this right now let me tell you yes 10:35 and then you guys can see the outside it's all dark but I
really want to get this done okay as we all know when we have y to the first power times y to a
second power of course we can just add the powers up together and we get y to the third power but
have you ever thought about what if we have y prime times y double prime what are we going to
get oh well why don't we put down Y triple prime this would be so cool isn't it however you know
this right here is not always true but is true for some functions though such as the linear function
because the linear function which is like Y is equal to MX plus B if you differentiate
that twice the second tier of the zero and the third rope needs also zero and you end up with
you were to you go to zero that's kind of cool but it's kind of boring that cannot review so in
this video we're going to find out more functions so that your first derivative times the second
derivative will be their third the robot here so check this out and yes I did do this on the
other day but it was too long you can just do the following right here right so here we go first of
all I'll just write this down right here for you guys the goal is to find Y and let's say
Y is a function of X and we want that the first derivative times the second derivative to be the
third derivative and that's C and in fact I will take a substitution because we have a third water
nonlinear differential equation I will let you reduce this a little bit in fact if you don't
do that it's actually okay if you just kind of be careful with the notation it's up to you by
the way I will begin by saying let you to be my first derivative of Y so I'll just say u equal Y
Prime like this well this right here will be okay we have to you times y double prime of course you
put prime here put prime here that's legit so you get u prime right here and then that will
be equal to and then actually leave some space you don't know why why triple prime you put two primes
here you put two price there and you get u double prime here very nice that's legit and now what can
we do we all the choices that you see you have u times u prime doesn't this remind you of the
Chen Lu questions yeah you had the thing about the roti of what function will end out to be u
times its derivative in fact we can just integrate this right now and the truth is we can integrate
both sides with respect to X because you here's a function of X and also this is a nice form because
you will see that this right here hope it's going to be just one half u square let me tell you guys
why first of all if you differentiate this could a twin front you're just kidding you right and then
minus one of course for the exponent and don't forget the chandu you have to multiply by the
derivative of U which is exactly you prime time so when you integrate this it's very nice to get
that very nice and then I'm going to put down a constant here let me just say plus c1 and on
the right hand side when you integrate a second derivative you can be first the robot here so I
would just say this is equal to u prime and I don't need to put down plus T however because
if you do put down plus c2 you can just - into opposite which is still going to be constant so
that's that typical to you and now notice u prime is what this right here is nicely equal to D u DX
because U is a function of X u prime is just this right if you differentiate this you just
differentiate that pretty much and on the left hand side this is probably to do let me fatter on
one half I think it can't be easier so I would do that and right here inside here we have u Square
and from C 1 F at Aldi 1/2 I should get plus 2 c1 but that's just another constant so let me
just put up plus c2 like this and here we have a separable differential equation so let me divide
this on both sides and let me multiply that the example sides however I want to work with the use
on the left hand side so when you tip at this right here I'll put it down as 1 over u square
plus C 2 and then the T you'll right next to it and then if I multiply the example sides I will
have the 1 half with the DX like this so this is wonderful and as usual we can integrate both
sides so let me just put on integration here and the question here and yes this time I remember to
put on the integral sign after the equal sign but to integrate both sides integrate both sides yeah
I was tired on the other day in fact I'm still tired this right now let me tell you yes 10:35
and then you guys can see the outside is all dark but I really want to get this done because of the
video I make anyway we continue here's the deal to integrate this integral we have to consider
three cases case 1 let's say C 2 yes 0 right so we have this be the integral of 1 over u squared
plus 0 and then D U and this is equal to that which is the integral 1 over 2 DX and we are
just going to deal with this little baby integrals when you integrate that you get negative 1 over
u and that's what about a constant on the right hand side this is going to be 1/2 X plus let me
use these three because I have to see 2 over there already so that's what we have and now
I'm going to isolate you and this is how we'll do it so here we go I'm going to take the reciprocal
so you have the 1 over and now multiplied by the negative both sides and then you will get 1 for
this right and then you have to multiply the top and bottom by two so you can carry off the little
complex fraction so overall you will get negative two on the top over X and then when you multiply
two with C streets so you can just put down a seaport like this and you is of course dy/dx
and when you put on the right DX you can multiply the example sides so another word dy is equal to
