/2))) For the particular solution: yp = C*cosh(2*t) + D*sin(2*t) Apply to original diffEQ, and match coefs: yp' = 2*C*sinh(2*t) + 2*D*cosh(2*t) yp" = 4*C*cosh(2*t) + 4*D*sinh(2*t) Thus: 4*C*cosh(2*t) + 4*D*sinh(2*t) + 2*C*sinh(2*t) + 2*D*cosh(2*t) + 2*C*cosh(2*t) + 2*D*sinh(2*t) = 5*cosh(2*t) 4*C + 2*D + 2*C = 5 4*D + 2*C + 2*D = 0 solution for C&D: C = 15/16, D = -5/16 Thus, the solution is: y = e^(-t/2) * (A*cos(t*sqrt(7)/2)) + B*sin(t*sqrt(7)/2))) + 15/16*cosh(2*t) - 5/16*sinh(2*t)