negative two over X plus C for DX and then we can integrate both sides and we get the left hand side
being Y and then the right hand side is negative 2 natural log of absolute value of x plus C 4 and
then don't forget another constant plus C 5 this right here if it differentiate once then twice
and three times you get that really cool right so I'm going to let you guys do the differentiation
on you all so now case 2 that's the case when C 2 is greater than zero but I'm not going to do a
general case please don't ask me that so I will just pick an easy number for C - okay I'll just
say pick C 2 to be 1 positive 1 very nice and with that said you are going to have the integral of 1
over u square plus 1 the U and this is equal to the integral of 1 over 2 DX like this and
then when you integrate this you'll get interest tangent and then when you integrate that you are
going to get 1/2 X and then that's a plus C 6 and as I said you get the interest engine let's take
the original tension on both sides so you get the original tangent of this guy and then you
have to you on the left hand side and the use of course dy/dx so I can just put that down dy
here and then DX here and I can just integrate both sides right here for you guys so what does
that you will get Y being ok when you integrate tangent you can natural log absolute value of C
cut off the input which is one-half X plus C 6 like this but you have to do the reverse chandu
right either robot here up 1/2 X plus Z 6 is 1/2 divided by 1/2 which is x 2 like that so you're
done with integration so India input on + C 7 here is another function that when you differentiate 1
2 3 times you'll get that really cool I like to ask my students to differentiate see see see see
7 to be 0 differentiate this part in all three times and then have them verify that which is
really cool now now say well now see I'm in case 3 that's the case when C - yes less than 0 and of
course please don't make me do the general case I will just say pick C 2 to be negative 1 nice
number ok so we had to deal with the integral of 1 over u square minus 1 U and then we have
the integral 1/2 DX like this this time though when you integrate this of course you get 1/2
X and we have C I now see 8 all right here's the deal when you integrate this you actually
end up with negative inverse hyperbolic tangent the reason it's impressed because this is being
flipped okay so it's the negative version so I'm going to multiply by negative on both sides and I
will get negative here and I'm just going to say this is gonna be seen I like that okay hopefully
you guys don't mind and then I will have to you by self after I take the original hyperbolic tangent
on both sides like this and then from here I'm going to just integrate both sides well this is
d u DX this is dy DX oh my god there are other things so just put on dy/dx finally like that
and now integrate both sides and then right here okay y equals okay hyperbolic tangent it's like
think of Akash you two goofs up up to you whatever you into it but it's really similar to that except
for you have natural log absolute value and you have cache of the input which is negative 1/2
X plus the night like this but again you do the reverse chandu divided by the derivative inside
which is going to be dividing by negative two like this so you have pretty much of this and
by the way this is supposed to be cosh like this and then plus the 10 you can put a negative right
here and then you do the reciprocal so this is actually the same as let me just write this down
again y equals to natural log absolute value with hyperbolic secant of negative 1/2 X plus C 9 oh my
god so finish yeah like this another function right here and if you do partial fractions
seriously just too young you're not going to the partial fractions but here's a little interesting
part where is the linear functions yeah where's the linear functions didn't I say the linear
functions will work as well that we're all the linear functions here is a special case I'll say
this is like the special cases right here where we have to go back to the denominators right here
you see we have 1 over u squared u right in this case we have to set u square equals zero
and then you'll find a student that's the missing students so first what u squared equals 0 well of
course this means U is equal 0 and U is what you will swipe ramp meaning you have y prime
equals 0 and the review what function will give you 0 of course constant functions so from here
you know why it's equal to a constant so I haven't used these also trusty so first one right here ok
second special case the missing solution from here I have to do care you square plus one equals zero
like this but this right here is good because it's not possible with real numbers so don't
worry okay right number three you're going to look at this guy and then set equal to zero so
you have u square minus one equals zero of course u equals 1 or u equals negative 1 from here you
know this is I say y prime equal 1 and the other one is y prime equal negative 1 this right here
if you integrate both sides integrate both sides of course you get y equals you get X plus let me
say this is Z 11 this is Z 12 okay and then on the other hand you can also have y equals
negative x plus these 13 and technically you can put on some coefficient in front of the X and the
reason you don't see that is because I picked it off to be negative 1 just a nice number so
hopefully you guys can see so here we go you have a lot of solutions for these guys whoa I am done
and hopefully you guys all liked this video and please give me a like and you guys like math be
sure you guys subscribe thank you guys so much and as always that's all I mean that's it bye [Music]